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ELECTRICAL CIRCUITS & FIELDS

YEAR 2013 MCQ 1.1

ONE MARK

Consider a delta connection of resistors and its equivalent star connection as shown below. If all elements of the delta connection are scaled by a factor k , k > 0 , the elements of the corresponding star equivalent will be scaled by a factor of

(A) k2

(B) k

A I D

(C) 1/k MCQ 1.2

MCQ 1.3

MCQ 1.5

k

The flux density at a point in space is given by Bv = 4xavx + 2kyavy + 8avz Wb/m2 . The value of constant k must be equal to (A) - 2 (B) - 0.5 (C) + 0.5 (D) i+ . n2

O N

.co a i A single-phase load is supplied by a osingle-phase voltage source. If the current d n . flowing from the load to the source w is 10+ - 150cA and if the voltage at the w load terminal is 100+60cV , w then the (A) (B) (C) (D)

MCQ 1.4

(D)

load load load load

absorbs absorbs delivers delivers

real real real real

power power power power

and and and and

delivers absorbs delivers absorbs

reactive reactive reactive reactive

power power power power

A source vs ^ t h = V cos 100pt has an internal impedance of ^4 + j3h W . If a purely resistive load connected to this source has to extract the maximum power out of the source, its value in W should be (A) 3 (B) 4 (C) 5 (D) 7 The transfer function

(A) 0.5s + 1 s+1

V2 ^s h of the circuit shown below is V1 ^s h

(B) 3s + 6 s+2

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(C) s + 2 s+1

Page 2

(D) s + 1 s+2

YEAR 2013

TWO MARKS

MCQ 1.6

A dielectric slab with 500 mm # 500 mm cross-section is 0.4 m long. The slab is subjected to a uniform electric field of Ev = 6avx + 8avy kV/mm . The relative permittivity of the dielectric material is equal to 2. The value of constant e0 is 8.85 # 10-12 F/m . The energy stored in the dielectric in Joules is (A) 8.85 # 10-11 (B) 8.85 # 10-5 (C) 88.5 (D) 885

MCQ 1.7

Three capacitors C1 , C2 and C 3 whose values are 10 mF , 5 mF , and 2 mF respectively, have breakdown voltages of 10 V, 5 V and 2 V respectively. For the interconnection shown below, the maximum safe voltage in Volts that can be applied across the combination, and the corresponding total charge in mC stored in the effective capacitance across the terminals are respectively,

(A) 2.8 and 36 (C) 2.8 and 32 MCQ 1.8

O N

A I D

(B)n7 and 119 o.i c . ia (D) 7 and 80

d no . w source voltage VS = 100+53.13c V In the circuit shown below, ifwthe w

then the Thevenin’s equivalent voltage in Volts as seen by the load resistance RL is

(A) 100+90c (C) 800+90c

(B) 800+0c (D) 100+60c

YEAR 2012 MCQ 1.9

ONE MARK

In the circuit shown below, the current through the inductor is

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(B) - 1 A 1+j

2 A 1+j (C) 1 A 1+j (A)

MCQ 1.10

Page 3

(D) 0 A

The impedance looking into nodes 1 and 2 in the given circuit is

(A) 50 W (C) 5 kW

A I D

O N

no w.

n o.i100 W (B) c . ia

d

(D) 10.1 kW

ww

(s2 + 9) (s + 2) (s + 1) (s + 3) (s + 4) is excited by sin (wt). The steady-state output of the system is zero at (A) w = 1 rad/s (B) w = 2 rad/s (C) w = 3 rad/s (D) w = 4 rad/s

MCQ 1.11

A system with transfer function G (s) =

MCQ 1.12

The average power delivered to an impedance (4 - j3) W by a current 5 cos (100pt + 100) A is (A) 44.2 W (B) 50 W (C) 62.5 W (D) 125 W

MCQ 1.13

In the following figure, C1 and C2 are ideal capacitors. C1 has been charged to 12 V before the ideal switch S is closed at t = 0. The current i (t) for all t is

(A) zero (B) a step function (C) an exponentially decaying function (D) an impulse function

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YEAR 2012 MCQ 1.14

TWO MARKS

If VA - VB = 6 V then VC - VD is

(A) - 5 V (C) 3 V MCQ 1.15

Page 4

(B) 2 V (D) 6 V

Assuming both the voltage sources are in phase, the value of R for which maximum power is transferred from circuit A to circuit B is

(A) 0.8 W (C) 2 W

A I D

O N

(B) 1.4 W (D) 2.8 W

.in

co ia.

Common Data for Questions 16oand d 17 :

.n

With 10 V dc connected at port ww A in the linear nonreciprocal two-port network w shown below, the following were observed : (i) 1 W connected at port B draws a current of 3 A (ii) 2.5 W connected at port B draws a current of 2 A

MCQ 1.16

With 10 V dc connected at port A, the current drawn by 7 W connected at port B is (A) 3/7 A (B) 5/7 A (C) 1 A (D) 9/7 A

MCQ 1.17

For the same network, with 6 V dc connected at port A, 1 W connected at port B draws 7/3 A. If 8 V dc is connected to port A, the open circuit voltage at port B is (A) 6 V (B) 7 V (C) 8 V (D) 9 V

Statement for Linked Answer Questions 18 and 19 :

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Page 5

In the circuit shown, the three voltmeter readings are V1 = 220 V, V2 = 122 V, V3 = 136 V .

MCQ 1.18

MCQ 1.19

The power factor of the load is (A) 0.45 (C) 0.55

(B) 0.50 (D) 0.60

If RL = 5 W , the approximate power consumption in the load is (A) 700 W (B) 750 W (C) 800 W (D) 850 W

A I D

YEAR 2011 MCQ 1.20

ONE MARK

The r.m.s value of the current i (t) in the circuit shown below is (B) 1 A (A) 1 A 2 2 (C) 1 A

O N

(D)

in co.

2A

.

ia od

n

. ww

w

MCQ 1.21

The voltage applied to a circuit is 100 2 cos (100pt) volts and the circuit draws a current of 10 2 sin (100pt + p/4) amperes. Taking the voltage as the reference phasor, the phasor representation of the current in amperes is (B) 10 - p/4 (A) 10 2 - p/4 (C) 10 + p/4

MCQ 1.22

(D) 10 2 + p/4

In the circuit given below, the value of R required for the transfer of maximum power to the load having a resistance of 3 W is

(A) zero (C) 6 W

(B) 3 W (D) infinity

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YEAR 2011 MCQ 1.23

TWO MARKS

A lossy capacitor Cx , rated for operation at 5 kV, 50 Hz is represented by an equivalent circuit with an ideal capacitor C p in parallel with a resistor R p . The value C p is found to be 0.102 μF and value of R p = 1.25 MW . Then the power loss and tan d of the lossy capacitor operating at the rated voltage, respectively, are (A) 10 W and 0.0002 (B) 10 W and 0.0025 (C) 20 W and 0.025

MCQ 1.24

Page 6

(D) 20 W and 0.04

A capacitor is made with a polymeric dielectric having an er of 2.26 and a dielectric breakdown strength of 50 kV/cm. The permittivity of free space is 8.85 pF/m. If the rectangular plates of the capacitor have a width of 20 cm and a length of 40 cm, then the maximum electric charge in the capacitor is (A) 2 μC (B) 4 μC (C) 8 μC (D) 10 μC

Common Data questions: 25 & 26

A I D

The input voltage given to a converter is vi = 100 2 sin (100pt) V The current drawn by the converter is ii = 10 2 sin (100pt - p/3) + 5 2 sin (300pt + p/4) + 2 2 sin (500pt - p/6) A MCQ 1.25

MCQ 1.26

O N

The input power factor of the converter is (A) 0.31 .in0.44 o(B) c . (C) 0.5 dia (D) 0.71

.no w The active power drawn by the w converter is w (A) 181 W (B) 500 W (C) 707 W

(D) 887 W

Common Data questions: 27 & 28 An RLC circuit with relevant data is given below.

MCQ 1.27

The power dissipated in the resistor R is (A) 0.5 W (B) 1 W (C)

MCQ 1.28

(D) 2 W

2W

The current IC in the figure above is (A) - j2 A

(B) - j 1 A 2

(C) + j 1 A 2

(D) + j2A

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YEAR 2010 MCQ 1.29

ONE MARK

The switch in the circuit has been closed for a long time. It is opened at t = 0. At t = 0+ , the current through the 1 mF capacitor is

(A) 0 A (C) 1.25 A MCQ 1.30

(B) 1 A (D) 5 A

As shown in the figure, a 1 W resistance is connected across a source that has a load line v + i = 100 . The current through the resistance is

A I D

(A) 25 A (C) 100 A YEAR 2010 MCQ 1.31

Page 7

(B) 50 A (C) 200 A

O N

.in

co ia.

TWO MARKS

If the 12 W resistor draws a current of od1 A as shown in the figure, the value of n . resistance R is w

ww

(A) 4 W (C) 8 W MCQ 1.32

(B) 6 W (D) 18 W

The two-port network P shown in the figure has ports 1 and 2, denoted by terminals (a,b) and (c,d) respectively. It has an impedance matrix Z with parameters denoted by Zij . A 1 W resistor is connected in series with the network at port 1 as shown in the figure. The impedance matrix of the modified two-port network (shown as a dashed box ) is

Z11 + 1 (A) e Z21 Z11 + 1 (C) e Z21

Z12 + 1 Z22 + 1o Z12 Z22 o

Z11 + 1 Z12 (B) e Z21 Z22 + 1o Z11 + 1 Z12 (D) e Z21 + 1 Z22 o

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Page 8

YEAR 2009 MCQ 1.33

ONE MARK

The current through the 2 kW resistance in the circuit shown is

(A) 0 mA (C) 2 mA MCQ 1.34

(B) 1 mA (D) 6 mA

How many 200 W/220 V incandescent lamps connected in series would consume the same total power as a single 100 W/220 V incandescent lamp ? (A) not possible (B) 4 (C) 3 (D) 2

A I D

YEAR 2009 MCQ 1.35

TWO MARKS

In the figure shown, all elements used are ideal. For time t < 0, S1 remained closed and S2 open. At t = 0, S1 is opened and S2 is closed. If the voltage Vc2 n across the capacitor C2 at t = 0 is zero, cthe o.i voltage across the capacitor combi. nation at t = 0+ will be dia

O N

no

. ww

w

(A) 1 V (C) 1.5 V MCQ 1.36

The equivalent capacitance of the input loop of the circuit shown is

(A) 2 mF (C) 200 mF MCQ 1.37

(B) 2 V (D) 3 V

(B) 100 mF (D) 4 mF

For the circuit shown, find out the current flowing through the 2 W resistance. Also identify the changes to be made to double the current through the 2 W resistance.

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(A) (5 A; PutVS = 30 V) (C) (5 A; Put IS = 10 A)

Page 9

(B) (2 A; PutVS = 8 V) (D) (7 A; Put IS = 12 A)

Statement for Linked Answer Question 38 and 39 :

MCQ 1.38

For the circuit given above, the Thevenin’s resistance across the terminals A and B is (A) 0.5 kW (B) 0.2 kW (C) 1 kW (D) 0.11 kW

MCQ 1.39

For the circuit given above, the Thevenin’s voltage across the terminals A and B is n o.i0.25 V (A) 1.25 V (B) c . dia(D) 0.5 V (C) 1 V o n

A I D

O N

.

w ww

YEAR 2008 MCQ 1.40

The number of chords in the graph of the given circuit will be

(A) 3 (C) 5 MCQ 1.41

ONE MARK

(B) 4 (D) 6

The Thevenin’s equivalent of a circuit operation at w = 5 rads/s, has Voc = 3.71+ - 15.9% V and Z0 = 2.38 - j0.667 W . At this frequency, the minimal realization of the Thevenin’s impedance will have a (A) resistor and a capacitor and an inductor (B) resistor and a capacitor (C) resistor and an inductor (D) capacitor and an inductor

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YEAR 2008 MCQ 1.42

TWO MARKS

The time constant for the given circuit will be

(A) 1/9 s (C) 4 s MCQ 1.43

(B) 1/4 s (D) 9 s

The resonant frequency for the given circuit will be

(A) 1 rad/s (C) 3 rad/s MCQ 1.44

Page 10

(B) 2 rad/s (D) 4 rad/s

A I D

Assuming ideal elements in the circuit shown below, the voltage Vab will be

O N

no w.

.in

co ia.

d

ww

(A) - 3 V (C) 3 V

(B) 0 V (D) 5 V

Statement for Linked Answer Question 38 and 39. The current i (t) sketched in the figure flows through a initially uncharged 0.3 nF capacitor.

MCQ 1.45

The charge stored in the capacitor at t = 5 ms , will be

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(A) 8 nC (C) 13 nC

Page 11

(B) 10 nC (D) 16 nC

MCQ 1.46

The capacitor charged upto 5 ms, as per the current profile given in the figure, is connected across an inductor of 0.6 mH. Then the value of voltage across the capacitor after 1 ms will approximately be (A) 18.8 V (B) 23.5 V (C) - 23.5 V (D) - 30.6 V

MCQ 1.47

In the circuit shown in the figure, the value of the current i will be given by

(A) 0.31 A (C) 1.75 A MCQ 1.48

MCQ 1.49

(B) 1.25 A (D) 2.5 A

A I D

Two point charges Q1 = 10 mC and Q2 = 20 mC are placed at coordinates (1,1,0) and (- 1, - 1, 0) respectively. The total electric flux passing through a plane z = 20 will be (A) 7.5 mC (B) 13.5 mC (C) 15.0 mC (D) 22.5 mC

O N

n o.i c . A capacitor consists of two metal platesiaeach 500 # 500 mm2 and spaced 6 mm d apart. The space between the metal noplates is filled with a glass plate of 4 mm . w thickness and a layer of paper wwof 2 mm thickness. The relative primitivities of

the glass and paper are 8 and 2 respectively. Neglecting the fringing effect, the capacitance will be (Given that e0 = 8.85 # 10 - 12 F/m ) (A) 983.3 pF (B) 1475 pF (C) 637.7 pF (D) 9956.25 pF MCQ 1.50

A coil of 300 turns is wound on a non-magnetic core having a mean circumference of 300 mm and a cross-sectional area of 300 mm2. The inductance of the coil corresponding to a magnetizing current of 3 A will be (Given that m0 = 4p # 10 - 7 H/m) (A) 37.68 mH (B) 113.04 mH (C) 3.768 mH (D) 1.1304 mH YEAR 2007

MCQ 1.51

Divergence of the vector field V (x, y, z) =- (x cos xy + y) it + (y cos xy) jt + (sin z2 + x2 + y2) kt is (A) 2z cos z2 (B) sin xy + 2z cos z2 (C) x sin xy - cos z (D) None of these YEAR 2007

MCQ 1.52

ONE MARK

TWO MARKS

The state equation for the current I1 in the network shown below in terms of

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Page 12

the voltage VX and the independent source V , is given by

MCQ 1.53

(A) dI1 =- 1.4VX - 3.75I1 + 5 V (B) dI1 = 1.4VX - 3.75I1 - 5 V dt 4 dt 4 (C) dI1 =- 1.4VX + 3.75I1 + 5 V (D) dI1 =- 1.4VX + 3.75I1 - 5 V dt 4 dt 4 The R-L-C series circuit shown in figure is supplied from a variable frequency voltage source. The admittance - locus of the R-L-C network at terminals AB for increasing frequency w is

A I D

O N

no w.

.in

co ia.

d

ww

MCQ 1.54

In the circuit shown in figure. Switch SW1 is initially closed and SW2 is open. The inductor L carries a current of 10 A and the capacitor charged to 10 V with polarities as indicated. SW2 is closed at t = 0 and SW1 is opened at t = 0 . The current through C and the voltage across L at (t = 0+) is

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(A) 55 A, 4.5 V (C) 45 A, 5.5 A MCQ 1.55

Page 13

(B) 5.5 A, 45 V (D) 4.5 A, 55 V

In the figure given below all phasors are with reference to the potential at point ''O'' . The locus of voltage phasor VYX as R is varied from zero to infinity is shown by

A I D

MCQ 1.56

MCQ 1.57

O N

no w.

.in

co ia.

d

A 3 V DC supply with an internal a passive nonww resistance of 2 W supplies 2 linear resistance characterized by the relation VNL = INL . The power dissipated in the non linear resistance is (A) 1.0 W (B) 1.5 W (C) 2.5 W (D) 3.0 W The matrix A given below in the node incidence matrix of a network. The columns correspond to branches of the network while the rows correspond to nodes. Let V = [V1V2 .....V6]T denote the vector of branch voltages while I = [i1 i2 .....i6]T that of branch currents. The vector E = [e1 e2 e3 e4]T denotes the vector of node voltages relative to a common ground. R1 1 1 0 0 0 V S W S 0 -1 0 -1 1 0 W S- 1 0 0 0 - 1 - 1W S W S 0 0 -1 1 0 1 W T X Which of the following statement is true ? (A) The equations V1 - V2 + V3 = 0,V3 + V4 - V5 = 0 are KVL equations for the network for some loops (B) The equations V1 - V3 - V6 = 0,V4 + V5 - V6 = 0 are KVL equations for the network for some loops (C) E = AV (D) AV = 0 are KVI equations for the network

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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia MCQ 1.58

Page 14

A solid sphere made of insulating material has a radius R and has a total charge Q distributed uniformly in its volume. What is the magnitude of the electric field intensity, E , at a distance r (0 < r < R) inside the sphere ? 1 Qr 4pe0 R3 Q (C) 1 2 4pe0 r

3 Qr 4pe0 R3 QR (D) 1 4pe0 r3

(A)

(B)

Statement for Linked Answer Question 59 and 60. An inductor designed with 400 turns coil wound on an iron core of 16 cm2 cross sectional area and with a cut of an air gap length of 1 mm. The coil is connected to a 230 V, 50 Hz ac supply. Neglect coil resistance, core loss, iron reluctance and leakage inductance, (m0 = 4p # 10 - 7 H/M) MCQ 1.59

MCQ 1.60

The current in the inductor is (A) 18.08 A (C) 4.56 A

A I D

The average force on the core to reduce the air gap will be (A) 832.29 N (B) 1666.22 N (C) 3332.47 N (D) 6664.84 N YEAR 2006

MCQ 1.61

(B) 9.04 A (D) 2.28 A

O N

.in

d In the figure the current source is o n1+0 A, R = 1 W , the impedances are . ZC =- j W and ZL = 2j W . The wwThevenin equivalent looking into the circuit w across X-Y is

(A)

2 +0 V, (1 + 2j) W

(C) 2+45% V, (1 + j) W

(B) 2+45% V, (1 - 2j) W (D)

2 +45% V, (1 + j) W

YEAR 2006 MCQ 1.62

ONE MARK

co ia.

TWO MARKS

The parameters of the circuit shown in the figure are Ri = 1 MW R0 = 10 W, A = 106 V/V If vi = 1 mV , the output voltage, input impedance and output impedance respectively are

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(A) 1 V, 3, 10 W (C) 1 V, 0, 3 MCQ 1.63

Page 15

(B) 1 V, 0, 10 W (D) 10 V, 3, 10 W

In the circuit shown in the figure, the current source I = 1 A , the voltage source V = 5 V, R1 = R2 = R3 = 1 W, L1 = L2 = L3 = 1 H, C1 = C2 = 1 F

The currents (in A) through R3 and through the voltage source V respectively will be (A) 1, 4 (B) 5, 1 (C) 5, 2 (D) 5, 4 MCQ 1.64

The parameter type and the matrix representation of the relevant two port parameters that describe the circuit shown are

A I D

O N

0 (A) z parameters, = 0 0 (C) h parameters, = 0 MCQ 1.65

0 0G 0 0G

ww

no w.

.in

d

co ia.

1 (B) h parameters, = 0 1 (D) z parameters, = 0

0 1G 0 1G

The circuit shown in the figure is energized by a sinusoidal voltage source V1 at a frequency which causes resonance with a current of I .

The phasor diagram which is applicable to this circuit is

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MCQ 1.66

Page 16

An ideal capacitor is charged to a voltage V0 and connected at t = 0 across an ideal inductor L. (The circuit now consists of a capacitor and inductor alone). If we let w0 = 1 , the voltage across the capacitor at time t > 0 is given by LC (A) V (B) V cos (w t) 0

0

(C) V0 sin (w0 t)

(D) V0 e

0

- w0t

cos (w0 t)

MCQ 1.67

An energy meter connected to an immersion heater (resistive) operating on an AC 230 V, 50 Hz, AC single phase source reads 2.3 units (kWh) in 1 hour. The heater is removed from the supply and now connected to a 400 V peak square wave source of 150 Hz. The power in kW dissipated by the heater will be (A) 3.478 (B) 1.739 (C) 1.540 (D) 0.870

MCQ 1.68

Which of the following statement holds for the divergence of electric and magnetic flux densities ? n (A) Both are zero o.i c . diabut non zero for time varying densities. (B) These are zero for static densities o n w. density (C) It is zero for the electricwflux w (D) It is zero for the magnetic flux density

A I D

O N

YEAR 2005 MCQ 1.69

In the figure given below the value of R is

(A) 2.5 W (C) 7.5 W MCQ 1.70

(B) 5.0 W (D) 10.0 W

The RMS value of the voltage u (t)= 3 + 4 cos (3t) is (A) 17 V (B) 5 V (C) 7 V

MCQ 1.71

ONE MARK

(D) (3 + 2 2 ) V

For the two port network shown in the figure the Z -matrix is given by

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MCQ 1.72

Z1 Z1 + Z2 (A) = Z1 + Z2 Z2 G

Z1 Z1 (B) = Z1 + Z2 Z2 G

Z1 Z2 (C) = Z2 Z1 + Z2 G

Z1 Z1 (D) = Z1 Z1 + Z2 G

In the figure given, for the initial capacitor voltage is zero. The switch is closed at t = 0 . The final steady-state voltage across the capacitor is

A I D

(A) 20 V (C) 5 V MCQ 1.73

Page 17

(B) 10 V (D) 0 V

If Ev is the electric intensity, 4 (4 # Ev) is equal to (A) Ev (B) Ev n (C) null vector (D).iZero YEAR 2005

O N

no

. ww

co

. dia

w

TWO MARKS

Statement for Linked Answer Question 74 and 75. A coil of inductance 10 H and resistance 40 W is connected as shown in the figure. After the switch S has been in contact with point 1 for a very long time, it is moved to point 2 at, t = 0 . MCQ 1.74

If, at t = 0+ , the voltage across the coil is 120 V, the value of resistance R is

(A) 0 W (C) 40 W

(B) 20 W (D) 60 W

MCQ 1.75

For the value as obtained in (a), the time taken for 95% of the stored energy to be dissipated is close to (A) 0.10 sec (B) 0.15 sec (C) 0.50 sec (D) 1.0 sec

MCQ 1.76

The RL circuit of the figure is fed from a constant magnitude, variable frequen-

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Page 18

cy sinusoidal voltage source Vin . At 100 Hz, the Rand L elements each have a voltage drop mRMS .If the frequency of the source is changed to 50 Hz, then new voltage drop across R is

5u (B) 8 RMS 8u (C) (D) 5 RMS For the three-phase circuit shown in the figure is given by (A)

MCQ 1.77

(A) 1 : 1 :

A I D

O N

3

(C) 1 : 1 : 0 MCQ 1.78

no w.

.in1 : 1 : 2 (B)

d

co ia.

(D) 1 : 1 : 3/2

The circuit shown in the figure ww is in steady state, when the switch is closed at t = 0 .Assuming that the inductance is ideal, the current through the inductor at t = 0+ equals

(A) 0 A (C) 1 A MCQ 1.79

2u 3 RMS 3u 2 RMS the ratio of the currents IR: IY : IB

(B) 0.5 A (D) 2 A

In the given figure, the Thevenin’s equivalent pair (voltage, impedance), as seen at the terminals P-Q, is given by

(A) (2 V, 5 W) (C) (4 V, 5 W)

(B) (2 V, 7.5 W) (D) (4 V, 7.5 W)

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Page 19

The charge distribution in a metal-dielectric-semiconductor specimen is shown in the figure. The negative charge density decreases linearly in the semiconductor as shown. The electric field distribution is as shown in

A I D

O N

no w.

.in

co ia.

d

ww

YEAR 2004 MCQ 1.81

The value of Z in figure which is most appropriate to cause parallel resonance at 500 Hz is

(A) 125.00 mH (C) 2.0 mF MCQ 1.82

ONE MARK

(B) 304.20 mF (D) 0.05 mF

A parallel plate capacitor is shown in figure. It is made two square metal plates of 400 mm side. The 14 mm space between the plates is filled with two layers of dielectrics of er = 4 , 6 mm thick and er = 2 , 8 mm thick. Neglecting fringing of fields at the edge the capacitance is

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(A) 1298 pF (C) 354 pF MCQ 1.83

(B) 944 pF (D) 257 pF

The inductance of a long solenoid of length 1000 mm wound uniformly with 3000 turns on a cylindrical paper tube of 60 mm diameter is (A) 3.2 mH (B) 3.2 mH (C) 32.0 mH (D) 3.2 H YEAR 2004

MCQ 1.84

TWO MARKS

In figure, the value of the source voltage is

A I D

O N

(A) 12 V (C) 30 V MCQ 1.85

(B)n24 V o.i c . ia (D) 44 V

d no . are w 20wW, 20 W and 10 W respectively. The resistances w

In figure, Ra , Rb and Rc R1 , R2 and R 3 in W of an equivalent star-connection are

(A) 2.5, 5, 5 (C) 5, 5, 2.5 MCQ 1.86

Page 20

(B) 5, 2.5, 5 (D) 2.5, 5, 2.5

In figure, the admittance values of the elements in Siemens are YR = 0.5 + j0, YL = 0 - j1.5, YC = 0 + j0.3 respectively. The value of I as a phasor when the voltage E across the elements is 10+0% V

(A) 1.5 + j0.5

(B) 5 - j18

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(C) 0.5 + j1.8 MCQ 1.87

(D) 5 - j12

In figure, the value of resistance R in W is

(A) 10 (C) 30 MCQ 1.88

(B) 20 (D) 40

In figure, the capacitor initially has a charge of 10 Coulomb. The current in the circuit one second after the switch S is closed will be

A I D

(A) 14.7 A (C) 40.0 A MCQ 1.89

Page 21

(B) 18.5 A (D) 50.0 A

O N

The rms value of the current in a wire which carries a d.c. current of 10 A and a sinusoidal alternating current of peak value.in 20 A is o c . (A) 10 A (B) 14.14 A dia o (C) 15 A (D) 17.32 A .n

w

ww

MCQ 1.90

0.9 0.2 The Z-matrix of a 2-port network as given by = 0.2 0.6G The element Y22 of the corresponding Y-matrix of the same network is given by (A) 1.2 (B) 0.4 (C) - 0.4 (D) 1.8 YEAR 2003

MCQ 1.91

Figure Shows the waveform of the current passing through an inductor of resistance 1 W and inductance 2 H. The energy absorbed by the inductor in the first four seconds is

(A) 144 J (C) 132 J MCQ 1.92

ONE MARK

(B) 98 J (D) 168 J

A segment of a circuit is shown in figure vR = 5V, vc = 4 sin 2t .The voltage vL is

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Page 22

given by

(A) 3 - 8 cos 2t (C) 16 sin 2t MCQ 1.93

(B) 32 sin 2t (D) 16 cos 2t

In the figure, Z1 = 10+ - 60%, Z2 = 10+60%, Z3 = 50+53.13% . Thevenin impedance seen form X-Y is

(A) 56.66+45% (C) 70+30%

O N

A I D

% (B) in60+30

no

. co(D) . 34.4+65% a i d

MCQ 1.94

Two conductors are carrying forward and return current of +I and - I as w. w w shown in figure. The magnetic field intensity H at point P is

MCQ 1.95

(A) I Y (B) I X pd pd (C) I Y (D) I X 2pd 2pd Two infinite strips of width w m in x -direction as shown in figure, are carrying forward and return currents of +I and - I in the z - direction. The strips are separated by distance of x m. The inductance per unit length of the configuration is measured to be L H/m. If the distance of separation between the strips in snow reduced to x/2 m, the inductance per unit length of the configuration is

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(A) 2L H/m (C) L/2 H/m

(B) L/4 H/m (D) 4L H/m

YEAR 2003 MCQ 1.96

A I D

O N

no w.

n H (B).i5.30 o c . ia(D) 1.32 H

d

In figure, the potential difference w between points P and Q is

w

(A) 12 V (C) - 6 V MCQ 1.98

TWO MARKS

In the circuit of figure, the magnitudes of VL and VC are twice that of VR . Given that f = 50 Hz , the inductance of the coil is

(A) 2.14 mH (C) 31.8 mH MCQ 1.97

Page 23

(B) 10 V (D) 8 V

Two ac sources feed a common variable resistive load as shown in figure. Under the maximum power transfer condition, the power absorbed by the load resistance RL is

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(A) 2200 W (C) 1000 W MCQ 1.99

(B) 1250 W (D) 625 W

In figure, the value of R is

(A) 10 W (C) 24 W MCQ 1.100

A I D

O N

no w.

ww

n o.i c . ia (B) 4 V

d

(D) 8 V

The h-parameters for a two-port network are defined by E1 h11 h12 I1 = I G = =h h G =E G 2 21 22 2 For the two-port network shown in figure, the value of h12 is given by

(A) 0.125 (C) 0.625 MCQ 1.102

(B) 18 W (D) 12 W

In the circuit shown in figure, the switch S is closed at time (t = 0). The voltage across the inductance at t = 0+ , is

(A) 2 V (C) - 6 V MCQ 1.101

Page 24

(B) 0.167 (D) 0.25

A point charge of +I nC is placed in a space with permittivity of 8.85 # 10 - 12 F/m as shown in figure. The potential difference VPQ between two points P and Q at distance of 40 mm and 20 mm respectively fr0m the point charge is

(A) 0.22 kV

(B) - 225 V

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(C) - 2.24 kV

Page 25

(D) 15 V

MCQ 1.103

A parallel plate capacitor has an electrode area of 100 mm2, with spacing of 0.1 mm between the electrodes. The dielectric between the plates is air with a permittivity of 8.85 # 10 - 12 F/m. The charge on the capacitor is 100 V. The stored energy in the capacitor is (A) 8.85 pJ (B) 440 pJ (C) 22.1 nJ (D) 44.3 nJ

MCQ 1.104

A composite parallel plate capacitor is made up of two different dielectric material with different thickness (t1 and t2 ) as shown in figure. The two different dielectric materials are separated by a conducting foil F. The voltage of the conducting foil is

(A) 52 V (C) 67 V

(B) 60 V (D) 33 V

MCQ 1.105

MCQ 1.106

MCQ 1.107

IA

D O

YEAR 2002

ONE MARK

A current impulse, 5d (t), is forced through a capacitor C . The voltage , vc (t), in . o across the capacitor is given by c ia.(B) 5u (t) - C d (A) 5t .no w 5u (t) w (C) 5 t (D) w C C The graph of an electrical network has N nodes and B branches. The number of links L, with respect to the choice of a tree, is given by (A) B - N + 1 (B) B + N (C) N - B + 1 (D) N - 2B - 1

N

Given a vector field F, the divergence theorem states that (A)

# F : dS = # 4: FdV S

(C)

# F # dS = # 4: FdV S

(B)

V

V

# F : dS = # 4# FdV S

(D)

V

# F # dS = # 4: FdV S

V

MCQ 1.108

Consider a long, two-wire line composed of solid round conductors. The radius of both conductors. The radius of both conductors is 0.25 cm and the distance between their centres is 1 m. If this distance is doubled, then the inductance per unit length (A) doubles (B) halves (C) increases but does not double (D) decreases but does not halve

MCQ 1.109

A long wire composed of a smooth round conductor runs above and parallel to the ground (assumed to be a large conducting plane). A high voltage exists

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Page 26

between the conductor and the ground. The maximum electric stress occurs at (A) The upper surface of the conductor (B) The lower surface of the conductor. (C) The ground surface. (D) midway between the conductor and ground. YEAR 2002 MCQ 1.110

TWO MARKS

A two port network shown in Figure, is described by the following equations I1 = Y11 E1 + Y12 E2 I1 = Y21 E1 + Y22 E2

A I D

The admittance parameters, Y11, Y12, Y21 and Y22 for the network shown are (A) 0.5 mho, 1 mho, 2 mho and 1 mho respectively (B) 13 mho, - 16 mho, - 16 mho and 13 mho respectively

MCQ 1.111

(C) 0.5 mho, 0.5 mho, 1.5 mho and 2 mho respectively (D) - 2 mho, - 3 mho, 3 mho and 25 mhoinrespectively 5 7 7 co. . a i In the circuit shown in Figure, whatdvalue of C will cause a unity power factor o n . at the ac source ? w

O N ww

(A) 68.1 mF (C) 0.681 mF

(B) 165 mF (D) 6.81 mF

MCQ 1.112

A series R-L-C circuit has R = 50 W ; L = 100 mH and C = 1 mF . The lower half power frequency of the circuit is (A) 30.55 kHz (B) 3.055 kHz (C) 51.92 kHz (D) 1.92 kHz

MCQ 1.113

A 10 V pulse of 10 ms duration is applied to the circuit shown in Figure, assuming that the capacitor is completely discharged prior to applying the pulse, the peak value of the capacitor voltage is

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(A) 11 V (C) 6.32 V MCQ 1.114

Page 27

(B) 5.5 V (D) 0.96 V

In the circuit shown in Figure, it is found that the input voltage (vi ) and current i are in phase. The coupling coefficient is K = M , where M is the L1 L2 mutual inductance between the two coils. The value of K and the dot polarity of the coil P-Q are

(A) K = 0.25 and dot at P (C) K = 0.25 and dot at Q MCQ 1.115

Consider the circuit shown in Figure If the frequency of the source is 50 Hz, then a value of t0 which results in a transient free response is

(A) 0 ms (C) 2.71 ms MCQ 1.116

(B) K = 0.5 and dot at P (C) K = 0.5 and dot at Q

A I D

O N ww

no w.

.in

d

co ia.

(B) 1.78 ms (D) 2.91 ms

In the circuit shown in figure, the switch is closed at time t = 0 . The steady state value of the voltage vc is

(A) 0 V (C) 5 V

(B) 10 V (D) 2.5 V

Common data Question for Q. 117-118* : A constant current source is supplying 10 A current to a circuit shown in figure. The switch is initially closed for a sufficiently long time, is suddenly opened at t=0

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MCQ 1.117

MCQ 1.118

The inductor current iL (t) will be (A) 10 A (C) 10e- 2t A

Page 28

(B) 0 A (D) 10 (1 - e- 2t) A

What is the energy stored in L, a long time after the switch is opened (A) Zero (B) 250 J (C) 225 J (D) 2.5 J

Common Data Question for Q. 119-120* : An electrical network is fed by two ac sources, as shown in figure, Given that Z1 = (1 - j) W , Z2 = (1 + j) W and ZL = (1 + j0) W .

MCQ 1.119

MCQ 1.120

MCQ 1.121

A I D

O N

.in o c . *Thevenin voltage and impedance across ia terminals X and Y d o (A) 0 V, (2 + 2j) W (B) 60 V, 1 W .n w (C) 0 V, 1 W (D) 30 V, (1 + j) W ww *Current iL through load is (A) 0 A (C) 0.5 A

respectively are

(B) 1 A (D) 2 A

*In the resistor network shown in figure, all resistor values are 1 W. A current of 1 A passes from terminal a to terminal b as shown in figure, Voltage between terminal a and b is

(A) 1.4 Volt

(B) 1.5 Volt

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(C) 0 Volt

(D) 3 Volt

YEAR 2001 MCQ 1.122

Page 29

ONE MARK

In a series RLC circuit at resonance, the magnitude of the voltage developed across the capacitor (A) is always zero (B) can never be greater than the input voltage (C) can be greater than the input voltage, however it is 90% out of phase with the input voltage (D) can be greater than the input voltage, and is in phase with the input voltage.

MCQ 1.123

Two incandescent light bulbs of 40 W and 60 W rating are connected in series across the mains. Then (A) the bulbs together consume 100 W (B) the bulbs together consume 50 W (C) the 60 W bulb glows brighter (D) the 40 bulb glows brighter

MCQ 1.124

A unit step voltage is applied at t = 0 to a series RL circuit with zero initial conditions. (A) It is possible for the current to be oscillatory.

A I D

O N

n

(B) The voltage across the resistor at t = c0o+.iis zero. ia.the steady state is zero. d (C) The energy stored in the inductor in no . w (D) The resistor current eventually w falls to zero.

w

MCQ 1.125

Given two coupled inductors L1 and L2 , their mutual inductance M satisfies (L + L2) (A) M = L21 + L22 (B) M > 1 2 (C) M >

MCQ 1.126

L1 L 2

L1 L 2

A passive 2-port network is in a steady-state. Compared to its input, the steady state output can never offer (A) higher voltage (B) lower impedance (C) greater power (D) better regulation YEAR 2001

MCQ 1.127

(D) M #

TWO MARKS

Consider the star network shown in Figure The resistance between terminals A and B with C open is 6 W, between terminals B and C with A open is 11 W, and between terminals C and A with B open is 9 W. Then

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(A) (B) (C) (D)

Page 30

RA = 4 W, RB = 2 W, RC = 5 W RA = 2 W, RB = 4 W, RC = 7 W RA = 3 W, RB = 3 W, RC = 4 W RA = 5 W, RB = 1 W, RC = 10 W

MCQ 1.128

A connected network of N > 2 nodes has at most one branch directly connecting any pair of nodes. The graph of the network (A) Must have at least N branches for one or more closed paths to exist (B) Can have an unlimited number of branches (C) can only have at most N branches (D) Can have a minimum number of branches not decided by N

MCQ 1.129

A 240 V single-phase ac source is connected to a load with an impedance of 10+60% W . A capacitor is connected in parallel with the load. If the capacitor suplies 1250 VAR, the real power supplied by the source is (A) 3600 W (B) 2880 W (C) 240 W (D) 1200 W

A I D

Common Data Questions Q.130-131*:

For the circuit shown in figure given values are R = 10 W , C = 3 mF , L1 = 40 mH, L2 = 10 mH and M = 10 mH

O N

no w.

.in

co ia.

d

ww

MCQ 1.130

MCQ 1.131

The resonant frequency of the circuit is (B) 1 # 105 rad/sec A) 1 # 105 rad/sec 3 2 (D) 1 # 105 rad/sec (C) 1 # 105 rad/sec 9 21 The Q-factor of the circuit in Q.82 is (A) 10 (B) 350 (C) 101 (D) 15

MCQ 1.132

Given the potential function in free space to be V (x) = (50x2 + 50y2 + 50z2) volts, the magnitude (in volts/metre) and the direction of the electric field at a point (1,-1,1), where the dimensions are in metres, are (B) 100/ 3 ; (it - tj + kt) (A) 100; (it + tj + kt) (C) 100 3 ; [(- it + tj - kt) / 3 ] (D) 100 3 ; [(- it - tj - kt) / 3 ]

MCQ 1.133

The hysteresis loop of a magnetic material has an area of 5 cm2 with the scales given as 1 cm = 2 AT and 1 cm = 50 mWb. At 50 Hz, the total hysteresis loss is. (A) 15 W (B) 20 W (C) 25 W (D) 50 W

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Page 31

The conductors of a 10 km long, single phase, two wire line are separated by a distance of 1.5 m. The diameter of each conductor is 1 cm. If the conductors are of copper, the inductance of the circuit is (A) 50.0 mH (B) 45.3 mH (C) 23.8 mH (D) 19.6 mH ***********

A I D

O N

no w.

.in

co ia.

d

ww

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SOLUTION

SOL 1.1

Page 32

Option (B) is correct. In the equivalent star connection, the resistance can be given as Rb Ra Ra + Rb + Rc Ra Rc RB = Ra + Rb + Rc Rb Rc RA = Ra + Rb + Rc RC =

So, if the delta connection components Ra , Rb and Rc are scaled by a factor k then ^k Rb h^k Rc h RAl = kRa + kRb + kRc

A I D

2 Rb Rc =k k Ra + Rb + Rc

= k RA Hence, it is also scaled by a factor k SOL 1.2

O N

Option (A) is correct. .in o c . Given, flux density ia d v v v o B = 4x a x + 2.ky n a y + 8 avz w Since, magnetic flux density w is always divergence less. w i.e., d$Bv = 0 So, for given vector flux density, we have d$Bv = 4 + 2k + 0 = 0 or, k =- 2

SOL 1.3

Option (B) is correct. Consider the voltage source and load shown in figure

We obtain the power delivered by load as Pdelivered = I L* VL = ^10 + 150ch^10 60ch = 100 210c = 1000 cos 210c + j1000 sin 210c =- 866.025 - j500 As both the reactive and average power (real power) are negative so, power is

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Page 33

absorbed by load. i.e., load absorbs real as well as reactive power. SOL 1.4

Option (C) is correct. For the purely resistive load, maximum average power is transferred when 2 2 RL = RTh + XTh where RTh + jXTh is the equivalent thevinin (input) impedance of the circuit. i.e., RL = 42 + 32 = 5W

SOL 1.5

Option (D) is correct. For the given capacitance, C = 100mF in the circuit, we have the reactance. XC = 1 sc 1 = s # 100 # 10-6 4 = 10 s So, 10 4 + 10 4 V2 ^s h = 4 s V1 ^s h 10 + 10 4 + 10 4 s s = s+1 s+2 Option (B) is correct. Energy density stored in a dielectric medium is obtained as

SOL 1.6

A I D

O N

wE = 1 e E 2 J/m2 2

.in

co ia.

d

The electric field inside the dielectric .no will be same to given field in free space w only if the field is tangentialwtowthe interface 2 So, wE = 1 2e0 ^ 62 + 82 h # 106 /mm2 2 Therefore, the total stored energy is WE =

#W v

E

dv

= e0 100 # 106 /mm2 # ^500 # 500h mm2 # ^0.4h = e0 # 100 # 106 # 0.4 # 25 # 10 4 = 8.85 # 10-12 # 1013 = 88.5 J SOL 1.7

Option (C) is correct.

Consider that the voltage across the three capacitors C1 , C2 and C 3 are V1 , V2 and V3 respectively. So, we can write V2 = C 3 ....(1) V3 C2

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Page 34

Since, Voltage is inversely proportional to capacitance Now, given that C1 = 10 mF ; ^V1hmax = 10V C2 = 5 mF ; ^V2hmax = 5 V C 3 = 2 mF ; ^V3hmax = 2V So, from Eq (1) we have V2 = 2 5 V3 for ^V3hmax = 2 We obtain, V2 = 2 # 2 = 0.8 volt < 5 5 i.e., V2 < ^V2hmax Hence, this is the voltage at C2 . Therefore, V3 = 2 volt V2 = 0.8 volt and V1 = V2 + V3 = 2.8 volt Now, equivalent capacitance across the terminal is Ceq = C2 C 3 + C1 C2 + C3 = 5 # 2 + 10 5+2 80 = mF 7 n Equivalent voltage is (max. value) o.i c . Vmax = V1 = 2.8 ia d o So, charge stored in the effective.ncapacitance is w Q = Ceq Vw max

A I D

O N w

= b 80 l # ^2.8h 7 = 32 mC

SOL 1.8

Option (C) is correct. For evaluating the equivalent thevenin voltage seen by the load RL , we open the circuit across it (also if it consist dependent source). The equivalent circuit is shown below

As the circuit open across RL so I2 = 0 or, j40I2 = 0 i.e., the dependent source in loop 1 is short circuited. Therefore, ^ j4h Vs VL1 = j4 + 3

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Page 35

j40 100 53.13c j4 + 3 40 90c = 100 53.13c 5 53.13c = 800 90c

VTh = 10 VL1 =

SOL 1.9

Option (C) is correct.

A I D

Applying nodal analysis at top node. V1 + 1 0c V1 + 1 0c = 1 0c + 1 j1

O N

SOL 1.10

in

V1 (j 1 + 1) + j 1 + 1 0c = j1 .co. ia d 1 o V1 = 1 + j 1 w.n ww 1 V1 + 1 0c - 1 + j + 1 j Current = I1 = = = 1 A j1 j1 (1 + j) j 1 + j Option (A) is correct. We put a test source between terminal 1, 2 to obtain equivalent impedance

ZTh = Vtest Itest By applying KCL at top right node Vtest + Vtest - 99I = I b test 9 k + 1k 100 Vtest + Vtest - 99I = I b test 10 k 100 But Ib =- Vtest =-Vtest 9k + 1k 10k

...(i)

Substituting Ib into equation (i), we have Vtest + Vtest + 99Vtest = I test 10 k 100 10 k

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SOL 1.11

Page 36

100Vtest + Vtest = I test 10 # 103 100 2Vtest = I test 100 Vtest ZTh = Itest = 50 W Option (C) is correct. (s2 + 9) (s + 2) G (s) = (s + 1) (s + 3) (s + 4) (- w2 + 9) (jw + 2) G (jw) = (jw + 1) (jw + 3) (jw + 4) The steady state output will be zero if G (jw) = 0 -w 2 + 9 = 0 w = 3 rad/s

SOL 1.12

Option (B) is correct. In phasor form

A I D

Z = 4 - j3 Z = 5 - 36.86cW I = 5 100c A Average power delivered. n Pavg. = 1 I 2 Z cos q = 1 # 25 # o5.icos 36.86c = 50 W c 2 2 . a

O N

Alternate method:

SOL 1.13

ww

w

.no

di

Z = (4 - j3) W I = 5 cos (100pt + 100) A 2 Pavg = 1 Re $ I Z . = 1 # Re "(5) 2 # (4 - j3), = 1 # 100 = 50 W 2 2 2 Option (D) is correct. The s -domain equivalent circuit is shown as below.

vc (0) /s v (0) = c 1 + 1 1 + 1 C1 s C 2 s C1 C 2 I (s) = b C1 C2 l (12 V) C1 + C 2 I (s) =

vC (0) = 12 V

I (s) = 12Ceq Taking inverse Laplace transform for the current in time domain,

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i (t) = 12Ceq d (t) SOL 1.14

Page 37

(Impulse)

Option (A) is correct. In the given circuit,

VA - VB = 6 V So current in the branch, IAB = 6 = 3 A 2 We can see, that the circuit is a one port circuit looking from terminal BD as shown below

For a one port network current entering one terminal, equals the current leaving the second terminal. Thus the outgoing current from A to B will be equal to the incoming current from D to C as shown i.e. IDC = IAB = 3 A

A I D

O N

no w.

.in

co ia.

d

ww

The total current in the resistor 1 W will be I1 = 2 + IDC = 2+3 = 5A So, VCD = 1 # (- I1) =- 5 V SOL 1.15

(By writing KCL at node D )

Option (A) is correct. We obtain Thevenin equivalent of circuit B .

Thevenin Impedance :

ZTh = R

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Thevenin Voltage : VTh = 3 0c V Now, circuit becomes as

I1 = 10 - 3 2+R Power transfer from circuit A to B P = (I 12) 2 R + 3I1 Current in the circuit,

2 = :10 - 3D R + 3 :10 - 3D = 49R 2 + 21 2+R 2+R (2 + R) (2 + R) 49R + 21 (2 + R) = = 42 + 70R2 (2 + R) 2 (2 + R) 2 dP = (2 + R) 70 - (42 + 70R) 2 (2 + R) = 0 dR (2 + R) 4 (2 + R) [(2 + R) 70 - (42 + 70R) 2] = 0 .in0 140 + 70R - 84 - 140Rco= . dia56 = 70R o .n w R = 0.8 W w

A I D

SOL 1.16

O N w

Option (C) is correct. When 10 V is connected at port A the network is

Now, we obtain Thevenin equivalent for the circuit seen at load terminal, let Thevenin voltage is VTh, 10 V with 10 V applied at port A and Thevenin resistance is RTh .

IL =

VTh,10 V RTh + RL

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For RL = 1 W , IL = 3 A VTh,10 V RTh + 1 For RL = 2.5 W , IL = 2 A V 2 = Th,10 V RTh + 2.5 Dividing above two 3 = RTh + 2.5 2 RTh + 1 3=

...(i)

...(ii)

3RTh + 3 = 2RTh + 5 RTh = 2 W Substituting RTh into equation (i) VTh,10 V = 3 (2 + 1) = 9 V Note that it is a non reciprocal two port network. Thevenin voltage seen at port B depends on the voltage connected at port A. Therefore we took subscript VTh,10 V . This is Thevenin voltage only when 10 V source is connected at input port A. If the voltage connected to port A is different, then Thevenin voltage will be different. However, Thevenin’s resistance remains same. Now, the circuit is

A I D

For RL = 7 W , SOL 1.17

O N IL =

no w.

.in

co ia.

d

VTh,10 V w = 9 = 1A 2 + RwL 2 + 7

Option (B) is correct. Now, when 6 V connected at port A let Thevenin voltage seen at port B is VTh,6 V . Here RL = 1 W and IL = 7 A 3

VTh, 6 V = RTh # 7 + 1 # 7 = 2 # 7 + 7 = 7 V 3 3 3 3 This is a linear network, so VTh at port B can be written as VTh = V1 a + b where V1 is the input applied at port A. We have V1 = 10 V , VTh,10 V = 9 V 9 = 10a + b When V1 = 6 V , VTh, 6 V = 9 V 7 = 6a + b Solving (i) and (ii) a = 0.5 , b = 4

...(i) ...(ii)

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Thus, with any voltage V1 applied at port A, Thevenin voltage or open circuit voltage at port B will be So, VTh, V = 0.5V1 + 4 For V1 = 8 V (open circuit voltage) VTh,8 V = 0.5 # 8 + 4 = 8 = Voc 1

SOL 1.18

Option (A) is correct. By taking V1, V2 and V3 all are phasor voltages. V1 = V2 + V3 Magnitude of V1, V2 and V3 are given as V1 = 220 V , V2 = 122 V , V3 = 136 V Since voltage across R is in same phase with V1 and the voltage V3 has a phase difference of q with voltage V1 , we write in polar form V1 = V2 0c + V3 q V1 = V2 + V3 cos q + jV3 sin q V1 = (V2 + V3 cos q) + jV3 sin q V1 =

(V2 + V3 cos q) 2 + (V2 sin q) 2

A I D

220 = (122 + 136 cos q) 2 + (136 sin q) 2 By solving, power factor cos q = 0.45 SOL 1.19

Option (B) is correct. Voltage across load resistance

O N

.in o c . VRL = V3 cos q = 136ia# 0.45 = 61.2 V d Power absorbed in RL no . 2 ww (61.2) 2 V RL PL = w= - 750 W RL

SOL 1.20

5

Option (B) is correct. The frequency domains equivalent circuit at w = 1 rad/ sec .

Since the capacitor and inductive reactances are equal in magnitude, the net impedance of that branch will become zero. Equivalent circuit

Current

i (t) = sin t = (1 sin t) A 1W

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rms value of current i rms = 1 A 2 SOL 1.21

Option (D) is correct. Voltage in time domain v (t) = 100 2 cos (100pt) Current in time domain i (t) = 10 2 sin (100pt + p/4) Applying the following trigonometric identity sin (f) = cos (f - 90c) So,

i (t) = 10 2 cos (100pt + p/4 - p/2) = 10 2 cos (100pt - p/4)

In phasor form, SOL 1.22

I = 10 2 - p/4 2

Option (A) is correct.

O N

A I D

Power transferred to the load

.in

co ia.

1o0d 2 R n R.th + RL l L w w

P = I 2 RL = b

For maximum power transferwRth , should be minimum. Rth = 6R = 0 6+R R =0 Note: Since load resistance is constant so we choose a minimum value of Rth SOL 1.23

Option (C) is correct.

2 (5 103) 2 Power loss = V rated = # = 20 W Rp 1.25 # 106 For an parallel combination of resistance and capacitor 1 1 tan d = = 1 = 0.025 = 40 wC p R p 2p # 50 # 1.25 # 0.102

SOL 1.24

Option (C) is correct. Charge Q = CV = e0 er A V = (e0 er A)V d d

C = e0 er A d

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Q = Q max We have e0 = 8.85 # 10-14 F/cm , er = 2.26 , A = 20 # 40 cm2 V = 50 103 kV/cm # d Maximum electrical charge on the capacitor V = V when b d l = 50 kV/cm d max Thus, SOL 1.25

Q = 8.85 # 10-14 # 2.26 # 20 # 40 # 50 # 103 = 8 mC

Option (C) is correct. vi = 100 2 sin (100pt) V Fundamental component of current ii = 10 2 sin (100pt - p/3) A 1

Input power factor I1 (rms) cos f1 Irms I1 (rms) is rms values of fundamental component of current and Irms is the rms value of converter current. 10 pf = cos p/3 = 0.44 102 + 52 + 22 Option (B) is correct. Only the fundamental component of current contributes to the mean ac input power. The power due to the harmonic components of current is zero. in p/3 = 500 W o.cos So, Pin = Vrms I1rms cos f1 = 100 #.c10 pf =

SOL 1.26

SOL 1.27

A I D

O N w

P = Vs Is cos p/4 = 1 # SOL 1.28

ia

d Option (B) is correct. no . Power delivered by the source wwwill be equal to power dissipated by the resistor. 2 cos p/4 = 1 W

Option (D) is correct.

IC = Is - I RL = =

2 p /4 -

2 - p /4

2 $^cos p/4 + j sin p/4h - ^cos p/4 - j sin p/4h.

= 2 2 j sin p/4 = 2j SOL 1.29

Option (B) is correct. For t < 0 , the switch was closed for a long time so equivalent circuit is

Voltage across capacitor at t = 0 vc (0) = 5 = 4 V 4#1 Now switch is opened, so equivalent circuit is

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For capacitor at t = 0+ vc (0+) = vc (0) = 4 V current in 4 W resistor at t = 0+ , i1 =

vc (0+) =1A 4

so current in capacitor at t = 0+ , ic (0+) = i1 = 1 A SOL 1.30

Option (B) is correct. Thevenin equivalent across 1 X resistor can be obtain as following Open circuit voltage vth = 100 V (i = 0) Short circuit current (vth = 0 ) isc = 100 A So, Rth = vth = 100 = 1 W isc 100

A I D

Equivalent circuit is

O N

n i = 100 = 50 A co.i 1+1 . a

SOL 1.31

Option (B) is correct. The circuit is

w

.no

ww

di

Current in R W resistor is i = 2-1 = 1 A Voltage across 12 W resistor is So, SOL 1.32

VA = 1 # 12 = 12 V i = VA - 6 = 12 - 6 = 6 W 1 R

Option (C) is correct.

V1 = Z11 I1 + Z12 I2 V2 = Z21 I1 + Z22 I2 Here, I1 = I l1, I2 = I l2

V l1 = Zl11 I l1 + Zl12 I l2 V l2 = Zl21 I l1 + Zl22 I l2

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When R = 1 W is connected V l1 = V1 + I l1 # 1 = V1 + I1 V l1 = Z11 I1 + Z12 I2 + I1 V l1 = (Z11 + 1) I1 + Z12 I2 So, Zl11 = Z11 + 1 Zl12 = Z12 Similarly for output port V l2 = Zl21 I l1 + Zl22 I l2 = Zl21 I1 + Zl22 I2 So, Zl21 = Z21 , Zl22 = Z22 Z11 + 1 Z12 Z-matrix is Z => Z21 Z22H SOL 1.33

Option (A) is correct.

In the bridge

A I D

O N

.in

od

.n R1 R 4 = R 2 R 3 = 1 w w So it is a balanced bridge w I = 0 mA SOL 1.34

co ia.

Option (D) is correct. Resistance of the bulb rated 200 W/220 V is 2 (220) 2 R1 = V = = 242 W 200 P1 Resistance of 100 W/220 V lamp is 2 (220) 2 RT = V = = 484 W 100 P2 To connect in series RT = n # R1 484 = n # 242 n =2

SOL 1.35

Option (D) is correct. For t < 0 , S1 is closed and S2 is opened so the capacitor C1 will charged upto 3 volt. VC1 (0) = 3 Volt Now when switch positions are changed, by applying charge conservation Ceq VC (0+) = C1 VC (0+) + C2 VC (0+) (2 + 1) # 3 = 1 # 3 + 2 # VC (0+) 1

1

2

2

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9 = 3 + 2VC (0+) VC (0+) = 3 Volt 2

2

SOL 1.36

Option (A) is correct.

Applying KVL in the input loop v1 - i1 (1 + 1) # 103 - 1 (i1 + 49i1) = 0 jw C v1 = 2 # 103 i1 + 1 50i1 jw C 1 Input impedance, Z1 = v1 = 2 # 103 + i1 jw (C/50) 100 nF Equivalent capacitance, Ceq = C = = 2 nF 50 50 SOL 1.37

SOL 1.38

A I D

Option (B) is correct. Voltage across 2 X resistor, VS = 2 V Current, I2W = VS = 4 = 2 A 2 2 n o.i To make the current double we have to take c . dia VS = 8 V o n

O N

w.

Option (B) is correct. ww To obtain equivalent Thevenin circuit, put a test source between terminals AB

Applying KCL at super node VP - 5 + VP + VS = I S 2 2 1 VP - 5 + VP + 2VS = 2IS 2VP + 2VS = 2Is + 5 VP + VS = IS + 2.5 VP - VS = 3VS & VP = 4VS So, 4VS + VS = IS + 2.5 5VS = IS + 2.5 VS = 0.2IS + 0.5 For Thevenin equivalent circuit

...(1)

...(2)

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VS = IS Rth + Vth By comparing (2) and (3), Thevenin resistance Rth = 0.2 kW SOL 1.39

Option (D) is correct. From above Vth = 0.5 V

SOL 1.40

Option (A) is correct. No. of chords is given as l = b-n+1 b " no. of branches n " no. of nodes l " no. of chords

...(3)

A I D

b = 6, n = 4 l = 6 - 4 + 1= 3 SOL 1.41

Page 46

Option (A) is correct. Impedance Zo = 2.38 - j0.667 W n Constant term in impedance indicates that o.ithere is a resistance in the circuit. c . ia Assume that only a resistance and d capacitor are in the circuit, phase difference o n in Thevenin voltage is given asw. ww) (Due to capacitor) q =- tan- 1 (wCR j Zo = R wC 1 = 0.667 So, wC

O N

and

R = 2.38 W q =- tan- 1 b 1 # 2.38 l =- 74.34c =[ 15.9c 0.667

given Voc = 3.71+ - 15.9c So, there is an inductor also connected in the circuit SOL 1.42

Option (C) is correct. Time constant of the circuit can be calculated by simplifying the circuit as follows

Ceq = 2 F 3 Equivalent Resistance

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Req = 3 + 3 = 6 W t = Req Ceq = 6 # 2 = 4 sec 3 Option (C) is correct. Impedance of the circuit is 1 R 1 - jwCR R Z = jwL + 1jwC = jw L + # 1 + jwCR 1 - jwCR j wC + R Time constant SOL 1.43

R (1 - jwCR) jwL (1 + w2 C2 R2) + R - jwCR2 = 1 + w2 C2 R2 1 + w2 C2 R2 j [wL (1 + w2 C2 R2) - wCR2] R = + 1 + w2 C2 R2 1 + w2 C2 R2 For resonance Im (Z) = 0 So, wL (1 + w2 C2 R2) = wCR2 L = 0.1 H, C = 1 F, R = 1 W So, w # 0.1 [1 + w2 (1) (1)] = w (1) (1) 2 = jw L +

1 + w2 = 10 & SOL 1.44

A I D

O N w=

9 = 3 rad/sec

Option (A) is correct. By applying KVL in the circuit

no w.

.in

co ia.

d

Vab - 2i + 5 = 0 ww i = 1 A, Vab = 2 # 1 - 5 =- 3 Volt SOL 1.45

Option (C) is correct. Charge stored at t = 5 m sec 5

Q =

# i (t) dt

=area under the curve

0

Q =Area OABCDO =Area (OAD)+Area(AEB)+Area(EBCD) = 1#2#4+1#2#3+3#2 2 2

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= 4 + 3 + 6 = 13 nC SOL 1.46

Option (D) is correct. Initial voltage across capacitor Q V0 = o = 13 nC = 43.33 Volt 0.3 nF C When capacitor is connected across an inductor it will give sinusoidal esponse as where At t = 1 m sec ,

SOL 1.47

vc (t) = Vo cos wo t 1 wo = 1 = -9 LC 0.3 # 10 # 0.6 # 10- 3 = 2.35 # 106 rad/sec vc (t) = 43.33 cos (2.35 # 106 # 1 # 10- 6) = 43.33 # (- 0.70) =- 30.44 V

Option (B) is correct. By writing node equations at node A and B Va - 5 + Va - 0 = 0 1 1 2Va - 5 = 0 Va = 2.5 V Similarly Vb - 4Vab ++Vb - 0 = 0 3 1 Vb - 4 (Va - Vb) + Vb = 0 3

O N

A I D .in

co ia.

Vb - 4 (2.5 - Vb) + 3Vb = 0 o d .n w 8Vb - 10 = 0 ww Vb = 1.25 V Current i = Vb = 1.25 A 1 SOL 1.48

Option ( ) is correct.

SOL 1.49

Option (B) is correct. Here two capacitance C1 and C2 are connected in series, so equivalent capacitance is Ceq = C1 C2 C1 + C 2

- 12 -6 # 500 # 10 C1 = e0 er1 A = 8.85 # 10 # 8 # 500 d1 4 # 10- 3

= 442.5 # 10- 11 F - 12 -6 # 500 # 10 C2 = e0 er2 A = 8.85 # 10 # 2 # 500 d2 2 # 10- 3 = 221.25 # 10- 11 F - 11 - 11 Ceq = 442.5 # 10 - 11 # 221.25 # 10- 11 = 147.6 # 10- 11 442.5 # 10 + 221.25 # 10

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- 1476 pF SOL 1.50

Option (B) is correct. Circumference no. of turns Cross sectional area

l = 300 mm n = 300 A = 300 mm2

Inductance of coil

L =

4p # 10- 7 # (300) 2 # 300 # 10- 6 m0 n2 A = l (300 # 10- 3)

= 113.04 mH SOL 1.51

Option (A) is correct. Divergence of a vector field is given as Divergence = 4: V In cartesian coordinates 4 = 2 it + 2 tj + 2 kt 2x 2y 2z So 4: V = 2 6- (x cos xy + y)@ + 2 6(y cos xy)@ + 2 6(sin z2 + x2 + y2)@ 2x 2y 2z =- x (- sin xy) y + y (- sin xy) x + 2z cos z2 = 2z cos z2

SOL 1.52

A I D

Option (A) is correct. Writing KVL for both the loops V - 3 (I1 + I2) - Vx - 0.5 dI1 = 0 dt V - 3I1 - 3I2 - Vx - 0.5 dI1 = 0 dt In second loop

O N

no w.

.in

co ia.

...(1)

d

- 5I2 + 0.2Vx + 0w.5wdI1 = 0 dt

SOL 1.53

...(2) I2 = 0.04Vx + 0.1 dI1 dt Put I2 from eq(2) into eq(2) V - 3I1 - 3 :0.04Vx + 0.1 dI1 D - Vx - 0.5 dI1 = 0 dt dt 0.8 dI1 =- 1.12Vx - 3I1 + V dt dI1 =- 1.4V - 3.75I + 5 V x 1 4 dt Option (A) is correct. Impedance of the given network is Z = R + j b wL - 1 l wC 1 AdmittanceY = 1 = Z R + j b wL - 1 l wC R - j b wL - 1 l R - j b wL - 1 l wC wC 1 = = # 2 1 1 2 R + j b wL R - j b wL R + b wL - 1 l l l wC wC wC j b wL - 1 l wC R = 2 2 R 2 + b wL - 1 l R 2 + b wL - 1 l wC wC

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Page 50

= Re (Y) + Im (Y) Varying frequency for Re (Y) and Im (Y) we can obtain the admittance-locus.

SOL 1.54

Option (D) is correct. At t = 0+ , when switch positions are changed inductor current and capacitor voltage does not change simultaneously So at t = 0+ vc (0+) = vc (0-) = 10 V iL (0+) = iL (0-) = 10 A The equivalent circuit is

A I D

O N

.in

co ia.

od

Applying KCL .n w vL (0+) vL (0+) - vc (0+)ww + = iL (0+) = 10 10 10

2vL (0+) - 10 = 100 Voltage across inductor at t = 0+ vL (0+) = 100 + 10 = 55 V 2

SOL 1.55

So, current in capacitor at t = 0+ v (0+) - vc (0+) iC (0+) = L = 55 - 10 = 4.5 A 10 10 Option (B) is correct. In the circuit

VX = V+0c Vy - 2V+0c + (Vy) jwC = 0 R

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Vy (1 + jwCR) = 2V+0c Vy = 2V+0c 1 + jwCR VYX = VX - VY = V VYX = V - 2V =- V VYX = V - 0 = V

R " 0, R " 3, SOL 1.56

2V 1 + jwCR

Option (A) is correct. The circuit is

Applying KVL 2 = VNL 3 - 2 # I NL 2 2 3 - 2I NL = I NL 2 = 3 & INL = 1 A 3I NL VNL = (1) 2 = 1 V So power dissipated in the non-linear resistance P = VNL INL = 1 # 1 = 1oW .in

SOL 1.57

SOL 1.58

A I D

O N

Option (C) is correct. In node incidence matrix b1 b 2 b 3 b 4 b 5 b 6 V R n1 S 1 1 1 0 0 0 W n2S 0 - 1 0 - 1 1 0 W n 3SS- 1 0 0 0 - 1 - 1WW n 4S 0 0 - 1 1 0 1 W X T In option (C)

.c

ia od

n

. ww

w

E = AV R V S1 1 1 0 0 0W S 0 -1 0 -1 1 0 W T T 8e1 e2 e 3 e 4B = S- 1 0 0 0 - 1 - 1W8V1 V2 -- V6B S W S 0 0 -1 1 0 1 W X R V TR V Se1W S V1 + V2 + V3 W Se2W S- V2 - V4 + V5W Se W = S- V - V - V W which is true. 5 6W S 3W S 1 Se 4W S- V3 + V4 + V6W T X T X Option (A) is correct. Assume a Gaussian surface inside the sphere (x < R)

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From gauss law y = Q enclosed =

# D : ds = Q enclosed

3 Q 4 pr3 = Qr # 3 4 3 R3 3 pR 3 # D : ds = Qr3 R Q r Qr3 D # 4p r 2 = 3 = pe0 R3 4 R

Q enclosed =

So, or SOL 1.59

SOL 1.60

SOL 1.61

a D = e0 E

A I D

Option (D) is correct. Inductance is given as

4p # 10- 7 # (400) 2 # (16 # 10- 4) m0 N2 A = L = = 321.6 mH (1 # 10- 3) l V = IXL = 230 ` XL = 2pfL n 2pfL o.i c . 230 o dia = .n321.6 # 10- 3 = 2.28 A 2 # 3.14 # 50w# Option (A) is correct. ww Energy stored is inductor E = 1 LI2 = 1 # 321.6 # 10- 3 # (2.28) 2 2 2 Force required to reduce the air gap of length 1 mm is F = E = 0.835- 3 = 835 N l 1 # 10 Option (D) is correct. Thevenin voltage:

O N

Vth = I (R + ZL + ZC ) = 1+0c [1 + 2j - j] = 1 (1 + j) = Thevenin impedance:

2 +45% V

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Zth = R + ZL + ZC = 1 + 2j - j = (1 + j) W SOL 1.62

Option (A) is correct. In the given circuit

Output voltage

SOL 1.63

vo = Avi = 106 # 1 mV = 1 V Input impedance Zi = vi = vi = 3 0 ii Output impedance Zo = vo = Avi = Ro = 10 W io io n o.i Option (D) is correct. c . diasources, so all inductors behaves as short All sources present in the circuit are o DC .n w circuit and all capacitors as open circuit ww Equivalent circuit is

A I D

O N

Voltage across R 3 is 5 = I1 R 3 5 = I1 (1) I1 = 5 A By applying KCL, current through voltage source

(current through R 3 )

1 + I 2 = I1 I2 = 5 - 1 = 4 A SOL 1.64

Option () is correct. Given Two port network can be described in terms of h-parametrs only.

SOL 1.65

Option (A) is correct. At resonance reactance of the circuit would be zero and voltage across inductor and capacitor would be equal

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Page 54

VL = VC At resonance impedance of the circuit Current

Z R = R1 + R 2 IR = V1 +0c R1 + R 2

V2 = IR R2 + j (VL - VC ) V2 = V1 +0c R2 R1 + R 2 Voltage across capacitor VC = 1 # IR = 1 # VR +0c = VR + - 90c R1 + R 2 jw C jw C wC (R1 + R2) So phasor diagram is Voltage

SOL 1.66

A I D

Option (B) is correct. This is a second order LC circuit shown below

O N

no w.

.in

co ia.

d

Capacitor current is given w asw dv (t) iC (t) = C c dt Taking Laplace transform

IC (s) = CsV (s) - V (0), V (0) "initial voltage Current in inductor iL (t) = 1 # vc (t) dt L V (s) IL (s) = 1 L s for t > 0 , applying KCL(in s-domain) IC (s) + IL (s) = 0 V (s) =0 CsV (s) - V (0) + 1 L s 1 2 :s + LCs D V (s) = Vo V (s) = Vo 2 s 2 , s + w0 Taking inverse Laplace transformation v (t) = Vo cos wo t , t > 0 SOL 1.67

a w20 = 1 LC

Option (B) is correct. Power dissipated in heater when AC source is connected

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Page 55

2

P = 2.3 kW = V rms R 2.3 # 103 =

(230) 2 R

R = 23 W (Resistance of heater) Now it is connected with a square wave source of 400 V peak to peak Power dissipated is 2 P = V rms , Vp - p = 400 V & Vp = 200 V R

SOL 1.68

(200) 2 = = 1.739 kW 23 Option (D) is correct. From maxwell’s first equation

Vrms = Vp =200 (for square wave)

4: D = rv r 4: E = v e (Divergence of electric field intensity is non-Zero) Maxwell’s fourth equation

A I D

4: B = 0 (Divergence of magnetic field intensity is zero) SOL 1.69

Option (C) is correct. Current in the circuit

O N I =

Or SOL 1.70

100 = 8 .no w R+5 ww 60 R = = 7.5 W 8

32 +

(4) 2 = 2

9 + 8 = 17 V

Option (D) is correct. Writing KVL in input and output loops V1 - (i1 + i2) Z1 = 0 V1 = Z1 i1 + Z1 i2 Similarly

SOL 1.72

d

(given)

Option (A) is correct. Rms value is given as mrms =

SOL 1.71

in 100 = 8.cAo. R + (10 || 10) ia

...(1)

V2 - i2 Z2 - (i1 + i2) Z1 = 0 ...(2) V2 = Z1 i1 + (Z1 + Z2) i2 From equation (1) and (2) Z -matrix is given as Z1 Z1 Z => Z1 Z1 + Z2H Option (B) is correct. In final steady state the capacitor will be completely charged and behaves as an open circuit

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SOL 1.73

Page 56

Steady state voltage across capacitor vc (3) = 20 (10) = 10 V 10 + 10 Option (D) is correct. We know that divergence of the curl of any vector field is zero 4 (4 # E) = 0

SOL 1.74

Option (A) is correct. When the switch is at position 1, current in inductor is given as

A I D

120 = 2 A 20 + 40 At t = 0 , when switch is moved to position 1,inductor current does not change simultaneously so iL (0-) =

O N

no w.

.in

co ia.

d

ww

iL (0+) = iL (0-)=2 A Voltage across inductor at t = 0+ vL (0+) = 120 V By applying KVL in loop 120 = 2 (40 + R + 20) 120 = 120 + R R = 0W SOL 1.75

Option (C) is correct. Let stored energy and dissipated energy are E1 and E2 respectively. Then Current i 22 = E2 = 0.95 E1 i 12 i2 = 0.95 i1 = 0.97i1 Current at any time t, when the switch is in position (2) is given by R

60

i (t) = i1 e- L t = 2e- 10 t = 2e- 6t After 95% of energy dissipated current remaining in the circuit is i = 2 - 2 # 0.97 = 0.05 A

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So, SOL 1.76

Page 57

0.05 = 2e- 6t t . 0.50 sec

Option (C) is correct. At f1 = 100 Hz, voltage drop across R and L is mRMS V (jw L) mRMS = Vin .R = in 1 R + jw1 L R + jw1 L So, R = w1 L At f2 = 50 Hz, voltage drop across R mlRMS = Vin .R R + jw2 L mRMS R + jw2 L = = R + jw1 L mlRMS =

w12 L2 + w22 L2 , w12 L2 + w12 L2

=

w12 + w22 = 2w12 8m 5 RMS

mlRMS = SOL 1.77

R2 + w22 L2 R2 + w12 L2 R = w1 L f 12 + f 22 2f 12

=

(100) 2 + (50) 2 = 2 (100) 2

5 8

A I D

Option (A) is correct. In the circuit

I B = IR +0c + Iy +120c

Since so,

O N

I B2 = I R2 + I y2 + 2IR Iy cos b 120 nc l = I R2 + I y2 + IR Iy i . 2 co I R = Iy ia.

d

o 3I R2 I B2 = I R2 + I R2 + I.R2n= IB =

w

w 3 IRw=

3 Iy

IR: Iy: IB = 1: 1: 3 SOL 1.78

Option (C) is correct. Switch was opened before t = 0 , so current in inductor for t < 0

iL (0-) = 10 = 1 A 10 Inductor current does not change simultaneously so at t = 0 when switch is closed current remains same iL (0+) = iL (0-)=1 A SOL 1.79

Option (A) is correct. Thevenin voltage:

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Nodal analysis at P Vth - 4 + Vth = 0 10 10 2Vth - 4 = 0 Vth = 2 V Thevenin resistance:

Rth = 10 W || 10 W = 5 W SOL 1.80

Option (A) is correct. Electric field inside a conductor (metal) is zero. In dielectric charge distribution os constant so electric field remains constant from x1 to x2 . In semiconductor electric field varies linearly with charge density.

SOL 1.81

Option (D) is correct. Resonance will occur only when Z is capacitive, in parallel resonance condition, suseptance of circuit should be zero. 1 + jw C = 0 jw L in

O N

1 - w2 LC = 0

o.

.c dia

1 (resonant .no frequency) w LC ww 1 C = 12 = = 0.05 m F 2 wL 4 # p # (500) 2 # 2 Option (D) is correct. Here two capacitor C1 and C2 are connected in series so equivalent Capacitance is Ceq = C1 C2 C1 + C 2 w=

SOL 1.82

A I D

8.85 # 10- 12 # 4 (400 # 10- 3) 2 C1 = e0 er1 A = d1 6 # 10- 3 Similarly

- 12 16 # 10- 2 = 94.4 10- 11 F = 8.85 # 10 # 4 # # 6 # 10- 3

8.85 # 10- 12 # 2 # (400 # 10- 3) 2 C2 = e0 er2 A = d2 8 # 10- 3

SOL 1.83

- 12 16 # 10- 12 = 35.4 10- 11 F = 8.85 # 10 # 2 # # -3 8 # 10 - 11 - 11 Ceq = 94.4 # 10 # 35.4 #-10 = 25.74 # 10- 11 - 257 pF (94.4 + 35.4) # 10 11 Option (C) is correct. Inductance of the Solenoid is given as m N2 A L = 0 l

Where

A " are of Solenoid

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Page 59

l " length 4p # 10- 7 # (3000) 2 # p (30 # 10- 3) 2 L = = 31.94 # 10- 3 H (1000 # 10- 3) - 32 mH SOL 1.84

Option (C) is correct. In the circuit

VA = (2 + 1) # 6 = 18 Volt 2 = E - VA 6 2 = E - 18 6

Voltage So,

E = 12 + 18 = 30 V SOL 1.85

SOL 1.86

A I D

Option (A) is correct. Delta to star (T - Y) conversions is given as Rb Rc R1 = = 10 # 10 = 2.5 W 20 + 10 + 10 Ra + Rb + Rc Ra Rc R2 = = 20 # 10o.in= 5 W c 10 20 + 10 + Ra + Rb + Rc ia. d o Ra Rb R3 = = .n20 # 10 = 5 W + 10 + 10 Ra + Rb + Rc ww20

O N

Option (D) is correct. For parallel circuit

w

I = E = EYeq Zeq Yeq " Equivalent admittance of the circuit Yeq = YR + YL + YC = (0.5 + j0) + (0 - j1.5) + (0 + j0.3) = 0.5 - j1.2 So, current I = 10 (0.5 - j1.2) = (5 - j12) A SOL 1.87

Option (B) is correct. In the circuit

100 10R 100 (10 || R) = # b 10 R 10 +Rl 10 + (10 || R) f 10 + p 10 + R 1000 R 50 R = = 100 + 20R 5+R Current in R W resistor Voltage

VA =

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Page 60

2 = VA R 2=

R = 20 W

or SOL 1.88

50R R (5 + R)

Option (A) is correct. Since capacitor initially has a charge of 10 coulomb, therefore Q 0 = Cvc (0) vc (0) " initial voltage across capacitor 10 = 0.5vc (0) vc (0) = 10 = 20 V 0.5 When switch S is closed, in steady state capacitor will be charged completely and capacitor voltage is vc (3) = 100 V At any time t transient response is t

vc (t) = vc (3) + [vc (0) - vc (3)] e- RC t

vc (t) = 100 + (20 - 100) e- 2 # 0.5 = 100 - 80e- t Current in the circuit i (t) = C dvc = C d [100 - 80e- t] dt dt

A I D

O N

= C # 80e- t = 0.5 # 80e- t = 40e- t

At t = 1 sec, SOL 1.89

.in

.co a i A i (t) = 40e- 1 = 14.71 d no . w Option (D) is correct. ww Total current in the wire I = 10 + 20 sin wt 102 +

Irms = SOL 1.90

300 = 17.32 A

Option (D) is correct. From Z to Y parameter conversion Y11 Y12 Z11 Z12 - 1 >Y Y H = >Z Z H 21 22 21 22 So,

SOL 1.91

(20) 2 = 100 + 200 = 2

0.6 - 0.2 Y11 Y12 >Y Y H = 0.150 >- 0.2 0.9 H 12 22 Y22 = 0.9 = 1.8 0.50

Option (C) is correct. Energy absorbed by the inductor coil is given as t

EL =

# Pdt 0

Where power

P = VI = I bL dI l dt t

So,

EL =

dt # LIb dI dt l

0

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Page 61

For0 # t # 4 sec 4

dt # Ib dI dt l

EL = 2

0

a dI = 3, 0 # t # 2 , * dt 0 2 = 0, 2 < t < 4 2 = 6 # I.dt =6(area under the curve i (t) - t ) 2

4

# I (3) dt + 2 # I (0) dt

=2

0

= 6 # 1 # 2 # 6 = 36 J 2 Energy absorbed by 1 W resistor is t

ER =

# I2 Rdt 0

=

4

2

# (3t)

2

I = 3t, 0 # t # 2 ) = 6A 2 # t # 4

# (6) 2 dt

# 1dt +

2

0 3 2

4 = 9 # :t D + 36[t]2 = 24 + 72 =96 J 3 0 Total energy absorbed in 4 sec E = EL + ER = 36 + 96 = 132 J

SOL 1.92

Option (B) is correct. Applying KCL at center node

O N

A I D

no w.

.in

co ia.

d

ww

iL = iC + 1 + 2 iL = iC + 3 iC =- C dvc =- 1 d [4 sin 2t] dt dt

SOL 1.93

=- 8 cos 2t so (current through inductor) iL =- 8 cos 2t + 3 Voltage across inductor vL = L diL = 2 # d [3 - 8 cos 2t] = 32 sin 2t dt dt Option (A) is correct. Thevenin impedance can be obtain as following

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Page 62

Zth = Z 3 + (Z1 || Z2) Given that

Z1 = 10+ - 60c = 10 c

1-

2

3j

m = 5 (1 -

3 j)

1+

3j m = 5 (1 + 3 j) 2 3 + 4j Z 3 = 50+53.13c = 50 b = 10 (3 + 4j) 5 l Z2 = 10+60c = 10 c

So,

5 (1 - 3j) 5 (1 + 3 j) 5 (1 - 3 j) + 5 (1 + 3 j) 25 (1 + 3) = 30 + 40j + 10 = 40 + 40j = 10 (3 + 4j) + 10

Zth = 10 (3 + 4j) +

Zth = 40 2 +45c W SOL 1.94

SOL 1.95 SOL 1.96

Option (A) is correct. Due to the first conductor carrying + I current, magnetic field intensity at point P is H 1 = I Y (Direction is determined using right hand rule) 2pd Similarly due to second conductor carrying - I current, magnetic field intensity is H 2 = - I (- Y) = I Y 2pd 2pd Total magnetic field intensity at point P. H = H1 + H 2 = I Y + I Y = I Y 2pd 2pd n pd

A I D

O N

Option ( ) is correct.

.i

co ia.

od

Option (C) is correct. .n w Given that magnitudes of V wLwand VC are twice of VR VL = VC = 2VR (Circuit is at resonance) Voltage across inductor VL = iR # jwL Current iR at resonance % iR = 5+0 = 5 = 1 A 5 R so,

SOL 1.97

VL = wL = 2VR wL = 2 # 5 2 # p # 50 # L = 10 L = 10 = 31.8 mH 314

VR = 5 V, at resonance

Option (C) is correct. Applying nodal analysis in the circuit At node P 2 + VP - 10 + VP = 0 2 8 16 + 4VP - 40 + VP = 0 5VP - 24 = 0 VP = 24 Volt 5 At node Q

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2=

SOL 1.98

Page 63

VQ - 10 VQ - 0 + 6 4

24 = 3VQ - 30 + 2VQ 5VQ - 54 = 0 VQ = 54 V 5 Potential difference between P-Q VPQ = VP - VQ = 24 - 54 =- 6 V 5 5 Option (D) is correct. First obtain equivalent Thevenin circuit across load RL

Thevenin voltage Vth - 110+0c + Vth - 90+0c 0 = 6 + 8j 6 + 8j

A I D

2Vth - 200+0c = 0 Vth = 100+0c V Thevenin impedance

O N

no w.

.in

co ia.

d

ww

Zth = (6 + 8j) W || (6 + 8j) W = (3 + 4j) W For maximum power transfer RL = Zth =

32 + 42 = 5 W

Power in load 2 P = ieff RL

SOL 1.99

2

2 (100) 100 5 = 625 Watt 5 = 80 # 3 + 4j + 5 # Option (D) is correct. By applying mesh analysis in the circuit

P =

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Page 64

I1 = 10 A, I2 =- 5 A Current in 2 W resistor I2W = I1 - (- I2) = 10 - (- 5) = 15 A So, voltage VA = 15 # 2 = 30 Volt Now we can easily find out current in all branches as following

A I D

O N

Current in resistor R is 5 A n 5 = 100 - 40 o.i c R . ia d o .Wn R = 60 = 12 5 ww SOL 1.100

w

Option (B) is correct. Before t = 0 , the switch was opened so current in inductor and voltage across capacitor for t < 0 is zero vc (0-) = 0 , iL (0-) = 0 at t = 0 , when the switch is closed, inductor current and capacitor voltage does not change simultaneously so vc (0+) = vc (0-) = 0 , iL (0+) = iL (0-) = 0 At t = 0+ the circuit is

Simplified circuit

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SOL 1.101

Page 65

Voltage across inductor (at t = 0+ ) vL (0+) = 10 # 2 = 4 Volt 3+2 Option (D) is correct. Given that E1 = h11 I1 + h12 E2 and I2 = h21 I1 + h22 E2 Parameter h12 is given as h12 = E1 E2 I = 0 (open circuit) 1

At node A E A - E1 + E A - E 2 + E A = 0 2 2 4

A I D

5EA = 2E1 + 2E2 Similarly E1 - E A + E1 = 0 2 2

O N 2E1 = EA

From (1) and (2)

.in

co ia.

...(1)

...(2)

od n . 5 (2E1) = 2E w1 + 2E2 4 ww 8E1 = 2E2 h12 = E1 = 1 4 E2

SOL 1.102

Option (B) is correct. VPQ = VP - VQ =

KQ KQ OP OQ

9 10- 9 - 9 # 109 # 1 # 10- 9 = 9 # 10 # 1 # -3 40 # 10 20 # 10- 3 = 9 # 103 : 1 - 1 D =- 225 Volt 40 20

SOL 1.103

SOL 1.104

Option (D) is correct. Energy stored in Capacitor is E = 1 CV2 2 - 12 -6 # 10 = 8.85 # 10- 12 F C = e0 A = 8.85 # 10 # 100 3 d 0.1 # 10 E = 1 # 8.85 # 10- 12 # (100) 2 = 44.3 nJ 2 Option (B) is correct. The figure is as shown below

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Page 66

The Capacitor shown in Figure is made up of two capacitor C1 and C2 connected in series. C1 = e0 er1 A , C2 = e0 er2 A t1 t2 Since C1 and C2 are in series charge on both capacitor is same. Q1 = Q 2 C1 (100 - V) = C2 V (Let V is the voltage of foil) e0 er1 A (100 - V) = e0 er2 A V t1 t2 3 (100 - V) = 4 V 0.5 1

A I D

300 - 3V = 2V 300 = 5V & V = 60 Volt SOL 1.105

Option (D) is correct. Voltage across capacitor is given by

O N

vc (t) = 1 C SOL 1.106

3

# i (t) dt

-3

Option (C) is correct. No. of links is given by

= 1 C

3

ia

d .no

w

ww

# 5d (.tc)odt.in= C5 # u (t)

-3

L = N-B+1 SOL 1.107

Option (A) is correct. Divergence theorem states that the total outward flux of a vector field F through a closed surface is same as volume integral of the divergence of F

# F $ ds s

SOL 1.108

=

# (4: F) dv V

Option (C) is correct. The figure as shown below

Inductance of parallel wire combination is given as ml L = 0 ln b d l p r Where

l " Length of wires d " Distance between wires r " Radius

L \ ln d So when d is double, inductance increase but does not double.

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Page 67

SOL 1.109

Option (B) is correct. Since distance from ground to lower surface is less than from ground to upper surface so electric stress is maximum at lower surface.

SOL 1.110

Option (B) is correct. Writing node equation for the circuit

I 1 = E1 - E A 2 I2 = E2 - EA 2

and At node A E A - E1 + E A + E A - E 2 = 0 2 2 2 3EA = E1 + E2

A I D

From eqn(1)

(E + E2) I 1 = 1 E1 - 1 1 2 2 3 I 1 = 1 E1 - 1 E 2 3 6 .in o c ( E E ) + . Similarly I2 = 1 E2 - 1 1 di2a 2 2 n3o w. 1 1 w I2 =- Ew + E 6 1 3 2 From (2) and (3) admittance parameters are

O N

...(1)

...(2)

...(3)

[Y11 Y12 Y21 Y22] = [1/3 - 1/6 - 1/6 1/3] SOL 1.111

Option (A) is correct. Admittance of the given circuit Y (w) = jwC + 1 ZL So,

ZL = 30+40c = 23.1 + j19.2 W 23.1 - j19.2 1 Y (w) = j2p # 50 # C + 23.1 + j19.2 # 23.1 - j19.2 23.1 - j19.2 = j (100p) C + 902.25 = 23.1 + j :(100p) C - 19.2 D 902.25 902.25

For unity power factor Im [Y (w)] = 0 100 # 3.14 # C = 19.2 902.25 C - 68.1 mF SOL 1.112

Option (B) is correct. In series RLC circuit lower half power frequency is given by following relations

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Page 68

w1 L - 1 =- R w1 C 1 =- 50 2p # f1 (1 # 10- 6) f1 = 3.055 kHz

(2p # f1 # 100 # 10- 6) -

SOL 1.113

Option (C) is correct. Since initial charge across capacitor is zero, voltage across capacitor at any time t is given as t

vc (t) = 10 (1 - e- t ) Time constant t = Req C = (10 kW || 1 kW) # C = b 10 l kW # 11 nF = 10 # 10- 6 sec = 10 m sec 11 t

So, vc (t) = 10 (1 - e- 10 m sec ) Pulse duration is 10 msec, so voltage across capacitor will be maximum at t = 10 m sec 10 m sec

vc (t = 10 m sec) = 10 (1 - e- 10 m sec ) = 10 (1 - e- 1) = 6.32 Volt SOL 1.114

A I D

Option (C) is correct. Since voltage and current are in phase so equivalent inductance is Leq L1 + L2 ! 2M 8 + 8 ! 2M 16 - 2M M Coupling Coefficient

= 12 H = 12 M " Mutual Inductance n o.i = 12 c . ia = 12 (Dot isnoatd position Q) . = 2 H ww

O N K =

w

2 = 0.25 8#8

SOL 1.115

Option ( ) is correct.

SOL 1.116

Option (C) is correct. In steady state there is no voltage drop across inductor (i.e. it is short circuit) and no current flows through capacitors (i.e. it is open circuit) The equivalent circuit is

So, SOL 1.117

vc (3) = 10 # 1=5 Volt 1+1

Option (C) is correct. When the switch was closed before t = 0 , the circuit is

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Page 69

Current in the inductor iL (0-) = 0 A When the switch was opened at t = 0 , equivalent circuit is

In steady state, inductor behaves as short circuit and 10 A current flows through it

A I D

iL (3) = 10 A Inductor current at any time t is given by

O N

- t iL (t) = iL (3) + 6iL (0) - iL (o 3.i)n @e L

SOL 1.118

R

.c - i5at d 10 = 10 + (0 - 10)oe = 10 (1 - e- 2t) A n w. Option (B) is correct. w w Energy stored in inductor is E = 1 Li2 = 1 # 5 # (10) 2 = 250 J 2 2

SOL 1.119

Option (C) is correct. To obtain Thevenin’s equivalent, open the terminals X and Y as shown below,

By writing node equation at X Vth - V1 + Vth - V2 = 0 Z1 Z2 V1 = 30+45c = 30 (1 + j) 2 V2 = 30+ - 45c = 30 (1 - j) 2 So, Vth - 30 (1 + j) Vth - 30 (1 - j) 2 2 + =0 1-j 1+j

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Page 70

2Vth - 30 (1 + j) 2 - 30 (1 - j) 2 = 0 2 2 Vth = 0 Volt Thevenin’s impedance

Zth = Z1 || Z2 = (1 - j) || (1 + j) = SOL 1.120

Option (A) is correct. Drawing Thevenin equivalent circuit across load :

So, current SOL 1.121

(1 - j) (1 + j) = 1W (1 - j) + (1 + j)

iL = 0 A

A I D

O N

Option (A) is correct. n o.i two wheatstone bridge connected in In the circuit we can observe that there.care ia therefore both the bridge are balanced dsame, parallel. Since all resistor values are o n and no current flows through diagonal arm. So the equivalent circuit is w.

ww

We can draw the circuit as

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Page 71

From T - Y conversion

A I D

Now the circuit is

O N

no w.

.in

co ia.

d

ww

SOL 1.122

VAB = 1 # 14 = 1.4 Volt 10 Option (C) is correct. In a series RLC circuit, at resonance, current is given as i = Vs +0c , VS " source voltage R So, voltage across capacitor at resonance

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Page 72

1 # Vs +0c jw C R Vc = Vs + - 90c wCR Vc =

Voltage across capacitor can be greater than input voltage depending upon values of w, C and R but it is 90c out of phase with the input SOL 1.123

Option (D) is correct. Let resistance of 40 W and 60 W lamps are R1 and R2 respectively a P \ 12 R P1 = R2 P2 R1 R2 = 40 R1 60 R2 < R1 40 W bulb has high resistance than 60 W bulb, when connected in series power is P1 = I2 R1

A I D

P2 = I2 R2 a R1 > R2 , So P1 > P2 Therefore, 40 W bulb glows brighter SOL 1.124

O N

Option (B) is correct. n o.i in following figure Series RL circuit with unit step input is.cshown

w

ww

ia

d .no

1, t > 0 u (t) = ) 0, otherwise Initially inductor current is zero i (0+) = 0 When unit step is applied, inductor current does not change simultaneously and the source voltage would appear across inductor only so voltage across resistor at t = 0+ vR (0+) = 0 SOL 1.125

Option (D) is correct. For two coupled inductors M = K L1 L2 Where K " coupling coefficient 0 2 Which of the following is true ? (A) x (t) has finite energy because only finitely many coefficients are non-zero (B) x (t) has zero average value because it is periodic (C) The imaginary part of x (t) is constant (D) The real part of x (t) is even MCQ 2.27

A I D

The z-transform of a signal x [n] is given by 4z - 3 + 3z - 1 + 2 - 6z2 + 2z3 It is applied to a system, with a transfer function H (z) = 3z - 1 - 2 Let the output be y [n]. Which of the following inis true ? . o c (A) y [n] is non causal with finite support ia. d (B) y [n] is causal with infinite support .no w (C) y [n] = 0; n > 3 ww (D) Re [Y (z)] z = e =- Re [Y (z)] z = e ji

O N

- ji

Im [Y (z)] z = e = Im [Y (z)] z = e ; - p # q < p ji

YEAR 2008

- ji

ONE MARK

MCQ 2.28

The impulse response of a causal linear time-invariant system is given as h (t). Now consider the following two statements : Statement (I): Principle of superposition holds Statement (II): h (t) = 0 for t < 0 Which one of the following statements is correct ? (A) Statements (I) is correct and statement (II) is wrong (B) Statements (II) is correct and statement (I) is wrong (C) Both Statement (I) and Statement (II) are wrong (D) Both Statement (I) and Statement (II) are correct

MCQ 2.29

A signal e - at sin (wt) is the input to a real Linear Time Invariant system. Given K and f are constants, the output of the system will be of the form Ke - bt sin (vt + f) where (A) b need not be equal to a but v equal to w (B) v need not be equal to w but b equal to a (C) b equal to a and v equal to w (D) b need not be equal to a and v need not be equal to w

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Page 82

YEAR 2008 MCQ 2.30

TWO MARKS

A system with x (t) and output y (t) is defined by the input-output relation : y (t) =

- 2t

#- 3x (t) dt

The system will be (A) Casual, time-invariant and unstable (B) Casual, time-invariant and stable (C) non-casual, time-invariant and unstable (D) non-casual, time-variant and unstable MCQ 2.31

MCQ 2.32

MCQ 2.33

MCQ 2.34

A signal x (t) = sinc (at) where a is a real constant ^sinc (x) = px h is the input to a Linear Time Invariant system whose impulse response h (t) = sinc (bt) , where b is a real constant. If min (a, b) denotes the minimum of a and b and similarly, max (a, b) denotes the maximum of a and b, and K is a constant, which one of the following statements is true about the output of the system ? (A) It will be of the form Ksinc (gt) where g = min (a, b) (B) It will be of the form Ksinc (gt) where g = max (a, b) (C) It will be of the form Ksinc (at) (D) It can not be a sinc type of signal sin (px)

A I D

Let x (t) be a periodic signal with time period T , Let y (t) = x (t - t0) + x (t + t0) for some t0 . The Fourier Series coefficients of y (t) are denoted by bk . If bk = 0 for all odd k , then t0 can be equal to (A) T/8 .inT/4 o(B) c . (C) T/2 dia (D) 2T

O N

.no w ofwa real system. When a signal x [n] = (1 + j) n H (z) is a transfer function w

is the input to such a system, the output is zero. Further, the Region of convergence (ROC) of ^1 - 12 z - 1h H(z) is the entire Z-plane (except z = 0 ). It can then be inferred that H (z) can have a minimum of (A) one pole and one zero (B) one pole and two zeros (C) two poles and one zero D) two poles and two zeros z Given X (z) = with z > a , the residue of X (z) zn - 1 at z = a for n $ 0 2 (z - a) will be (B) an (A) an - 1 (D) nan - 1

(C) nan MCQ 2.35

Let x (t) = rect^t - 12 h (where rect (x) = 1 for - 12 # x #

1 2

and zero otherwise.

sin (px)

MCQ 2.36

If sinc (x) = px , then the FTof x (t) + x (- t) will be given by (B) 2 sinc` w j (A) sinc` w j 2p 2p (C) 2 sinc` w j cos ` w j (D) sinc` w j sin ` w j 2p 2 2p 2 Given a sequence x [n], to generate the sequence y [n] = x [3 - 4n], which one of the following procedures would be correct ?

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Page 83

(A) First delay x (n) by 3 samples to generate z1 [n], then pick every 4th sample of z1 [n] to generate z2 [n], and than finally time reverse z2 [n] to obtain y [n]. (B) First advance x [n] by 3 samples to generate z1 [n], then pick every 4th sample of z1 [n] to generate z2 [n], and then finally time reverse z2 [n] to obtain y [n] (C) First pick every fourth sample of x [n] to generate v1 [n], time-reverse v1 [n] to obtain v2 [n], and finally advance v2 [n] by 3 samples to obtain y [n] (D) First pick every fourth sample of x [n] to generate v1 [n], time-reverse v1 [n] to obtain v2 [n], and finally delay v2 [n] by 3 samples to obtain y [n] YEAR 2007 MCQ 2.37

ONE MARK

Let a signal a1 sin (w1 t + f) be applied to a stable linear time variant system. Let the corresponding steady state output be represented as a2 F (w2 t + f2). Then which of the following statement is true? (A) F is not necessarily a “Sine” or “Cosine” function but must be periodic with w1 = w2 . (B) F must be a “Sine” or “Cosine” function with a1 = a2 (C) F must be a “Sine” function with w1 = w2 and f1 = f2 (D) F must be a “Sine” or “Cosine” function with w1 = w2

MCQ 2.38

A I D

The frequency spectrum of a signal is shown in the figure. If this is ideally sampled at intervals of 1 ms, then the frequency spectrum of the sampled signal will n be o.i

O N

.c

ia od

n

. ww

w

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A I D

O N

YEAR 2007 MCQ 2.39

.in

co ia.

od

.n A signal x (t) is given by w 1, - T/4 < tw#w3T/4 x (t) = *- 1, 3T/4 < t # 7T/4 - x (t + T)

Page 84

TWO MARKS

Which among the following gives the fundamental fourier term of x (t) ? (B) p cos ` pt + p j (A) 4 cos ` pt - p j p T 4 4 2T 4 (C) 4 sin ` pt - p j (D) p sin ` pt + p j p T 4 4 2T 4

Statement for Linked Answer Question 41 and 41 : MCQ 2.40

A signal is processed by a causal filter with transfer function G (s) For a distortion free output signal wave form, G (s) must (A) provides zero phase shift for all frequency (B) provides constant phase shift for all frequency (C) provides linear phase shift that is proportional to frequency (D) provides a phase shift that is inversely proportional to frequency

MCQ 2.41

G (z) = az - 1 + bz - 3 is a low pass digital filter with a phase characteristics same as that of the above question if (A) a = b (B) a =- b (C) a = b(1/3)

(D) a = b(- 1/3)

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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia MCQ 2.42

Page 85

Consider the discrete-time system shown in the figure where the impulse response of G (z) is g (0) = 0, g (1) = g (2) = 1, g (3) = g (4) = g = 0

This system is stable for range of values of K (A) [- 1, 12 ] (B) [- 1, 1] 1 (C) [- 2 , 1] (D) [- 12 , 2] MCQ 2.43

If u (t), r (t) denote the unit step and unit ramp functions respectively and u (t) * r (t) their convolution, then the function u (t + 1) * r (t - 2) is given by (A) 12 (t - 1) u (t - 1) (B) 12 (t - 1) u (t - 2) (C)

MCQ 2.44

1 2

(t - 1) 2 u (t - 1)

(D) None of the above

X (z) = 1 - 3z - 1, Y (z) = 1 + 2z - 2 are Z transforms of two signals x [n], y [n] respectively. A linear time invariant system has the impulse response h [n] defined by these two signals as h [n] = x [n - 1] * y [n] where * denotes discrete time convolution. Then the output of the system for the input d [n - 1] (A) has Z-transform z - 1 X (z) Y (z) (B) equals d [n - 2] - 3d [n - 3] + 2d [n - 4] - 6d [n - 5]

A I D

O N

(C) has Z-transform 1 - 3z - 1 + 2z - 2 - 6z - 3 (D) does not satisfy any of the above threeo.in YEAR 2006

.c

w

. ww

n

ia od

ONE MARK

MCQ 2.45

The following is true (A) A finite signal is always bounded (B) A bounded signal always possesses finite energy (C) A bounded signal is always zero outside the interval [- t0, t0] for some t0 (D) A bounded signal is always finite

MCQ 2.46

x (t) is a real valued function of a real variable with period T . Its trigonometric Fourier Series expansion contains no terms of frequency w = 2p (2k) /T; k = 1, 2g Also, no sine terms are present. Then x (t) satisfies the equation (A) x (t) =- x (t - T) (B) x (t) = x (T - t) =- x (- t) (C) x (t) = x (T - t) =- x (t - T/2) (D) x (t) = x (t - T) = x (t - T/2)

MCQ 2.47

A discrete real all pass system has a pole at z = 2+30% : it, therefore (A) also has a pole at 12 +30% (B) has a constant phase response over the z -plane: arg H (z) = constant constant (C) is stable only if it is anti-causal (D) has a constant phase response over the unit circle: arg H (eiW) = constant

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YEAR 2006 MCQ 2.48

MCQ 2.49

Page 86

TWO MARKS

x [n] = 0; n < - 1, n > 0, x [- 1] =- 1, x [0] = 2 is the input and y [n] = 0; n < - 1, n > 2, y [- 1] =- 1 = y [1], y [0] = 3, y [2] =- 2 is the output of a discrete-time LTI system. The system impulse response h [n] will be (A) h [n] = 0; n < 0, n > 2, h [0] = 1, h [1] = h [2] =- 1 (B) h [n] = 0; n < - 1, n > 1, h [- 1] = 1, h [0] = h [1] = 2 (C) h [n] = 0; n < 0, n > 3, h [0] =- 1, h [1] = 2, h [2] = 1 (D) h [n] = 0; n < - 2, n > 1, h [- 2] = h [1] = h [- 1] =- h [0] = 3 n The discrete-time signal x [n] X (z) = n3= 0 3 z2n , where 2+n transform-pair relationship, is orthogonal to the signal n (A) y1 [n] ) Y1 (z) = n3= 0 ` 2 j z - n 3

/

denotes a

/ (B) y2 [n] ) Y2 (z) = /n3= 0 (5n - n) z - (2n + 1) (C) y3 [n] ) Y3 (z) = /n3=- 3 2 - n z - n (D) y4 [n] ) Y4 (z) = 2z - 4 + 3z - 2 + 1 MCQ 2.50

MCQ 2.51

A I D

A continuous-time system is described by y (t) = e - x (t) , where y (t) is the output and x (t) is the input. y (t) is bounded (A) only when x (t) is bounded (B) only when x (t) is non-negative (C) only for t # 0 if x (t) is bounded for t $.i0n o (D) even when x (t) is not bounded ia.c

O N

no w.

d

The running integration, given ww by y (t) =

t

#- 3 x (t') dt'

(A) has no finite singularities in its double sided Laplace Transform Y (s) (B) produces a bounded output for every causal bounded input (C) produces a bounded output for every anticausal bounded input (D) has no finite zeroes in its double sided Laplace Transform Y (s) YEAR 2005 MCQ 2.52

For the triangular wave from shown in the figure, the RMS value of the voltage is equal to

(A) (C) 1 3 MCQ 2.53

TWO MARKS

1 6

(B) (D)

1 3 2 3

2 The Laplace transform of a function f (t) is F (s) = 5s 2+ 23s + 6 as t " 3, f (t) s (s + 2s + 2) approaches

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MCQ 2.54

MCQ 2.55

(A) 3 (B) 5 (D) 3 (C) 17 2 The Fourier series for the function f (x) = sin2 x is (A) sin x + sin 2x (B) 1 - cos 2x (C) sin 2x + cos 2x (D) 0.5 - 0.5 cos 2x If u (t) is the unit step and d (t) is the unit impulse function, the inverse z -transform of F (z) = z +1 1 for k > 0 is (A) (- 1) k d (k)

(B) d (k) - (- 1) k

(C) (- 1) k u (k)

(D) u (k) - (- 1) k

YEAR 2004 MCQ 2.56

TWO MARKS

The rms value of the periodic waveform given in figure is

(A) 2 6 A (C) MCQ 2.57

4/3 A

A I D

O N ww

no w.

.in

d

co ia.

(B) 6 2 A (D) 1.5 A

The rms value of the resultant current in a wire which carries a dc current of 10 A and a sinusoidal alternating current of peak value 20 is (A) 14.1 A (B) 17.3 A (C) 22.4 A (D) 30.0 A YEAR 2002

MCQ 2.58

Page 87

ONE MARK

Fourier Series for the waveform, f (t) shown in Figure is

(A) 82 8sin (pt) + 1 sin (3pt) + 1 sin (5pt) + .....B 9 25 p (B) 82 8sin (pt) - 1 cos (3pt) + 1 sin (5pt) + .......B 9 25 p

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MCQ 2.59

(C) 82 8cos (pt) + 1 cos (3pt) + 1 cos (5pt) + .....B 9 25 p (D) 82 8cos (pt) - 1 sin (3pt) + 1 sin (5pt) + .......B 9 25 p Let s (t) be the step response of a linear system with zero initial conditions; then the response of this system to an an input u (t) is t t (B) d ; s (t - t) u (t) dt E (A) s (t - t) u (t) dt dt 0 0

#

(C) MCQ 2.60

#

#0 s (t - t); #0 u (t1) dt1Edt t

t

(D)

s"3

(C) Lim sY (s)

(D) Lim sY (s)

s"0

s"3

What is the rms value of the voltage waveform shown in Figure ?

(A) (200/p) V (C) 200 V YEAR 2001 MCQ 2.62

1

#0 [s (t - t)] 2 u (t) dt

Let Y (s) be the Laplace transformation of the function y (t), then the final value of the function is (A) LimY (s) (B) LimY (s) s"0

MCQ 2.61

Page 88

A I D

O N

no w.

n o.i c . ia (B) (100/p) V

d

ww

(D) 100 V

ONE MARK

Given the relationship between the input u (t) and the output y (t) to be y (t) =

t

#0 (2 + t - t) e- 3(t - t)u (t) dt ,

The transfer function Y (s) /U (s) is - 2s (A) 2e s+3 (C) 2s + 5 s+3

s+2 (s + 3) 2 (D) 2s + 72 (s + 3) (B)

Common data Questions Q.63-64* Consider the voltage waveform v as shown in figure

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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia MCQ 2.63

MCQ 2.64

The DC component of v is (A) 0.4 (C) 0.8

Page 89

(B) 0.2 (D) 0.1

The amplitude of fundamental component of v is (A) 1.20 V (B) 2.40 V (C) 2 V (D) 1 V ***********

A I D

O N

no w.

.in

co ia.

d

ww

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SOLUTION

SOL 2.1

Page 90

Option (A) is correct. Given, the maximum frequency of the band-limited signal fm = 5 kHz According to the Nyquist sampling theorem, the sampling frequency must be greater than the Nyquist frequency which is given as fN = 2fm = 2 # 5 = 10 kHz So, the sampling frequency fs must satisfy fs $ fN fs $ 10 kHz only the option (A) does not satisfy the condition therefore, 5 kHz is not a valid sampling frequency.

SOL 2.2

A I D

Option (A) is correct. Given, the signal

v ^ t h = 30 sin 100t + 10 cos 300t + 6 sin ^500t + p4 h So we have

O N

.in

w1 = 100 rad/s .co a i d w2 = 300 rad/s no . w w3 = 500 rad/s ww Therefore, the respective time periods are T1 = 2p = 2p sec w1 100 T2 = 2p = 2p sec w2 300 T3 = 2p sec 500

SOL 2.3

So, the fundamental time period of the signal is LCM ^2p, 2p, 2ph L.C.M. ^T1, T2 T3h = HCF ^100, 300, 500h or, T0 = 2p 100 Thus, the fundamental frequency in rad/sec is w0 = 2p = 100 rad/s 10 Option (C) is correct. If the two systems with impulse response h1 ^ t h and h2 ^ t h are connected in cascaded configuration as shown in figure, then the overall response of the system is the convolution of the individual impulse responses.

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Page 91

SOL 2.4

Option (C) is correct. For a system to be casual, the R.O.C of system transfer function H ^s h which is rational should be in the right half plane and to the right of the right most pole. For the stability of LTI system. All poles of the system should lie in the left half of S -plane and no repeated pole should be on imaginary axis. Hence, options (A), (B), (D) satisfies both stability and causality an LTI system. But, Option (C) is not true for the stable system as, S = 1 have one pole in right hand plane also.

SOL 2.5

Option (C) is correct. Given, the input x ^ t h = u ^t - 1h It’s Laplace transform is -s X ^s h = e s

A I D

The impulse response of system is given

h^t h = t u^t h n Its Laplace transform is o.i c . dia H ^s h = 12 o s .n w Hence, the overall response at wwthe output is

O N

SOL 2.6

Y ^s h = X ^s h H ^s h -s =e3 s its inverse Laplace transform is ^t - 1h2 y^t h = u ^t - 1h 2 Option (B) is correct. Given, the impulse response of continuous time system

h ^ t h = d ^t - 1h + d ^t - 3h From the convolution property, we know x ^ t h * d ^t - t 0h = x ^t - t 0h So, for the input x ^ t h = u ^ t h (Unit step fun n ) The output of the system is obtained as y^t h = u^t h * h^t h

= u ^ t h * 6d ^t - 1h + d ^t - 3h@ = u ^t - 1h + u ^t - 3h at t = 2

y ^2 h = u ^2 - 1h + u ^2 - 3h

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Page 92

=1 SOL 2.7

Option (C) is correct. n n x [n] = b 1 l - b 1 l u [n] 3 2 n -n n = b 1 l u [n] + b 1 l u [- n - 1] - b 1 l u (n) 3 3 2 Taking z -transform X 6z @ = =

3

/ n =- 3 3

/

n =- 3 3

1 n -n b 3 l z u [ n] +

1 -n b 2 l z u [ n] = n

3

3

/ n =- 3 3

/ b 13 l z

n=0

m

-n

+

-1

/ n =- 3

3

-

m=1

1 -n -n b3l z -

/ b 21z l

n

3

/ b 12 l z n

-n

n=0

Taking m =- n

n=0

1 44 2 44 3 II

14 42 4 43 I

n

n=0

/ b 31z l + / b 13 z l n

1 -n -n b 3 l z u [ - n - 1]

14 42 4 43 III

1 < 1 or z > 1 3 3z Series II converges if 1 z < 1 or z < 3 3 Series III converges if 1 < 1 or z > 1 2z 2 Region of convergence of X (z) will be intersection of above three So, ROC : 1 < z < 3 2 Option (D) is correct. n Using s -domain differentiation propertycof o.iLaplace transform. Series I converges if

SOL 2.8

If

O N f (t)

L

a.

Fo(dsi)

n

. ww

dF (s) ds 2s + 1 So, L [tf (t)] = - d ; 2 1 = ds s + s + 1E (s2 + s + 1) 2 Option (A) is correct. Convolution sum is defined as tf (tw )

SOL 2.9

A I D

L

-

y [n] = h [n] * g [n] = For causal sequence,

y [n] =

3

/ h [n] g [n - k] k =- 3

3

/ h [n] g [n - k] k=0

y [n] = h [n] g [n] + h [n] g [n - 1] + h [n] g [n - 2] + ..... For n = 0 ,

y [0] = h [0] g [0] + h [1] g [- 1] + ........... = h [ 0] g [ 0] g [- 1] = g [- 2] = ....0 ...(i) = h [ 0] g [ 0]

For n = 1,

y [1] = h [1] g [1] + h [1] g [0] + h [1] g [- 1] + .... = h [ 1] g [ 1] + h [ 1 ] g [ 0 ] 1 = 1 g [1] + 1 g [0] 2 2 2

1 h [1] = b 1 l = 1 2 2

1 = g [1] + g [0] g [1] = 1 - g [0]

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SOL 2.10

Page 93

y [0] 1 = =1 h [0] 1

From equation (i),

g [0] =

So,

g [1] = 1 - 1 = 0

Option (C) is correct. (2 cos w) (sin 2w) H (jw) = = sin 3w + sin w w w w We know that inverse Fourier transform of sin c function is a rectangular function.

A I D

So, inverse Fourier transform of H (jw) h (t) = h1 (t) + h2 (t) SOL 2.11

O N

Option (D) is correct. y (t) =

t

n o.i 1 1 c . h (0) = h1 (0) +dhi2a(0) = + = 1 2 2 no . w ww

# x (t) cos (3t) dt -3

Time invariance : Let, x (t) = d (t) y (t) =

t

# d (t) cos (3t) dt -3

= u (t) cos (0) = u (t)

For a delayed input (t - t 0) output is y (t, t 0) = Delayed output

t

# d (t - t ) cos (3t) dt -3

0

= u (t) cos (3t 0)

y (t - t 0) = u (t - t 0) y (t, t 0) ! y (t - t 0) System is not time invariant. Stability : Consider a bounded input x (t) = cos 3t y (t) =

#

t

-3

cos2 3t =

1 - cos 6t = 1 2 2 -3

#

t

# 1dt - 12 # cos 6t dt t

t

-3

-3

As t " 3, y (t) " 3 (unbounded) System is not stable. SOL 2.12

Option (D) is correct.

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f (t) = a 0 +

Page 94

3

/ (an cos wt + bn sin nwt)

n=1

The given function f (t) is an even function, therefore bn = 0 f (t) is a non zero average value function, so it will have a non-zero value of a 0 T/2 a 0 = 1 # f (t) dt (average value of f (t)) ^T/2h 0 • an is zero for all even values of n and non zero for odd n T an = 2 # f (t) cos (nwt) d (wt) T 0 • •

So, Fourier expansion of f (t) will have a 0 and an , n = 1, 3, 5f3 SOL 2.13

Option (A) is correct. x (t) = e-t Laplace transformation X (s) = 1 s+1

A I D

O N

y (t) = e-2t n o.i c 1 . Y (s) = s+2 dia o .n Convolution in time domain iswequivalent to multiplication in frequency domain.

ww

z (t) = x (t) ) y (t)

Z (s) = X (s) Y (s) = b 1 lb 1 l s+1 s+2 By partial fraction and taking inverse Laplace transformation, we get Z (s) = 1 - 1 s+1 s+2 z (t) = e-t - e-2t SOL 2.14

Option (D) is correct. f (t)

L

F1 (s)

L

e-st F1 (s) = F2 (s) F (s) F 1)(s) e-st F1 (s) F 1)(s) G (s) = 2 = F1 (s) 2 F1 (s) 2 e-sE F1 (s) 2 ) 2 = "a F1 (s) F 1 (s) = F1 (s) F1 (s) 2 = e-st Taking inverse Laplace transform f (t - t)

g (t) = L - 1 [e-st] = d (t - t) SOL 2.15

Option (C) is correct. h (t) = e-t + e-2t

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Page 95

Laplace transform of h (t) i.e. the transfer function H (s) = 1 + 1 s+1 s+2 For unit step input r (t) = m (t) or R (s) = 1 s Output, Y (s) = R (s) H (s) = 1 : 1 + 1 D s s+1 s+2 By partial fraction Y (s) = 3 - 1 - b 1 l 1 2s s + 1 s+2 2 Taking inverse Laplace e-2t u (t) y (t) = 3 u (t) - e-t u (t) 2 2 = u (t) 61.5 - e-t - 0.5e-2t@ SOL 2.16

Option (C) is correct. System is given as

A I D

2 (s + 1) R (s) = 1 s

H (s) = Step input

O N

Output Y (s) = H (s) R (s) = 2 b 1 l = 2 - 2 (s + 1) sin s (s + 1) o. Taking inverse Laplace transform a.c y (t) = (2 - 2e- t) u.n (to) w Final value of y (t), ww

di

yss (t) = lim y (t) = 2 t"3

Let time taken for step response to reach 98% of its final value is ts . So, 2 - 2e- ts = 2 # 0.98 0.02 = e- ts ts = ln 50 = 3.91 sec. SOL 2.17

Option (D) is correct. Period of x (t), T = 2p = 2 p = 2.5 sec 0.8 p w

SOL 2.18

Option (B) is correct. Input output relationship y (t) =

5t

#- 3x (t) dt,

t>0

Causality : y (t) depends on x (5t), t > 0 system is non-causal. For example t = 2 y (2) depends on x (10) (future value of input) Linearity :

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Output is integration of input which is a linear function, so system is linear. SOL 2.19

Option (A) is correct. Fourier series of given function x (t) = A0 +

3

/ an cos nw0 t + bn sin nw0 t

n=1

So,

a x (t) =- x (t) odd function A0 = 0 an = 0 T bn = 2 x (t) sin nw0 t dt T 0

#

T /2 T = 2= (1) sin nw0 t dt + (- 1) sin nw0 t dt G T 0 T /2 T /2 T = 2 =c cos nw0 t m - c cos nw0 t m G - nw0 0 - nw0 T/2 T = 2 6(1 - cos np) + (cos 2np - cos np)@ nw0 T = 2 61 - (- 1) n @ np 4 , n odd bn = * np 0 , n even So only odd harmonic will be present in x (t) For second harmonic component (n = 2) amplitude is zero. n

#

SOL 2.20

3

#- 3 SOL 2.21

A I D

O N

Option (D) is correct. By parsval’s theorem 1 3 X (w) 2 dw = 2p - 3

#

#

no w.

.i

co ia.

d

#- 3wxw2 (t) dt 3

X (w) 2 dw = 2p # 2 = 4p

Option (C) is correct. Given sequences x [n] = {1, - 1}, 0 # n # 1 y [n] = {1, 0, 0, 0, - 1}, 0 # n # 4 If impulse response is h [n] then y [ n] = h [ n] * x [ n] Length of convolution (y [n]) is 0 to 4, x [n] is of length 0 to 1 so length of h [n] will be 0 to 3. Let h [n] = {a, b, c, d} Convolution

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y [n] = {a, - a + b, - b + c, - c + d, - d} By comparing

So, SOL 2.22

a -a + b -b + c -c + d h [ n]

=1 =0 &b=a=1 =0 &c=b=1 =0 &d=c=1 = {1, 1, 1, 1} -

Option (D) is correct. We can observe that if we scale f (t) by a factor of 1 and then shift, we will get 2 g (t). First scale f (t) by a factor of 1 2 g1 (t) = f (t/2)

Shift g1 (t) by 3,

A I D

g (t) = g1 (t - 3) = f` t - 3 j n2

O N

no w.

.i

co ia.

d

ww

g (t) = f` t - 3 j 2 2 SOL 2.23

Option (C) is correct. g (t) can be expressed as g (t) = u (t - 3) - u (t - 5) By shifting property we can write Laplace transform of g (t) - 3s G (s) = 1 e - 3s - 1 e - 5s = e (1 - e - 2s) s s s

SOL 2.24

Option (D) is correct. L Let x (t) X (s) L y (t) Y (s) L h (t) H (s) So output of the system is given as Y (s) = X (s) H (s) Now for input So now output is

x (t - t)

L

e - st X (s)

h (t - t)

L

e- st H (s)

(shifting property)

Y' (s) = e - st X (s) $ e - ts H (s) = e - 2st X (s) H (s) = e - 2st Y (s)

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y' (t) = y (t - 2t) SOL 2.25

Option (B) is correct. Let three LTI systems having response H1 (z), H2 (z) and H 3 (z) are Cascaded as showing below

Assume H1 (z) = z2 + z1 + 1 (non-causal) H2 (z) = z3 + z2 + 1 (non-causal) Overall response of the system H (z) = H1 (z) H2 (z) H3 (z) H (z) = (z2 + z1 + 1) (z3 + z2 + 1) H3 (z) To make H (z) causal we have to take H3 (z) also causal. H3 (z) = z - 6 + z - 4 + 1

Let

= (z2 + z1 + 1) (z3 + z2 + 1) (z - 6 + z - 4 + 1) H (z) " causal

A I D

Similarly to make H (z) unstable atleast one of the system should be unstable. SOL 2.26

Option (C) is correct. Given signal 3

O N

x (t) =

/ak e j2pkt/T

in

Let w0 is the fundamental frequency of .signal co. x (t) k =- 3

x (t) =

3

/ak e

jkw0 t

k =- 3

ia

d .no

ww

w

a 2p = w0 T

x (t) = a - 2 e - j2w t + a - 1 e - jw t + a0 + a1 e jw t + a2 e j2w t 0

0

0

0

= (2 - j) e - 2jw t + (0.5 + 0.2j) e - jw t + 2j + 0

0

+ (0.5 - 0.2) e jw t + (2 + j) e j2w t 0

= 2 6e - j2w t + e j2w t @ + j 6e j2w t - e - j2w t @ + 0

0

0

0

0

0.5 6e jw t + e - jw t @ - 0.2j 6e+ jw t - e - jw t @ + 2j = 2 (2 cos 2w0 t) + j (2j sin 2w0 t) + 0.5 (2 cos w0 t) 0.2j (2j sin w0 t) + 2j = 6 4 cos 2w0 t - 2 sin 2w0 t + cos w0 t + 0.4 sin w0 t @ + 2j Im [x (t)] = 2 (constant) 0

SOL 2.27

0

0

0

Option (A) is correct. Z-transform of x [n] is X (z) = 4z - 3 + 3z - 1 + 2 - 6z2 + 2z3 Transfer function of the system H (z) = 3z - 1 - 2 Output Y (z) = H (z) X (z) Y (z) = (3z - 1 - 2) (4z - 3 + 3z - 1 + 2 - 6z2 + 2z3) = 12z -4 + 9z -2 + 6z -1 - 18z + 6z2 - 8z -3 - 6z -1 - 4 + 12z2 - 4z3 = 12z - 4 - 8z - 3 + 9z - 2 - 4 - 18z + 18z2 - 4z3

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Or sequence y [n] is y [n] = 12d [n - 4] - 8d [n - 3] + 9d [n - 2] - 4d [n] 18d [n + 1] + 18d [n + 2] - 4d [n + 3] y [n] = Y 0, n < 0 So y [n] is non-causal with finite support. SOL 2.28

Option (D) is correct. Since the given system is LTI, So principal of Superposition holds due to linearity. For causal system h (t) = 0 , t < 0 Both statement are correct.

SOL 2.29

Option (C) is correct. For an LTI system output is a constant multiplicative of input with same frequency. Here

input g (t) = e - at sin (wt)

output y (t) = Ke - bt sin (vt + f) Output will be in form of Ke - at sin (wt + f) So \= b, v = w SOL 2.30

Option (D) is correct. Input-output relation y (t) =

A I D

- 2t

#- 3x (t) dt

Causality : Since y (t) depends on x (- 2t), So it is non-causal. n Time-variance : o.i

O N

a.c i d y (t) = # x (t - t0) dt = Y y (to- t0) -3 .n w So this is time-variant. ww - 2t

Stability : Output y (t) is unbounded for an bounded input. For example Let

x (t) = e - t (bounded) y (t) =

SOL 2.31

#- 3e- t dt = 8 -e 1 B- 3 $ Unbounded - 2t

- t - 2t

Option (A) is correct. Output y (t) of the given system is y (t) = x (t) ) h (t) Or Y (jw) = X (jw) H (jw) Given that, x (t) = sinc (at) and h (t) = sinc (bt) Fourier transform of x (t) and h (t) are X (jw) = F [x (t)] = p rect` w j, - a < w < a a 2a H (jw) = F [h (t)] = p rect` w j, - b < w < b b 2b 2 Y (jw) = p rect` w j rect` w j ab 2a 2b

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So, Where And SOL 2.32

Y (jw) = K rect ` w j 2g g = min (a, b) y (t) = K sinc (gt)

Option (B) is correct. Let ak is the Fourier series coefficient of signal x (t) Given y (t) = x (t - t0) + x (t + t0) Fourier series coefficient of y (t) bk = e - jkwt ak + e jkwt ak bk = 2ak cos kwt0 bk = 0 (for all odd k ) kwt0 = p , k " odd 2 k 2p t0 = p 2 T For k = 1, t0 = T 4 0

0

A I D

O N

SOL 2.33

Option ( ) is correct.

SOL 2.34

Option (D) is correct.

no w.

.in

co ia.

d

ww

z , z >a (z - a) 2 Residue of X (z) zn - 1 at z = a is = d (z - a) 2 X (z) zn - 1 z = a dz z = d (z - a) 2 zn - 1 2 dz (z - a) z=a n-1 n d z = = nz z = a = nan - 1 dz z = a Option (C) is correct. Given signal x (t) = rect `t - 1 j 2 1, - 1 # t - 1 # 1 or 0 # t # 1 2 2 2 So, x (t) = * 0, elsewhere Similarly x (- t) = rect`- t - 1 j 2 1, - 1 # - t - 1 # 1 or - 1 # t # 0 2 2 2 x (- t) = * 0, elsewhere Given that

SOL 2.35

Page 100

X (z) =

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F [x (t) + x (- t)] = =

3

Page 101

3

#- 3 x (t) e- jwt dt + #- 3 x (- t) e- jwt dt 0

1

#0 (1) e- jwt dt + #- 1 (1) e- jwt dt 1

0

- jw t - jw t = ; e E + ; e E = 1 (1 - e - jw) + 1 (e jw - 1) - jw 0 - jw - 1 jw jw - j w /2 j w /2 =e (e jw/2 - e - jw/2) + e (e jw/2 - e - jw/2) jw jw

SOL 2.36

(e jw/2 - e - jw/2) (e - jw/2 + e jw/2) = jw = 2 sin ` w j $ 2 cos ` w j = 2 cos w sinc` w j w 2 2 2 2p Option (B) is correct. In option (A) z1 [n] = x [n - 3] z2 [n] = z1 [4n] = x [4n - 3] y [n] = z2 [- n] = x [- 4n - 3] = Y x [3 - 4n] In option (B)

A I D

z1 [n] = x [n + 3] z2 [n] = z1 [4n] = x [4n + 3] y [n] = z2 [- n] = x [- 4n + 3] In option (C)

O N

n v1 [n] = x [4n] o.i c . v2 [n] = v1 [- n] = x [- 4n] dia o n + 3)] = y [n] = v2 [n + 3] = x [- w 4 (.n Y x [3 - 4n] w In option (D) w v1 [n] = x [4n] v2 [n] = v1 [- n] = x [- 4n] y [n] = v2 [n - 3] = x [- 4 (n - 3)] = Y x [3 - 4n] SOL 2.37

Option ( ) is correct. The spectrum of sampled signal s (jw) contains replicas of U (jw) at frequencies ! nfs . Where n = 0, 1, 2....... 1 fs = 1 = = 1 kHz Ts 1 m sec

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SOL 2.38

Option (D) is correct. For an LTI system input and output have identical wave shape (i.e. frequency of input-output is same) within a multiplicative constant (i.e. Amplitude response is constant) So F must be a sine or cosine wave with w1 = w2

SOL 2.39

Option (C) is correct. Given signal has the following wave-form

A I D

O N

no w.

.in

co ia.

d

ww

Function x(t) is periodic with period 2T and given that x (t) =- x (t + T) (Half-wave symmetric) So we can obtain the fourier series representation of given function. SOL 2.40

Option (C) is correct. Output is said to be distortion less if the input and output have identical wave shapes within a multiplicative constant. A delayed output that retains input waveform is also considered distortion less. Thus for distortion less output, input-output relationship is given as y (t) = Kg (t - td ) Taking Fourier transform. Y (w) = KG (w) e - jwt = G (w) H (w) H (w) & transfer function of the system d

So, H (w) = Ke - jwt Amplitude response H (w) = K Phase response, qn (w) =- wtd For distortion less output, phase response should be proportional to frequency. d

SOL 2.41

Option (A) is correct. G (z) z = e = ae- jw + be- 3jw jw

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for linear phase characteristic a = b . SOL 2.42

Option (A) is correct. System response is given as G (z) H (z) = 1 - KG (z) g [n] = d [n - 1] + d [n - 2] G (z) = z - 1 + z - 2 (z - 1 + z - 2) = 2 z+1 -1 -2 z - Kz - K 1 - K (z + z ) For system to be stable poles should lie inside unit circle. So

H (z) =

z #1 z = K!

K2 + 4K # 1 K ! 2

K2 + 4K # 2

K2 + 4K # 2 - K K2 + 4K # 4 - 4K + K2 8K # 4 K # 1/2 SOL 2.43

A I D

Option (C) is correct. Given Convolution is,

h (t) = u (t + 1) ) r (t - 2) Taking Laplace transform on both sides,

O N

in

H (s) = L [h (t)] = L [u (t + 1)] ) L.c[ro(.t - 2)]

We know that, L [u (t)] = 1/s

and

ia

d .no

w w s 1 w L [u (t + 1)] = e c 2 m s L [r (t)] = 1/s2

L r (t - 2) = e - 2s c 12 m s s 1 So H (s) = ;e ` jE;e - 2s c 12 mE s s -s 1 H (s) = e c 3 m s Taking inverse Laplace transform h (t) = 1 (t - 1) 2 u (t - 1) 2 SOL 2.44

(Time-shifting property)

(Time-shifting property)

Option (C) is correct. Impulse response of given LTI system. h [ n ] = x [ n - 1] ) y [ n ] Taking z -transform on both sides. H (z) = z - 1 X (z) Y (z) We have X (z) = 1 - 3z - 1 and Y (z) = 1 + 2z - 2 So

a x [n - 1]

Z

z - 1 x (z)

H (z) = z - 1 (1 - 3z - 1) (1 + 2z - 2) Output of the system for input u [n] = d [n - 1] is , y (z) = H (z) U (z)

U [n]

Z

U (z) = z - 1

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So Y (z) = z - 1 (1 - 3z - 1) (1 + 2z - 2) z - 1 = z - 2 (1 - 3z - 1 + 2z - 2 - 6z - 3) = z - 2 - 3z - 3 + 2z - 4 - 6z - 5 Taking inverse z-transform on both sides we have output. y [n] = d [n - 2] - 3d [n - 3] + 2d [n - 4] - 6d [n - 5] SOL 2.45

Option (B) is correct. A bounded signal always possesses some finite energy. E =

t0

g (t) 2 dt < 3

#- t

0

SOL 2.46

Option (C) is correct. Trigonometric Fourier series is given as 3

/an cos nw0 t + bn sin nw0 t

x (t) = A0 +

n=1

Since there are no sine terms, so bn = 0 bn = 2 T0

#0

T0

= 2= T0

#0

= 2; T0

#T

x (t) sin nw0 t dt

T0 /2

x (t) sin nw0 t dt +

IA

D O

T0

0

0

T0 /2

0

x (T - t) sin nw0 (T - t) (- dt)+

#T /2 x (t) sin nw0 t dt E T

0

#T /2 x (T - t) sin n` 2TpcoT.i-n t j dt ++ #T /2 x (t) sin nw0 t dt E TO

N

= 2; T0

T

0

Where t = T - t & dt =- dt

= 2; T0

#T /2 x (t) sin nw0 t dt G

O

T

.

ia od

0

n (2np - nw0) dt+ # x (t) sin nw0 t dt E #T /2 x (T -wt).sin T /2

= 2 ;T0

T0

0

ww

T0

0

#T /2 x (T - t) sin (nw0 t) dt + + #T /2 x (t) sin nw0 t dt E T0

bn = 0 if x (t) = x (T - t) From half wave symmetry we know that if x (t) =- x`t ! T j 2 Then Fourier series of x (t) contains only odd harmonics. SOL 2.47

Option (C) is correct. Z -transform of a discrete all pass system is given as -1 ) H (z) = z - z-0 1 1 - z0 z It has a pole at z 0 and a zero at 1/z) 0. Given system has a pole at z = 2+30% = 2

( 3 + j) = ( 3 + j) 2

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system is stable if z < 1 and for this it is anti-causal. SOL 2.48

Option (A) is correct. According to given data input and output Sequences are x [n] = {- 1, 2}, - 1 # n # 0 y [n] = {- 1, 3, - 1, - 2}, - 1 # n # 2 If impulse response of system is h [n] then output y [n] = h [ n] ) x [ n] Since length of convolution (y [n]) is - 1 to 2, x [n] is of length - 1 to 0 so length of h [n] is 0 to 2. Let h [n] = {a, b, c} Convolution

A I D

O N

no w.

.in

co ia.

d

ww

So, a=1

y [n] = {- a, 2a - b, 2b - c, 2c} y [n] = {- 1, 3, - 1, - 2} -

SOL 2.49

2a - b = 3 & b =- 1 2a - c =- 1 & c =- 1 Impulse response h [n] = "1, - 1, - 1, Option ( ) is correct.

SOL 2.50

Option (D) is correct. Output y (t) = e - x (t) If x (t) is unbounded, x (t) " 3 y (t) = e - x (t) " 0 (bounded) So y (t) is bounded even when x (t) is not bounded.

SOL 2.51

Option (B) is correct. Given

y (t) =

t

# x (t') dt' -3

Laplace transform of y (t)

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Y (s) =

X (s) , has a singularity at s = 0 s t

# x (t') dt'

For a causal bounded input, y (t)= SOL 2.52

Page 106

is always bounded.

-3

Option (A) is correct. RMS value is given by 1 T

Vrms = Where

#0

T

V2 (t) dt

2 T `T j t, 0 # t # 2

V (t) = * 0, So

1 T

#0

T

V 2 (t) dt = 1 = T

SOL 2.53

#0

T /2

T /2

2t 2 ` T j dt +

#0

Option (A) is correct. By final value theorem

SOL 2.54

T

T /2

A I D

O N

lim f (t) = lim s F (s) = lim s

t"3

#T/2 (0) dt G

3 t2 dt = 43 ; t E T 3 0 3 = 43 # T = 1 6 24 T = 1 V 6

= 1 $ 42 T T

Vrms

T q H = 6B@-1 >P H 3 3 where 6B@ is obtained as R V 1 S 1 + 1 W X12 X23 X23 W S 6B@ = S - 1 1 + 1 W X23 X13 W S X23 T X 1 + 1 -1 2 -1 H=> H => -1 1 + 1 -1 2 Its inverse is obtained as .in o -1 c . 2 -1 a 6B@-1 = >1 - 2 H odi .n w 2 w1 0.1 H = 1 > w H> 3 + 1 2 - 0. 2 Therefore, q2 1 2 1 0. 1 >q H = 3 >+ 1 2H>- 0.2H 3 0 H = 1> 3 - 0.3 0 => - 0.1H

A I D

O N

i.e., and SOL 3.3

q2 = 0 q3 =- 0.1 rad

Option (C) is correct. From the above solution, we have P2 = 0.1 P3 =- 0.2 since, P1 + P2 + P3 = 0 (Where P1 is injection at bus 1) So, P1 - P2 - P3 =- 0.1 + 0.2 = 0.1 pu Now, the apparent power delivered to base is 3 2 2 ^100 # 10 h V S = = 100 R 6 = 100 # 10 VA

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Page 211

Therefore, the real power delivered by slack bus (bus 1) P = P1 S = ^0.1h # 100 # 106 = 10 # 106 watt = 10 MW SOL 3.4

Option (B) is correct. For bus admittance matrix, Y11 + (Y12 + yline) + Y13 = 0 - j13 + (j10 + yline) + j 5 = 0 yline =- j2 Magnitude of susceptance is + 2

SOL 3.5

Option (A) is correct. i1 (t) = Im sin (wt - f1) i2 (t) = Im cos (wt - f2) We know that, cos (q - 90c) = sin q So, i1 (t) can be written as i1 (t) = Im cos (wt - f1 - 90c) i2 (t) = Im cos (wt - f2) Now, in phasor form I1 = Im f1 + 90c I 2 = Im f 2 Current are balanced if I1 + I 2 = 0 .in Im f1 + 90c + Im f 2 = 0 a.co d+i cos f2 + j sin f2 = 0 o Im cos ^f1 + 90ch + jIm sin ^f1 + 90 c h n . wcw Im 8cos ^f1 + 90ch + j sin ^f1 +w90 hB + Im 6cos f 2 + j sin f 2@ = 0 Im 8cos ^f1 + 90ch + cos f 2B + jIm 8sin f 2 + sin ^f1 + 90chB = 0 cos ^f1 + 90ch + cos f2 = 0 cos ^f1 + 90ch =- cos f2 = cos ^p + f2h f1 + 90c = p + f2 or, f1 = p + f2 2 Option (A) is correct. Let penalty factor of plant G , is L1 given as 1 L1 = 1 - 2PL 2PG

A I D

O N

SOL 3.6

1

PL = 0.5PG2 2PL = 0.5 (2P ) = P G G 2PG 1 So, L1 = 1 - PG Penalty factor of plant G2 is 1 L2 = =1 1 - 2PL 2PG For economic power generation 1

1

1

2

2

2PL ca 2PG = 0 m 2

2

C1 # L1 = C 2 # L 2

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where C1 and C2 are the incremental fuel cost of plant G1 and G2 . So, (10000) b 1 l = 12500 # 1 1 - PG 4 = 1-P G 5 PG = 1 pu 5 It is an 100 MVA, so PG = 1 # 100 = 20 MW 5 2 Loss PL = 0.5 b 1 l = 1 pu 5 50 or PL = 1 # 100 = 2 MW 50 2

2

2

2

PL = PG + PG - PL 40 = 20 + P2 - 2 PG = 22 MW

Total power,

1

2

2

SOL 3.7

Option (C) is correct. For double line-to-ground (LLG ) fault, relation between sequence current is

A I D

I positive =-^I negative + I zeroh Gives values satisfy this relation, therefore the type of fault is LLG . SOL 3.8

Option (B) is correct. Complex power for generator

O N

SG = SD1 + SD2 = .1in+ 1 = 2 pu .co1 pu, so Power transferred from bus 1 to bus 2iais d V1 V2 sin no(q1 - q2) . w 1= ww X 1

= 1 # 1 sin (q1 - q2) 0.5

(Line is lossless)

V1 = V2 = 1 pu X = 0.5 pu

0.5 = sin (q1 - q2) q1 - q2 = 30c q2 = q1 - 30c =- 30c (q1 = 0c) So, V1 = 1 0c V V2 = 1 - 30c V 1 0c - 1 30c Current, I12 = V1 - V2 = = (1 - j 0.288) pu j 0.5 Z ) Current in SD is I2 , SD = V2 I2 1 = 1 - 30c I2) I2 = 1 - 30c pu Current in QG , IG = I2 - I12 = 1 - 30c - (1 - j 0.288) = 0.268 - 120c VAR rating of capacitor, 2

2

2

QC = V2 VG = 1 # 0.268 = 0.268 pu SOL 3.9

Option (D) is correct.

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Page 213

Total reactance, X = j1 + j 0.5 = j1.5 pu Critical angle is given as, dcr = cos-1 [(p - 2d0) sin d0 - cos d0] d0 " steady state torque angle. Steady state power is given as

...(i)

Pm = Pmax sin d0 E V X E V Pm = sin d0 X (1.5) (1) 0.5 = sin d0 1.5

Pmax =

where, So,

Pm = 0.5 pu

A I D

sin d0 = 0.5 d0 = 30c

d0 = 30c # p = 0.523 180c Substituting d0 into equation (i) .in dcr = cos-1 [(p - 2 # 0.523 .co) sin 30c - cos 30c] In radian,

O N

ia

= cos-1 [(2.095n)o(0d.5) - 0.866] w. ) - 79.6c = cos-1 (0w .1815

w

SOL 3.10

Option ( ) is correct

SOL 3.11

Option (A) is correct. Negative phase sequence relay is used for the protection of alternators against unbalanced loading that may arise due to phase-to-phase faults.

SOL 3.12

Option (C) is correct. Steady state stability or power transfer capability E V Pmax = X To improve steady state limit, reactance X should be reduced. The stability may be increased by using two parallel lines. Series capacitor can also be used to get a better regulation and to increase the stability limit by decreasing reactance. Hence (C) is correct option.

SOL 3.13

Option (A) is correct. We know that loss \ PG2 loss \ length Distance of load from G1 is 25 km Distance of load from G2 & G 3 is 75 km generally we supply load from nearest generator. So maximum of load should be supplied from G1 . But G2 & G 3 should be operated at same minimum generation.

SOL 3.14

Option (B) is correct.

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Page 214

Power angle for salient pole alternator is given by V sin f + Ia Xq tan d = t Vt cos f + Ia Ra Since the alternator is delivering at rated kVA rated voltage Ia = 1 pu Vt = 1 pu f = 0c sin f = 0, cos f = 1 Xq = 1 pu, Xd = 1.2 pu 1 0 + 1 (1) tan d = # 1+0 =1 d = 45c SOL 3.15

Option (B) is correct. The admittance diagram is shown below

here

A I D

O N

y10 =- 10j, y12 =- 5j, y.i23n= 12.5j, y 30 =- 10j

.co a i d Note: y23 is taken positive because noit is capacitive. . ww Y11 = y10 w + y12 =- 10j - 5j =- 15j

SOL 3.16

Y12 = Y21 =- y21 = 5j Y13 = Y31 =- y13 = 0 Y22 = y20 + y21 + y23 = 0 + (- 5j) + (12.5j) = 7.5j Y23 = Y32 =- y23 =- 12.5j Y33 = y 30 + y13 + y23 =- 10j + 0 + 12.5j = 2.5j So the admittance matrix is RY Y Y V R- 15j 5j 0 VW S 11 12 13W S Y = SY21 Y22 Y33W = S 5j 7.5j - 12.5j W SSY Y Y WW SS 0 - 12.5j 2.5j WW 33 32 31 X T X T Option (A) is correct. For generator G1 X mG = 0.25 # 100 = 0.1 pu 250 For generator G2 X mG = 0.10 # 100 = 0.1 pu 100 1

1

XL = XL = 0.225 # 10 = 2.25 W For transmission lines L1 and L2 2

1

2 Z base = kV base = 15 # 15 = 2.25 W 100 MVA base

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Page 215

X mL (pu) = 2.25 = 1 pu 2.25 X mL (pu) = 2.25 = 1 pu 2.25 So the equivalent pu reactance diagram 2

1

SOL 3.17

Option (D) is correct. We can see that at the bus 3, equivalent thevenin’s impedance is given by Xth = ^ j0.1 + j1.0h || ^ j0.1 + j1.0h = j1.1 || j1.1 = j0.55 pu

SOL 3.18

Fault MVA = Base MVA = 100 = 181.82 MVA 0.55 Xth Option (C) is correct. Given that, a a I > 0 so

SOL 3.19

A I D

I >0 VAB > 0 since it is Rectifier iO/P .n o c . I/P VCD > 0 since it is Inverter dia o current will flow in given direction. VAB > VCD , Than.n

O N w

ww Option (A) is correct. Given step voltage travel along lossless transmission line.

a Voltage wave terminated at reactor as.

By Applying KVL V + VL = 0 VL =- V VL =- 1 pu SOL 3.20

Option (A) is correct. Given two buses connected by an Impedance of (0 + j5) W The Bus ‘1’ voltage is 100+30c V and Bus ‘2’ voltage is 100+0c V We have to calculate real and reactive power supply by bus ‘1’

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Page 216

P + jQ = VI) = 100+30c ;100+30c - 100+0cE 5j = 100+30c [20+ - 60c - 20+ - 90c] = 2000+ - 30c - 2000+ - 60c P + jQ = 1035+15c real power P = 1035 cos 15c = 1000 W reactive power Q = 1035 sin 15c = 268 VAR SOL 3.21

SOL 3.22

Option (C) is correct. Given 3-f, 33 kV oil circuit breaker. Rating 1200 A, 2000 MVA, 3 sec Symmetrical breaking current Ib = ? Ib = MVA kA = 2000 = 34.99 kA - 35 kA 3 kV 3 # 33 Option (C) is correct. Given a stator winding of an alternator with high internal resistance fault as shown in figure

A I D

O N

no w.

.in

co ia.

d

ww

Current through operating coil

I1 = 220 # 5 A, I2 = 250 # 5 A 400 400 Operating coil current = I2 - I1 = (250 - 220) # 5/400 = 0.375 Amp SOL 3.23

Option (C) is correct. Zero sequence circuit of 3-f transformer shown in figure is as following:

No option seems to be appropriate but (C) is the nearest. SOL 3.24

Option (D) is correct. Given that A 50 Hz Generator is initially connected to a long lossless transmission line which is open circuited as receiving end as shown in figure. Due to ferranti effect the magnitude of terminal voltage does not change, and the

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Page 217

field current decreases. SOL 3.25

Option (B) is correct. Given : 3-f, 50 Hz, 11 kV distribution system, We have to find out e1, e2 = ? Equivalent circuit is as following

11 (6C) e1 = 3 = 11 # 6 = 3.46 kV 11 6C + 5C 3

SOL 3.26

e2 = 11 # 5 = 2.89 kV 11 3 Option (A) is correct. Given : 3-f, 50 Hz, 11 kV cable

A I D

C1 = 0.2 mF C2 = 0.4 mF Charging current IC per phase = ? Capacitance Per Phase C = 3C1 + C2

O N

.in

C = 3 # 0.2 +a0.4 .co= 1 mF i d w = 2pf .= no314 w 3 w V w Changing current IC = = V (wC) = 11 # 10 # 314 # 1 # 10- 6 XC 3 = 2 Amp

SOL 3.27

Option (B) is correct. Generator G1 and G2 XG1 = XG2 = X old # New MVA # b Old kV l New kV Old MVA 2 = j0.9 # 200 # b 25 l = j0.18 100 25 2 Same as XT1 = j0.12 # 200 # b 25 l = j0.27 90 25 2 XT2 = j0.12 # 200 # b 25 l = j0.27 90 25 X Line = 150 # 220 2 = j0.62 (220) The Impedance diagram is being given by as

2

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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia SOL 3.28

Option ( ) is correct.

SOL 3.29

Option (C) is correct. We know complex power

a

So SOL 3.30

S = P + jQ = VI (cos f + j sin f) = VIe jf I = S jf Ve Real Power loss = I2 R 2 2 1 PL = c S jf m R = S j2R # 2 V Ve e f PL \ 12 V

Page 218

2 a S j2R = Constant e f

Option (C) is correct. YBus matrix of Y-Bus system are given as R V S- 5 2 2.5 0 W S 2 - 10 2.5 0 W YBus = j S W S2.5 2.5 - 9 4 W 4 4 - 8W S0 T X element We have to find out the buses having shunt R V Sy11 y12 y13 y14W Sy21 y22 y23 y24W We know YBus = S W Sy 31 y 32 y 33 y 34W Sy 41 y 42 y 43 y 44W T X .in 5j Here y11 = y10 + y12 + y13 + yc14o=. ia y24 =- 10j y22 = y20 + y21 + yo23d+ .n w+ y 33 = y 30 +w y 31 y 32 + y 34 =- 9j w y 44 = y 40 + y 41 + y 42 + y 43 =- 8j y12 = y21 =- y12 = 2j y13 = y 31 =- y13 = 2.5j y14 = y 41 =- y14 = 0j y23 = y 32 =- y23 = 2.5j y24 = y 42 =- y24 = 4j So y10 = y11 - y12 - y13 - y14 =- 5j + 2j + 2.5j + 0j =- 0.5j y20 = y22 - y12 - y23 - y24 =- 10j + 2j + 2.5j + 4j =- 1.5j y 30 = y 33 - y 31 - y 32 - y 34 =- 9j + 2.5j + 2.5j + 4j = 0 y 40 = y 44 - y 41 - y 42 - y 43 =- 8j - 0 + 4j + 4j = 0 Admittance diagram is being made by as

A I D

O N

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Page 219

From figure. it is cleared that branch (1) & (2) behaves like shunt element. SOL 3.31

Option (B) is correct. We know that • Shunt Capacitors are used for improving power factor. • • •

SOL 3.32

Series Reactors are used to reduce the current ripples. For increasing the power flow in line we use series capacitor. Shunt reactors are used to reduce the Ferranti effect.

A I D

Option (C) is correct. We know that for different type of transmission line different type of distance relays are used which are as follows. in o.used Short Transmission line -Ohm reactance c . a direlay Medium Transmission Line -Reactance is used o .n w Long Transmission line -Mho w relay is used

O N w

SOL 3.33

Option (C) is correct. Given that three generators are feeding a load of 100 MW. For increased load power demand, Generator having better regulation share More power, so Generator -1 will share More power than Generator -2.

SOL 3.34

Option (A) is correct. Given Synchronous generator of 500 MW, 21 kV, 50 Hz, 3-f, 2-pole P.F = 0.9 , Moment of inertia M = 27.5 # 103 kg-m2 Inertia constant H = ? Generator rating in MVA G = P = 500 MW = 555.56 MVA 0.9 cos f N = 120 # f = 120 # 50 = 3000 rpm 2 pole 2 Stored K.E = 1 Mw2 = 1 M b 2pN l 2 2 60 = 1 # 27.5 # 103 # b 2p # 3000 l MJ 2 60 = 1357.07 MJ Stored K.E Inertia constant (H) = Rating of Generator (MVA) H = 1357.07 555.56

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Page 220

= 2.44 sec SOL 3.35

Option (D) is correct. Given for X to F section of phase ‘a’ Va -Phase voltage and Ia -phase current. Impedance measured by ground distance, Bus voltage = Va W Relay at X = Ia Current from phase 'a'

SOL 3.36

Option (D) is correct. For EHV line given data is Length of line = 300 km and b = 0.00127 S rad/km wavelength l = 2p = 2p = 4947.39 km 0.00127 b l % = 300 So 100 = 0.06063 # 100 4947.39 # l l % = 6.063 l

SOL 3.37

Option (B) is correct. For three phase transmission line by solving the given equation VRI V RDV V R(X - X ) 0 0 m WS aW S aW S s We get, 0 (Xs - Xm) 0 WSIbW SDVbW = S SSDV WW SS 0 0 (Xs + 2Xm)WWSSIcWW c XT X X T T Zero sequence Impedance = Xs + 2Xm = 48 n and Positive Sequence Impedance = Negative o.i Sequence Impedance c . =di(aXs - Xm) o .n = 15 w ww(2) By solving equation (1) and Zs or Xs = 26 and Zm or Xm = 11

A I D

O N

SOL 3.38

Option ( ) is correct.

SOL 3.39

Option (B) is correct. SIL has no effect of compensation So SIL = 2280 MW

SOL 3.40

Option (C) is correct. Given PG1 + PG2 = 250 MW C1 (PG1) = PG1 + 0.055PG12 and 4 C2 (PG2) = 3PG2 + 0.03PG22 from equation (2) dC1 = 1 + 0.11P G1 dPG1 dC2 = 3 + 0.06P and G2 dPG2 Since the system is loss-less dC1 = dC2 Therefore dPG1 dPG2 So from equations (3a) and (3b) We have 0.11PG1 - 0.06PG2 = 2 Now solving equation (1) and (4), we get

...(1)

...(2)

...(1) ...(2)

...(3a) ...(3b)

...(4)

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Page 221

PG1 = 100 MW PG2 = 150 MW SOL 3.41

Option (B) is correct. After connecting both the generators in parallel and scheduled to supply 0.5 Pu of power results the increase in the current. ` Critical clearing time will reduced from 0.28 s but will not be less than 0.14 s for transient stability purpose.

SOL 3.42

Option (D) is correct. Given that the each section has equal impedance. Let it be R or Z , then by using the formula line losses = / I2 R On removing (e1); losses = (1) 2 R + (1 + 2) 2 R + (1 + 2 + 5) 2 R = R + 9R + 64R = 74R Similarly, On removing e2 ;losses = 52 R + (5 + 2) 2 R + (5 + 2 + 1) 2 R = 138R lossess on removing e 3 = (1) 2 R + (2) 2 R + (5 + 2) 2 R = 1R + 4R + 49R = 54R on removing e 4 lossless = (2) 2 R + (2 + 1) 2 R + 52 R

A I D

= 4R + 9R + 25R = 38R So, minimum losses are gained by removing e 4 branch. SOL 3.43

O N

Option (A) is correct. n o.i c Given : V (t) = Vm cos (wt) . dia the fault o For symmetrical 3 - f fault, current after n

w. w w +

2 Vm cos (wt - a) Z At the instant of fault i.e t = t 0 , the total current i (t) = 0 i (t) = Ae

- (R/L) t

0 = Ae- (R/L) t + 0

`

2 Vm cos (wt - a) 0 Z

Ae- (R/L) t =- 2 Vm cos (wt 0 - a) Z Maximum value of the dc offset current 0

Ae- (R/L) t =- 2 Vm cos (wt 0 - a) Z For this to be negative max. 0

or

(wt 0 - a) = 0 t0 = a w

...(1)

Z = 0.004 + j0.04 Z = Z +a = 0.0401995+84.29c a = 84.29cor 1.471 rad.

and

From equation (1) t0 =

1.471 = 0.00468 sec (2p # 50) t 0 = 4.682 ms

SOL 3.44

Option (C) is correct.

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Since the fault ‘F’ is at mid point of the system, therefore impedance seen is same from both sides.

Z = 0.0201+84.29c 2 Z1 (Positive sequence) = Z = 0.0201+84.29c 2 also Z1 = Z2 = Z 0 (for 3-f fault) `

1+0c I f (pu) = 1+0c = Z1 0.0201+84.29c

So magnitude

If

(p.u.)

I f = 49.8 #

` Fault current SOL 3.45

= 49.8

Option (A) is correct. If fault is LG in phase ‘a’

100 = 7.18 kA 3 # 400

A I D

O N

no w.

.in

co ia.

d

ww

Z1 = Z = 0.0201+84.29c 2

and Then

Z2 = Z1 = 0.0201+84.29c Z 0 = 3Z1 = 0.0603+84.29c Ia /3 = Ia1 = Ia2 = Ia0 1.0+0c Z1 + Z 2 + Z 0 1. 0 = = 9.95 pu (0.0201 + 0.0201 + 0.0603)

Ia1 (pu) = and

SOL 3.46

SOL 3.47

Ia1

Fault Current I f = Ia = 3Ia1 = 29.85 pu 100 So Fault current I f = 29.85 # = 4.97 kA 3 # 400 Option (A) is correct. a Equal Phase shift of point A & B with respect to source from both bus paths. So the type of transformer Y-Y with angle 0c. Option (C) is correct. Given incremental cost curve

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Page 223

PA + PB = 700 MW For optimum generator PA = ? , PB = ? From curve, maximum incremental cost for generator A = 600 at 450 MW and maximum incremental cost for generator B = 800 at 400 MW minimum incremental cost for generator B = 650 at 150 MW

A I D

Maximim incremental cost of generation A is less than the minimum incremental constant of generator B. So generator A operate at its maximum load = 450 MW n for optimum generation. o.i c . ia PA = 450 MW od

O N

.n

w PB = (700 -w 450) w = 250 MW

SOL 3.48

Option (C) is correct. Here power sharing between the AC line and HVDC link can be changed by controlling the HVDC converter alone because before changing only grid angle we can change the power sharing between the AC line and HVDC link.

SOL 3.49

Option (B) is correct. We have to find out maximum electric field intensity at various points. Electric field intensity is being given by as follows

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Page 224

From figures it is cleared that at point Y there is minimum chances of cancelation. So maximum electric field intensity is at point Y. SOL 3.50

Option (D) is correct. To increase capacitive dc voltage slowly to a new steady state value first we have to make d =- ve than we have to reach its original value.

SOL 3.51

Option (B) is correct. Given that .045 2p # 50 1 1 Suspectance of Line = 1.2 pu & C = 2p # 50 # 1.2 Reactance of line

= 0.045 pu & L =

A I D

Velocity of wave propagation = 3 # 105 Km/sec Length of line l = ? We know velocity of wave propagation VX = l LC .in .45 o 1 1 5 c . l = VX LC = 3 i# 10 # 2p 50 # 1.2 a 2 50 p # # d

O N

no

= 185 Km w.

ww

SOL 3.52

Option (C) is correct. Due to the fault ‘F’ at the mid point and the failure of circuit-breaker ‘4’ the sequence of circuit-breaker operation will be 5, 6, 7, 3, 1, 2 (as given in options) (due to the fault in the particular zone, relay of that particular zone must operate first to break the circuit, then the back-up protection applied if any failure occurs.)

SOL 3.53

Option (A) is correct.

SOL 3.54

Option (C) is correct.

R V 1 - 1 W S 0 3 3 W RSiaVW S 1 W Si W R = [Van Vbn Vcn] SS- 1 0 b 3 3 W SS WW i S 1 W c - 1 0 WT X S S 3 W 3 T X By solving we get R = ;Van (ib - ic) + Vbn (ic - ia) + Vc (ia - ib)E 3 3 3 (ib - ic) R = 3 (VI) , where = I and Van = V 3

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Here

Since `

Page 225

P1 " power before the tripping of one ckt P2 " Power after tripping of one ckt P = EV sin d X Pmax = EV X here, [X2 = (0.1 + X) (pu)] P2 max = EX , X2

To find maximum value of X for which system does not loose synchronism P2 = Pm (shown in above figure) EV sin d = P ` m 2 X2 as Pm = 1 pu, E = 1.0 pu,V = 1.0 pu 1.0 # 1.0 sin 130c = 1 X2 .in & & & SOL 3.55

A I D

O N

.co

X2 = 0.77 dia o .n (0.1 + X) = 0.77 w w X = 0.67 w

Option (B) is correct. Given that

FP = KAFS

Rf V Rf V S aW S pW where, Phase component FP = SfbW, sequence component FS = SfnW SSf WW SSf WW c o X X T T R 1 1 1V W S and A = Sa2 a 1W SS a a2 1WW X T VP = KAVS ` 3 IP = KAIS and VS = Zl [IS ] R0.5 0 0 V S W where Zl = S 0 0.5 0 W SS 0 0 2.0WW T X We have to find out Z if VP = ZIP From equation (2) and (3) VP = KAZl [IS ] -1 VP = KAZlb A l I p K

...(1)

...(2) ...(3)

...(4)

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SOL 3.56

Page 226

...(5) VP = AZlA- 1 I p R 1 1 1V S W A = Sa2 a 1W SS a a2 1WW T X Adj A -1 A = A R 2V S1 a a W Adj A = S1 a2 a W S W S1 1 1 W T X A =1 3 R 2V S1 a a W A- 1 = 1 S1 a2 a W 3S W S1 1 1 W T X From equation (5) R 1 1 1VR0.5 0 0VR1 a a2V R 1 0.5 0.5V S W S S W S W W ...(6) Vp = 1 Sa2 a 1WS 0 0.5 0WS1 a2 a W I p = S0.5 1 0.5W I p 3S S W SS0.5 0.5 1 WW S a a2 1WWSS 0 0 2WWS1 1 1 W T T X X Comparing of equation (5)XTand (6) XT R 1 0.5 0.5V W S Z = S0.5 1 0.5W SS0.5 0.5 1 WW .in X T o c . Option (A) is correct. ia d o Given that the first two power system .n are not connected and separately loaded. w Now these are connected by wwshort transmission line. as P1 = P2 = Q1 = Q2 = 0 So here no energy transfer. The bus bar voltage and phase angle of each system should be same than angle difference is

A I D

O N

q = 30c - 20c = 10c SOL 3.57

Option (B) is correct. Given that, 230 V, 50 Hz, 3-f, 4-wire system P = Load = 4 kw at unity Power factor IN = 0 through the use of pure inductor and capacitor Than L = ?, C = ? a IN = 0 = IA + IB + IC Network and its Phasor is being as

...(1)

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Page 227

Here the inductor is in phase B and capacitor is in Phase C. We know P = VI 3 So Ia = P = 4 # 10 = 17.39 Amp. 230 V From equation (1) IA =- (IB + IC ) `

IA =-c IB #

`

IA = IB

Now

XC

and

XC

&

C XL

& So SOL 3.58

L

a Ib - Ic

3 +I 3 C # 2 2 m

3 IB = 3 IC - IC = 17.39 - 10 Amp 3 V 230 = = - 23 W 10 IC = 1 2pfC 1 = 139.02 mF = 1 = 2p # 50 # 23 2pfXC = V = 230 - 23 W = 2pfL 10 IL 23 = XL = = 72.95 mH 2p # 100 2pf

L = 72.95 mH in phase B C = 139.02 mF in phase C

O N

A I D .in

Option (A) is correct. .co a i d Maximum continuous power limit ofoits n prime mover with speed governor of 5% . droop. ww w Generator feeded to three loads of 4 MW each at 50 Hz. Now one load Permanently tripped ` f = 48 Hz If additional load of 3.5 MW is connected than f = ? a Change in Frequency w.r.t to power is given as drop out frequency Df = # Change in power rated power = 5 # 3.5 = 1.16% = 1.16 # 50 = 0.58 Hz 15 100 System frequency is = 50 - 0.58 = 49.42 Hz

SOL 3.59

Option (B) is correct. With the help of physical length of line, we can recognize line as short, medium and long line.

SOL 3.60

Option (A) is correct. For capacitor bank switching vacuum circuit breaker is best suited in view of cost and effectiveness.

SOL 3.61

Option (B) is correct. Ratio of operating coil current to restraining coil current is known as bias in biased differential relay.

SOL 3.62

Option (B) is correct. HVDC links consist of rectifier, inverter, transmission lines etc, where rectifier

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Page 228

consumes reactive power from connected AC system and the inverter supplies power to connected AC system. SOL 3.63

Option (C) is correct. Given ABCD constant of 220 kV line A = D = 0.94+10c, B = 130+730c, C = 0.001+900c, VS = 240 kV % voltage regulation is being given as - (VR) Full load (V ) %V.R. = R No Load # 100 VR (Full load) At no load IR = 0 (VR) NL = VS /A , (VR) Full load = 220 kV 240 - 220 0 %V.R. = .94 # 100 220 %V.R. = 16

SOL 3.64

Option ( ) is correct.

SOL 3.65

Option (B) is correct. Given that, Vab1 = X+q1 , Vab2 = Y+q2 , Phase to neutral sequence volt = ? First we draw phasor of positive sequence and negative sequence.

A I D

O N

no w.

.in

co ia.

d

ww

From figure we conclude that positive sequence line voltage leads phase voltage by 30c VAN1 = X+q1 - 30c VAN2 = 4+q2 + 30c SOL 3.66

SOL 3.67

SOL 3.68

Option (A) is correct. For system base value 10 MVA, 69 kV, Load in pu(Z new ) = ? (MVA) old kVnew 2 Z new = Z old # # b kVold l (MVA) new 2 Z new = 0.72 # 20 # b 69 l = 36 pu 10 13.8 Option (A) is correct. Unreliable convergence is the main disadvantage of gauss seidel load flow method. Option (C) is correct. Generator feeds power to infinite bus through double circuit line 3-f fault at middle of line. Infinite bus voltage(V ) = 1 pu Transient internal voltage of generator(E ) = 1.1 pu Equivalent transfer admittance during fault = 0.8 pu = 1/X delivering power(PS ) = 1.0 pu

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Page 229

Perior to fault rotor Power angle d = 30c, f = 50 Hz Initial accelerating power(Pa ) = ?

SOL 3.69

SOL 3.70

Pa = PS - Pm2 sin d = 1 - EV sin 30c = 1 - 1.1 # 1 # 1 = 0.56 pu 2 X 1/0.8 Option (B) is correct. If initial acceleration power = X pu Initial acceleration = ? Inertia constant = ? X (pu) # S 180 # 50 # X # S = a = Pa = M S#S SH/180F a = 1800X deg / sec2 Inertia const. = 1 = 0.056 18 Option (D) is correct. The post fault voltage at bus 1 and 3 are. Pre fault voltage. RV V R1+0cV S 1W S W VBus = SV2W = S1+0cW SSV WW SS1+0cWW 3 T X T X At bus 2 solid fault occurs Z (f) = 0 , r = 2 Fault current I f = Vr c = V2 c Zrr + Z f Z22 + Z f o.in a.c Z f = 1+0c =- 4j o di j0.24 .n

A I D

O N w

ww

Vi (f) V1 (f) V1 (f) V3 (f) V3 (f)

= Vi c (0) - Zir I (f), Vi c = Prefault voltage = Vi c - Z12 I f = 1+0c - j0.08 (- j4) = 1 - 0.32 = 0.68 pu = V3 c - Z 32 I f = 1+0c - j0.16 (- j4) = 1 - 0.64 = 0.36 pu

SOL 3.71

Option ( ) is correct.

SOL 3.72

Option (D) is correct. Rating of D-connected capacitor bank for unity p.f. PL = S cos f = 12 3 # 0.8 = 16.627 kW reactive power QL = S sin f = 12 3 # 0.6 = 12.47 kW For setting of unity p.f. we have to set capacitor bank equal to reactive power = 12.47 kW real power

SOL 3.73

Option (D) is correct. Given that pu parameters of 500 MVA machine are as following

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Page 230

M = 20 pu, X = 2 pu Now value of M and X at 100 MVA base are for inertia (M) (pu) new = (pu) old # old MVA new MVA (M pu) new = (M Pu) old # 500 = 20 # 5 = 100 pu 100 1 and for reactance (X ) (pu) new = (pu) old # new MVA old MVA (X pu) new = (X pu) old # 100 500 (X Pu) new = 2 # 1 = 0.4 pu 5 SOL 3.74

Option (D) is correct. 800 kV has Power transfer capacity = P At 400 kV Power transfer capacity = ? We know Power transfer capacity P = EV sin d X

A I D

P \ V2 SOL 3.75

SOL 3.76

So if V is half than Power transfer capacity is 1 of previous value. 4 Option (B) is correct. .in is governed by the switching over oline In EHV lines the insulation strength of c . ia voltages. od

O N

.n

w Option (A) is correct. ww For bulk power transmission over very long distance HVDC transmission preferably used.

SOL 3.77

Option (D) is correct. Parameters of transposed overhead transmission line XS = 0.4 W/km , Xm = 0.1 W/km + ve sequence reactance X1 = ? Zero sequence reactance X 0 = ? We know for transposed overhead transmission line. + ve sequence component X1 = XS - Xm = 0.4 - 0.1 = 0.3 W/km Zero sequence component X 0 = XS + 2Xm = 0.4 + 2 (0.1) = 0.6 W/km

SOL 3.78

Option (C) is correct. Industrial substation of 4 MW load = PL QC = 2 MVAR for load p.f. = 0.97 lagging If capacitor goes out of service than load p.f. = ? cos f tan f QL - QC PL QL - 2 4

= 0.97 = tan (cos- 1 0.97) = 0.25 = 0.25 = 0.25 & QL = 3 MVAR

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f = tan- 1 c

Page 231

QL = tan- 1 b 3 l = 36c 4 PL m

cos f = cos 36c = 0.8 lagging SOL 3.79

Option (D) is correct. Y22 = ? I1 = V1 Y11 + (V1 - V2) Y12 = 0.05V1 - j10 (V1 - V2) =- j9.95V1 + j10V2 I2 = (V2 - V1) Y21 + (V2 - V3) Y23 = j10V1 - j9.9V2 - j0.1V3 Y22 = Y11 + Y23 + Y2 =- j9.95 - j9.9 - 0.1j =- j19.95

SOL 3.80

Option (C) is correct. F1 = a + bP1 + cP 12 Rs/hour F2 = a + bP2 + 2cP 22 Rs/hour For most economical operation P1 + P2 = 300 MW then P1, P2 = ? We know for most economical operation 2F1 = 2F2 2P1 2P2 2cP1 + b = 4cP2 + b P1 = 2P2 P1 + P2 = 300 from eq (1) and (2)

O N

A I D .in

co ia.

...(1) ...(2)

od P1 = 200 MW , P2 = 100 .nMW

SOL 3.81

Option (B) is correct.

w

ww

V1 A B V2 We know that ABCD parameters > H = > I1 C DH >I1H , C = I1 B = V1 I2 V = 0 V2 I = 0 V1 Z 1 + Z2 In figure C = = 1 V1 Z2 Z Z1 + Z 2 # 2 1 or Z2 = 1 = = 40+ - 45c C 0.025+45c 2

SOL 3.82

2

Option (D) is correct. Given

Steady state stability Power Limit = 6.25 pu If one of double circuit is tripped than Steady state stability power limit = ? Pm1 = EV = 1 # 1 = 6.25 X 0.12 + X 2

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Page 232

1 = 6.25 0.12 + 0.5X & X = 0.008 pu If one of double circuit tripped than 1 Pm2 = EV = 1 # 1 = 0.12 + 0.08 X 0.12 + X Pm2 = 1 = 5 pu 0.2 SOL 3.83

SOL 3.84

Option (D) is correct. Given data Substation Level = 220 kV 3-f fault level = 4000 MVA LG fault level = 5000 MVA Positive sequence reactance: 4000 Fault current I f = 3 # 220 X1 = Vph /I f 220 3 = = 220 # 220 = 12.1 W 4000 4000 3 # 220 Option (B) is correct. Zero sequence Reactance X 0 = ? .in o 5000 c . If = a 3 # 220 o di .nI f 5000 w Ia1 = Ia2 =w w Ia0 = 3 = 3 3 # 220 220 V 3 X1 + X2 + X 0 = ph = 5000 Ia1 220 # 3 3 X1 + X2 + X 0 = 220 # 220 = 29.04 W 3 # 5000

A I D

O N

X1 = X2 = 12.1 W X 0 = 29.04 - 12.1 - 12.1 = 4.84 W SOL 3.85

Option (B) is correct. Instantaneous power supplied by 3-f ac supply to a balanced R-L load. P = Va Ia + Va Ib + Vc Ic = (Vm sin wt) Im sin (wt - f) + Vm sin (wt - 120c) Im sin (wt - 120c - f) + Vm sin (wt - 240c) Im sin (wt - 240c - f) = VI [cos f - cos (2wt - f) + cos f - cos (2wt - 240 - f) + cos f - cos (2wt + 240 - f)] ...(1) P = 3VI cos f equation (1) implies that total instantaneous power is being constant.

SOL 3.86

Option (C) is correct. In 3-f Power system, the rated voltage is being given by RMS value of line to

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Page 233

line voltage. SOL 3.87

Option (B) is correct.

In this figure the sequence is being given as RBY SOL 3.88

Option (C) is correct. In thermal power plants, the pressure in the working fluid cycle is developed by the help to feed water pump.

SOL 3.89

Option (A) is correct. Kaplan turbines are used for harnessing low variable waterheads because of high percentage of reaction and runner adjustable vanes.

SOL 3.90

Option (B) is correct. MHO relay is the type of distance relay which is used to transmission line protection. MHO Relay has the property of being inherently directional.

SOL 3.91

Option (C) is correct. Surge impedance of line is being given by as

A I D

L = 11 # 10- 3 = 306.88 W C 11.68 # 10-in9 Ideal power transfer capability co. . a i 2 d 2 o (800)n V P = = w. = 2085 MW 306 Z0 w .88

O N Z =

SOL 3.92

w

Option (D) is correct. Given that, Power cable voltage = 110 kV C = 125 nF/km Dielectric loss tangent = tan d = 2 # 10- 4 Dielectric power loss = ? dielectric power loss is given by P = 2V2 wC tan d = 2 (110 # 103) 2 # 2pf # 125 # 10- 9 # 2 # 10- 4 = 2 (121 # 108 # 2 # 3.14 # 50 # 250 # 10- 13) = 189 W/km

SOL 3.93

Option (A) is correct. Given data Lightening stroke discharge impulse current of I = 10 kA Transmission line voltage = 400 kV Impedance of line Z = 250 W Magnitude of transient over-voltage = ? The impulse current will be equally divided in both directions since there is equal distribution on both sides. Then magnitude of transient over-voltage is

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Page 234

V = IZ/2 = 10 # 103 # 250 2 = 1250 # 103 V = 1250 kV SOL 3.94

Option (C) is correct. The A, B, C, D parameters of line A = D = 0.936+0.98c B = 142+76.4c C = (- 5.18 + j914) 10- 6 W At receiving end PR = 50 MW , VR = 220 kV p.f = 0.9 lagging VS = ? Power at receiving end is being given by as follows VS VR A VR 2 PR = cos (b - d) cos (b - a) B B VS # 220 0.936 (220) 2 = cos (76.4c - d) cos 75.6c 142 142 ` VS cos (76.4 - d) = 50 # 142 + 0.936 # 220 # 0.2486 = 32.27 + 51.19 220

A I D

VS cos (76.4 - d) = 83.46 Same as QR = PR tan f = PR tan (cos- 1 f) = 50 tan (cos- 1 0.9) = 24.21 MW 2 VS VR A VR.in QR = sin (b - d) - a.co sin (b - a) B di B o n VS # 220 w. 0.936 # (220) 2 w = c d sin (76.4 ) sin 75.6c 142 w 142 (24.21) 142 + 0.936 # 220 # 0.9685 = VS sin (76.4c - d) 220 from equation (1) & (2)

O N VS

2

...(2)

= (215) 2 + (83.46) 2

VS = SOL 3.95

...(1)

53190.5716 = 230.63 kV

Option (B) is correct. A new generator of Eg = 1.4+30c pu XS = 1.0 pu, connected to bus of Vt Volt Existing Power system represented by thevenin’s equivalent as Eth = 0.9+0c, Zth = 0.25+90c, Vt = ?

From the circuit given E - Eth 1.212 + j7 - 0.9 I = g = 1.4+30c - 0.9+0c = Zth + XS j (1.25) j (1.25) 0.312 + j7 = = 0.56 - 0.2496j j (1.25)

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Page 235

Vt = Eg - IXS = 1.212 + j7 - (0.56 - 0.2496j) (j1) = 1.212 - 0.2496 + j (0.7 - 0.56) = 0.9624 + j0.14 Vt = 0.972+8.3c SOL 3.96

Option (C) is correct. Given that 3-f Generator rated at 110 MVA, 11 kV Xdm = 19% , Xdl= 26% XS = 130% , Operating at no load 3-f short circuit fault between breaker and transformer symmetrical Irms at breaker = ? We know short circuit current Isc = 1 = 1 =- j5.26 pu Xdm j0.19 rating MVA of generator Base current IB = 3 # kV of generator 6 IB = 110 # 10 3 3 # 11 # 10 IB = 5773.67 Amp Symmetrical RMS current = IB # Isc & Irms

SOL 3.97

A I D

= 5773.67 # 5.26 = 30369.50 Amp = 30.37 kA

O N

Option (A) is correct.

n o.i c . ia 2

+ ve sequence current Ia = 1 [Ia + adIb + a Ic] 3 no w. 1 w = w 3 [10+0c + 1+120c # 10+180c + 0] = 1 [10+0c + 10+300c] = 1 [10 + 5 - j8.66] 3 3 = 1 [15 - j8.66] = 17.32+ - 30c 3 3 = 5.78+ - 30c SOL 3.98

Option (D) is correct. Given data 500 MVA , 50 Hz, 3 - f generator produces power at 22 kV Generator " Y connected with solid neutral Sequence reactance X1 = X2 = 0.15 , X 0 = 0.05 pu Sub transient line current = ? E 1 Ia1 = = = 1 =- 2.857j j0.15 + j0.15 + j0.05 0.35j Z1 + Z 2 + Z 0 Now sub transient Line current Ia = 3Ia1 Ia = 3 (- 2.857j) =- 8.57j

SOL 3.99

Option (B) is correct. Given: 50 Hz, 4-Pole, 500 MVA, 22 kV generator p.f. = 0.8 lagging Fault occurs which reduces output by 40%. Accelerating torque = ? Power = 500 # 0.8 = 400 MW

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After fault,

Where

Page 236

Power = 400 # 0.6 = 240 MW a Pa = Ta # w Ta = Pa w w = 2pfmechanical fmechanical = felectrical # 2 = felectrical # 2 4 P Pa = 400 - 240 = 160 MW 160 Ta = 2 # p # 50/2 Ta = 1.018 MN

SOL 3.100

Option (D) is correct. Turbine rate speed N = 250 rpm To produce power at f = 50 Hz. No. of Poles P =? a N = 120 f P P = 120 f = 120 # 50 = 24 250 N P = 24 Poles

A I D

O N

SOL 3.101

Option (C) is correct. n o.i self GMD of conductor is increased c In case of bundled conductors, We know that . dia voltage of line depends upon GMD of and in a conductor critical disruptive o .n w conductor. Since GMD of conductor is increased this causes critical disruptive w w voltage is being reduced and if critical disruptive voltage is reduced, the corona loss will also be reduced.

SOL 3.102

Option (B) is correct. Given that no. of buses n = 300 Generator bus = 20 Reactive power support buses = 25 Fixed buses with Shunt Capacitor = 15 Slack buses (ns ) = 20 + 25 - 15 = 30 a Size of Jacobian Matrix is given as = 2 (n - ns) # 2 (n - ns) = 2 (300 - 30) # 2 (300 - 30) = 540 # 540

SOL 3.103

Option (B) is correct. Auxiliary component in HVDC transmission system are DC line inductor and reactive power sources.

SOL 3.104

Option (C) is correct. a Exchanged electrical power is being given as follows P = EV 6sin (d1 - d2)@ Xd Given that

...(1)

P " Power supply by generator = 0.5 pu

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Page 237

E " Voltage for rotar generator = 2.0 pu V " Voltage of motor rotor = 1.3 pu Xd = X eq = Reactance of generator + Reactance of motor + Recatance of connecting line Xd = 1.1 + 1.2 + 0.5 = 2.8 d1 - d2 = Rotor angle difference = ? from eq(1), 0.5 = 2 # 1.3 sin (d1 - d2) 2.8 & d1 - d2 = sin- 1 b 2.8 # 0.5 l 2. 6 & d1 - d2 = 32.58 SOL 3.105

Option (B) is correct. Time period between energization of trip circuit and the arc extinction on an opening operation is known as the interrupting time of Circuit breaker.

SOL 3.106

Option (B) is correct. Given that ABCD parameters of line as A = D = 0.9+0c, B = 200+90% W , C = 0.95 # 10 - 3 +90% S . at no-load condition,

A I D

Receiving end voltage (VR) = sending end voltage (VS ) ohmic value of reactor = ? We know VS = AVR + BIR n VS = VR o.i c . VR = AVR +oBI diaR .n VR (1 - A) = w BIwR w B VR = = 200+90c IR 1-A 1 - 0.9+0c VR = 2000+90c IR The ohmic value of reactor = 2000 W

O N

SOL 3.107

Option (A) is correct. Surge impedance of cable Z1 =

L; C

L = 0.4 mH/km, C = 0.5 mF /km

0.4 # 10- 3 = 28.284 0.5 # 10- 6 surge impedance of overhead transmission line Z2 = Z 3 = L ; L = 1.5 mm/km, C = 0.015 mF/km C =

1.5 # 10- 5 = 316.23 0.015 # 10- 6 Now the magnitude of voltage at junction due to surge is being given by as Vl = 2 # V # Z2 V = 20 kV Z 2 + Z1 Z2 = Z 3 =

3 = 2 # 20 # 10 # 316.23 316 + 28.284

= 36.72 kV

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Page 238

Option (D) is correct. Let that current in line is I amp than from figure current in line section PR is (I - 10) amp current in line section RS is (I - 10 - 20) = (I - 30) amp current in SQ Section is (I - 30 - 30) = (I - 60) amp Given that VP and VQ are such that VP - VQ = 3 V by applying KVL through whole line VP - VQ = (I - 10) 0.1 + (I - 30) 0.15 + (I - 60) # 0.2 & 3 = 0.45I - 17.5 I = 20.5 = 45.55 amp 0.45 Now the line drop is being given as = (I - 10) 0.1 + (I - 30) 0.15 + (I - 60) 0.2 = (33.55) 0.1 + (15.55) 0.15 + (14.45) 0.2 = 8.58 V The value of VP for minimum voltage of 220 V at any feeder is = 220 + Line voltage = 220 + 8.58 = 228.58 V

SOL 3.109

A I D

Option (D) is correct. Given Load Power = 100 MW n o.i c . VS = VR = 11 kV d2 ia o n j0.2 # (11) 2 p.u. #w (kV) . Impedance of line ZL = = = j0.242 W w 100 wMV VS VR sin d We know PL = X

O N

3 3 100 # 106 = 11 # 10 # 11 # 10 sin d 0.242 100 # 0.242 = sin d 121

d = sin- 1 (0.2) = 11.537c Reactive Power is being given by VS VR VR 2 QL = cos d X X 3

3

= 11 # 10 # 11 # 10 cos (11.537c) 0.242

(11 # 103) 2 0.242

6

SOL 3.110

= 121 # 10 [cos (11.537c) - 1] =- 10.1 MVAR 0.242 Option (B) is correct. Given the bus Impedance Matrix of a 4-bus Power System R V Sj0.3435 j0.2860 j0.2723 j0.2277W Sj0.2860 j0.3408 j0.2586 j0.2414W Z bus = S W Sj0.2723 j0.2586 j0.2791 j0.2209W Sj0.2277 j0.2414 j0.2209 j0.2791W T X bus 2 and reference Now a branch os j0.2 W is connected between

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Page 239

RZ V S ij W 1 ZB(New) = ZB (Old) Sh W Z g Z jnB Zij + Zb S W8 ji SZnjW X New element Zb = j0.2 W isT connected in jth and reference bus j = 2 , n = 4 so R V SZ12W SZ22W 1 Z Z Z Z Zij + Zb SSZ23WW 8 21 22 23 24B SZ24W R V T X Sj0.2860W Sj0.3408W 1 = S W8j0.2860 j0.3408 j0.2586 j0.2414B ...(1) 6j (0.3408) + j0.2@ Sj0.2586W Sj0.2414W T X Given that we are required to change only Z22, Z23 j2 (0.3408) 2 So in equation (1) Zl22 = = j0.2147 j (0.5408) j2 (0.3408) (0.2586) Zl23 = = j0.16296 0.5408

A I D

Z22(New) = Z22(Old) - Zl22 = j0.3408 - j0.2147 = j0.1260 Z23(New) = Z23 (Old) - Zl23 = j0.2586 - j0.16296 = j0.0956 SOL 3.111

Option (D) is correct. Total zero sequence impedance, + ve sequence impedance and - ve sequence in impedances co. Z0 Z1 Z2 Zn for L-G fault

O N

a.

= (Z 0) Line + (Z 0) Generatoro=dij0.04 + j0.3 = j0.34 pu .n w = (Z1) Line + (Z1) Generator = j0.1 + j0.1 = j0.2 pu ww = (Z2) Line + (Z2) Generator = j0.1 + j0.1 = j0.2 pu = j0.05 pu

Ia1 =

Ea 0.1 = j0.2 + j0.2 + j0.34 + j0.15 Z 0 + Z 1 + Z 2 + 3Z n

=- j1.12 pu generator MVA IB = = 3 generator kV Fault current

20 # 106 = 1750 Amp 3 # 6.6 # 103

I f = (3Ia) IB = 3 (- j1.12) (1750) =- j5897.6 Amp Neutral Voltage and

Vn = I f Zn Zn = ZB # Z pu (6.6) 2 0.05 = 0.1089 W = 20 # Vn = 5897.6 # 0.1089 = 642.2 V

SOL 3.112

Option (A) is correct. We know that Optimal Generation IC1 = IC2 , and P3 = 300 MW (maximum load) (Independent of load) IC 3 = 30

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20 + 0.3P1 = 30 + 0.4 P2 0.3P1 - 0.4P2 = 10 P1 + P2 + P3 = 700 P1 + P2 + 300 = 700 P1 + P2 = 400 From equation (1) and (2) P1 = 242.8 MW P2 = 157.14 MW

...(1)

...(2)

SOL 3.113

Option (A) is correct. For transmission line protection-distance relay For alternator protection-under frequency relay For bus bar protection-differential relay For transformer protection-Buchholz relay

SOL 3.114

Option (C) is correct.

A I D

O N

We know by equal area criteria PS (dm - d0) =

#d

C

dm

Page 240

.in

co ia.

Pmax sin ddd

d

o Pmax sin d0 (dm - d0) = Pmax [cos d.n 0 - cos dm] w Pmax = 2 ww P0 = Pmax sin d0 = 1 d0 = 30c dmax = 110c (given) Now from equation (1) 2 sin 30c (110 - 30) p = 2 [cos dc - cos 110c] 180 0.5 # 80p = cos dc + 0.342 180

...(1)

cos dc = 0.698 - 0.342 dc = 69.138c SOL 3.115

Option (D) is correct. a Both sides are granted So, Ia = Ea = 10+0c = 5+ - 90c 2j Za Ib = Eb = 10+ - 90c = 3.33+ - 180c 3j Zb Ic = Ec = 10+120c = 2.5+30c 4j Zc We know Ia = 1 [Ia + aIb + a2 Ic] 3 1

where a = 1+120c & a2 = 1+240c

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Page 241

Ia1 = 1 [5+ - 90c + 3.33+ ^- 180c + 120ch + 2.5+ ^240c + 30ch] 3 Ia1 = 1 [5+ - 90c + 3.33+ - 60c + 2.5+270c] 3 = 1 [- 5j + 1.665 - j2.883 - 2.5j] 3 = 1 [1.665 - j10.383] = 3.5+ - 80.89c 3 SOL 3.116

Option (B) is correct. Given data A balanced delta connected load = 8 + 6j = 2 V2 = 400 volt Improved Power Factor cos f2 = 0.9 f1 = tan- 1 ^6/8h = 36.85c f2 = cos- 1 (0.9) = 25.84c 400 = 40+ - 36.86c I = V = 400 = 8 + 6j 10+36.86c Z = 32 - j24 Since Power factor is Improved by connecting a Y-connected capacitor bank like as

A I D

O N

no w.

.in

co ia.

d

ww

Phasor diagram is being given by as follows

In figure

oa = I l cos f2 = I cos f1 I l cos 25.84c = 32 I l # 0.9 = 32 Il = 35.55 ac = 24 Amp. ab = I l sin f2 = 35.55 sin 25.84c

(ac = I sin f1)

ab = 15.49 Amp Ic = bc = ac - ab = 24 - 15.49 = 8.51 Amp KVAR of Capacitor bank = 3 # V # IC = 3 # 400 # 8.51 1000 1000 = 10.2 KVAR SOL 3.117

Option (B) is correct. Given power system with these identical generators

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Page 242

G1 has Speed governor G2 and G3 has dr0op of 5% When load increased, in steady state generation of G1 is only increased while generation of G2 and G3 are unchanged. SOL 3.118

Option (A) is correct. R1 , R2 -Distance Relay Zone-1 and Zone-2 setting for both the relays Correct setting for Zone-2 of relay R1 and R2 are given as TZ2 R = 0.6 sec, TZR = 0.3 sec a Fault at Zone-2, therefore firstly operated relay is R2 , so time setting of R2 is 0.3 sec and R1 is working as back up relay for zone-2, so time setting for R1 is 0.6 sec. 1

2

SOL 3.119

Option (B) is correct. The reactive power absorbed by the rectifier is maximum when the firing angle a = 30c.

SOL 3.120

Option (D) is correct. Given a power system consisting of two areas as shown connected by single tieline

A I D

O N

For load flow study when entering the network in data, the tie line data inadvertently . o left out. If load flow programme is run .with c this incomplete data than load flow diais specified. will not converge if only one slack obus

n

. ww

SOL 3.121

Option (D) is correct. w Given that XS = 0.2 pu Mid point voltage of transmission line = 0.98 pu VS = VR = 1 Steady state power transfer limit P = VS VR sin d = 1.1 sin 90c= 5 pu 0.2 XS

SOL 3.122

Option (B) is correct. We have to find out the thevenin’s equivalent zero sequence impedance Z 0 at point B.The zero sequence network of system can be drawn as follows

equivalent zero sequence impedance is being given as follows Z 0 = 0.1j + 0.05j + 0.07j + (3 # 0.25) Z 0 = 0.75 + j0.22 SOL 3.123

*Given data : ZC = 400 W (Characteristics Impedance)

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Page 243

b = 1.2 # 10- 3 rad/km (Propagation constant) l = 100 km (length of line) Pmax = ? If VS = 230 kV VS = VR cos (bl) + jZC sin (bl) IR VS = AVR + BIR A = cos bl = cos (1.2 # 10- 3 # 100) = 0.9928+0c B = jZC sin (bl) = j400 sin (1.2 # 10- 3 # 100) = j47.88 = 47.88+90c VS = 230 kV, l = 100 km Since it is a short line, so VS - VR = 230 kV again we know for transmission line the equation (Pr - Pr0) 2 + (Qr - Qr0) = Pr2 Where

...(1) 2

Pr0 =- AVR cos (b - a) MW B

A I D 2

Qr0 =- AVR sin (b - a) MW B Pr = VS VR MVA B

O N

and maximum power transferred is being given n by as

.i

Prm = Pr - Pr0a.co Pr

di o n VSwV.R = 230 # 230 =w 47.88 w B

Pr = 1104.84 MVA 2 Pr0 =- AVR cos (b - a) MW B =-

0.9928 # (230) 2 # cos (90c - 0) 47.88

Pr0 = 0 MW So maximum Power transferred Prm = Pr - Pr0 = 1104.84 MW SOL 3.124

*Given: two transposed 3-f line run parallel to each other. The equation for voltage drop in both side are given as R R V VR V S0.15 0.05 0.05 0.04 0.04 0.04WSIa1W SDVa1W S0.05 0.15 0.05 0.04 0.04 0.04WSIb1W SDVb1W S SDV W WS W S c1W = j S0.05 0.05 0.15 0.04 0.04 0.04WSIc1W S0.04 0.04 0.04 0.15 0.05 0.05WSIa2W SDVa2W S0.04 0.04 0.04 0.05 0.15 0.05WSIb2W SDVb2W S S W WS W S0.04 0.04 0.04 0.05 0.05 0.15WSIc2W SDVc2W T self and mutual zero sequence X XT impedance X We haveT to compute of the system i.e. Z 011, Z 012, Z 021, Z 022 in the following equation. DV01 = Z 011 I 01 + Z 021 I 02 DV02 = Z 021 I 01 + Z 022 I 02

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Page 244

We know that + ve , - ve and zero sequence Impedance can be calculated as respectively. Z1 = j (XS - Xm) Z2 = j (XS - Xm) Z 0 = j (XS + 2Xm) So zero sequence Impedance calculated as Z 011 Z 011 Z 012 Z 012 Z 022

SOL 3.125

= j (XS + 2Xm) = j [0.15 # 2 (0.05)] = 0.25j = Z 021 = j (XS + 2Xm) = Z 021 = j [0.15 + 2 (0.04)] = 0.23j = j (XS + 2Xm) = j [0.15 + 2 (0.01)] = 0.25j

XS = 0.15 , Xm = 0.05 XS = 0.15 , Xm = 0.04 XS = 0.15 , Xm = 0.05

*Given

A I D

O N

X = 0.2 pu n pu, H = 5 MJ/MVA For generator X' = 0.1 pu , El =.i1.0 o c . Mechanical Power Pm = 0.0 pu, diwaB = 2p # 50 rad/sec o Initially generator running on w open .n circuit, at switch closure d = 0 w wBw= dd = winit dt maximum winit = ? , so that generator pulls into synchronism We know that swing equation H d2 d = (P - P ) pu ......(1) m e pf dt2 E V sin d = 1.1 sin d = 3.33 sin d Pe = 0. 3 X From equation (1) 5 d2 d = 0 - 3.33 sin d 3.14 # 50 dt2 d2 d =- 104.72 sin d dt2 integrating on both side. dd = 104.72 cos d + d 0 dt d0 = 0 (given) w = dd dt For (winit) max = b dd l dt max dd b dt l

when cos d = 1

max

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SOL 3.126

SOL 3.127

Page 245

(winit) max = b dd l = 104.72 rad/sec dt max Option (C) is correct. A lossless radial transmission line with surge impedance loading has flat voltage profile and unity power factor at all points along it. Option (B) is correct. Given that 3-f transformer, 20 MVA, 220 kV(Y) - 33 kV(D) Xl = leakage Reactance = 12% X = reffered to LV in each phase = ? (LV side voltage) 2 Reactance of Leakage = 3# MVA Rating # = 3#

SOL 3.128

(33 kV) 2 0.12 = 19.6 W 20 MVA #

Option (D) is correct. Given 75 MVA, 10 kV synchronous generator Xd = 0.4 pu We have to find out (Xd ) new at 100 MVA, 11 kV

^kVhold 2 ^MVAhnew # H > H ^kVhnew ^MVAhold

A I D

(Xd ) new = (X d) old # >

(Xd ) new = 0.4 # b 10 l # 100 = 0.44 pu 11 75 Option (A) is correct. n o.i Given Y-alternator: 440 V, 50 Hz c . ia dcurrent Per phase Xs = 10 W , Capacitive Load I = 20 A o .n w For zero voltage regulation ww load p.f = ? Let Load Z = R + jX Zero voltage regulation is given so 2

SOL 3.129

O N

E Ph - IXs - I (R + jX) = 0 440 - 20 (j10) - 20 (R + jX) = 0 ...(1) 3 separating real and imaginary part of equation (1) 20R = 440 3 R = 22 3 and 20 (X + 10) = 440 3 22 X = - 10 = 4.68 3 3 4.68/ 3 - 1 4.68 q = tan- 1 X = tan- 1 f p = tan b 22 l R 22/ 3 and power factor cos q = cos b tan- 1 4.68 l 22 cos q = 0.82 SOL 3.130

Option (B) is correct. Given 240 V, 1-f AC source, Load Impedance Z = 10+60c W Capacitor is in parallel with load and supplies 1250 VAR

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Page 246

The real power P by source = ?

from figure current through load IL = I + IC I = V = 240 = 24+ - 60c Z 10+60c IC = VAR = 1250 = 5.20j 240 V IL = 24+ - 60c + 5.20j = 12 - 15.60j a apparent power S = VI = P + jQ = 240 (12 + 15.60j) = 2880 + 3744j = P + jQ Where P = Real Power , Q = Reactive Power P = 2880 W SOL 3.131

Option ( ) is correct.

SOL 3.132

Option (C) is correct. We have to find out maximum voltage location on line by applying KVL in the circuit VS - VR = 0.05j , where VS = 1 voltage at voltage at voltage at

O N

A I D

VR P1 P2 P3

in

= 1 - 0.05j co. . a i = VS = 1 pu . o d n . = 1 - 0w .1w j (by applying KVL) w = 1 - 0.1j + j0.15 (by applying KVL)

..(1) ...(2)

...(3) = 1 + 0.05j From equation (1), (2) and (3) it is cleared that voltage at P3 is maximum. SOL 3.133

Option (B) is correct. Given: two generators P1 = 50 (50 - f) P2 = 100 (51 - f) total load = 400 MW than f = ? P1 + P2 = 400 50 (50 - f) + 100 (51 - f) = 400 50 + 102 - 8 = 3f f = 48 Hz

SOL 3.134

*Given 132 kV transmission line connected to cable as shown in figure

Characteristics impedance of line and cable are 400 Wand 80 W 250 kV surge travels from A to B than (a) We have to calculate voltage surge at C.

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Page 247

(b) Reflected component of surge when reaches A. (c) Surge current in cable BC Vi = 250 kV , ZC1 = 400 W , ZC2 = 80 W (a) Voltage surge at C Vt = Z # ZC2 # Vi = 2 # 80 # 250 = 83.34 kV 400 + 80 ZC1 + ZC2 (b) Reflected voltage at A Vr = b ZC2 - ZC1 l Vi = 80 - 400 # 250 =- 166.67 kV 400 + 80 ZC2 + ZC1 (c) Surge current in cable BC It = Ii + Ir = Ii - aIi

SOL 3.135

= (1 - a) Ii , Here a = ZC2 - ZC1 ZC2 + ZC1 It = b1 - ZC2 - ZC1 l Vi = b1 + 320 l 250 480 400 ZC2 + ZC1 ZC1 = b1 + 4 l 25 = 1.04 kAmp 6 40 *We have to draw reactance diagram for given YBus matrix R V S- 6 2 2.5 0 W S 2 - 10 2.5 4 W YBus = j S W S2.5 2.5 - 9 4 W 4 4 - 8W S0 T X as in a It is 4 # 4 matrix (admittance matrix) R V co. . Sy11 y12 y13 y14W a i Sy21 y22 y23 y24W no d YBus = S W w. y y y y 34 33 32 31 w S wW Sy 41 y 42 y 43 y 44W T X Here diagonal elements

A I D

O N

...(1) y11 = y10 + y12 + y13 + y14 =- 6j ...(2) y22 = y20 + y21 + y23 + y24 =- 10j ...(3) y 33 = y 30 + y 31 + y 32 + y 34 =- 9j ...(4) y 44 = y 40 + y 41 + y 42 + y 43 =- 9j and diagonal elements y12 = y21 =- y12 = 2j _ b y13 = y 31 =- y13 = 2.5j b b y14 = y 41 =- y14 = 0j b .....(5) ` y23 = y 32 =- y23 = 2.5j b y24 = y 42 =- y24 = 4j b bb y 34 = y 34 = 4j a from equation (1) y10 = y11 - y12 - y13 - y14 =- 6j + 2j + 2.5j + 0j =- 1.5j Same as from equation (2) y20 = y22 - y21 - y23 - y24 =- 10j + 2j + 2.5j + 4j =- 1.5j from equation (3) y 30 = y 33 - y 31 - y 32 - y 34 =- 9j + 2.5j + 2.5j + 4j = 0 from equation (4) y 40 = y 44 - y 41 - y 42 - y 43 =- 8j + 0 + 4j + 4j = 0 Now we have to draw the reactance diagram as follows

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SOL 3.136

Page 248

*Given synchronous generator is connected to infinite bus through loss less double circuit line Pd = 1+30c pu sudden fault reduces the peak power transmitted to 0.5 pu after clearance of fault, peak power = 1.5 pu Critical clearing angle ( dcr ) = ?

A I D

O N

.in

no w.

co ia.

d

ww

d0 = 30c = 0.52 rad From equal area criteria

#d

dcr

0

Where

(PL11 - Pmax 11 sin d) dd =

#d

dmax

cr

(Pmax 111 sin d - Pm) dd

...(1)

dmax = p - sin- 1 b Pm l Pmax 111

dmax = p - 0.8729 = 2.41 rad By integrating equation (1) 6Pm d + Pmax 11 cos d@ddcrmax - Pmax 111 (cos dmax - cos dcr ) = 0 & Pm (dcr - ds) + Pmax 11 (cos dcr - cos d0) + Pm (dmax - dcr ) + Pmax 111 (cos dmax - cos dcr ) = 0 P (d - d0) - Pmax 11 cos d0 + Pmax 111 cos dmax & cos dcr = m max Pmax 111 - Pmax 11 1 (2.41 - 0.52) - 0.5 cos (0.52) + 1.5 cos (2.41) = 1.5 - 0.5 cos dcr = 0.35 dcr = cos- 1 0.35 = 1.21 rad SOL 3.137

*Given: L - G fault on unloaded generator Z 0 = j0.15 , Z1 = j0.25 , Z2 = j0.25 pu, Zn = j0.05 pu Vprefault = 1 pu

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Page 249

If = ? Fault Current I f = 3Ia1 =

3Vprefault 3#1 = Z1 + Z2 + Z 0 + 3Zn (j0.25 + j0.25 + j0.15) + 3 (j.05)

3 =- 3.75j 0.80j Sequence network is being drawn as follows =

SOL 3.138

A I D

O N

*Given power system has two generator n 2 o.i c Generator - 1; C1 = 0.006P G1 + 8PG1 + 350 . ia d 2 o Generator - 2; C2 = 0.009P G2 +.n 7PG2 + 400 w # PG1 # 650 MW Generator Limits are 100wMW w 50 MW # PG2 # 500 MW PG1 + PG2 = 600 MW , PG1, PG2 = ? For optimal generation We know for optimal Generation 2C1 = 2C2 2PG1 2PG2 2C1 = 0.012P + 8 G1 2PG1 2C2 = 0.018P + 7 G2 2PG2 from equation (1) 0.012PG1 + 8 = 0.018PG2 + 7 0.012PG1 - 0.018PG2 =- 1 PG1 + PG2 = 600 From equation (2)

...(1)

...(2) ...(3)

0.012PG1 - 0.018 (600 - PG1) =- 1 & 0.03PG1 = 9.8 & PG1 = 326.67 MW PG2 = 600 - PG1 = 600 - 326.67 = 273.33 MW ***********

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3

CONTROL SYSTEMS

YEAR 2013 MCQ 3.1

ONE MARK

The Bode plot of a transfer function G ^s h is shown in the figure below.

A I D

O N

in 1 rad/s and 10 rad/s respectiveThe gain _20 log G ^s h i is 32 dB and - 8 dBo.at c . ^s h is ly. The phase is negative for all w. Then iaG d .no (B) 392.8 (A) 39.8 w s s w w (C) 32 (D) 322 s s

MCQ 3.2

Assuming zero initial condition, the response y ^ t h of the system given below to a unit step input u ^ t h is

(A) u ^ t h 2 (C) t u ^ t h 2 YEAR 2013 MCQ 3.3

(B) tu ^ t h (D) e-t u ^ t h TWO MARKS

Y ^s h The signal flow graph for a system is given below. The transfer function U ^s h for this system is

GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia

s+1 5s2 + 6s + 2 (C) 2 s + 1 s + 4s + 2

s+1 s 2 + 6s + 2 (D) 2 1 5s + 6s + 2

(A)

MCQ 3.4

Page 252

(B)

w ^s h The open-loop transfer function of a dc motor is given as = 10 . Va ^s h 1 + 10s When connected in feedback as shown below, the approximate value of Ka that will reduce the time constant of the closed loop system by one hundred times as compared to that of the open-loop system is

(A) 1 (C) 10

O N

A I D (B) 5 (D) 100

.in

co ia.

Common Data Questions: 5 & 6nod

w.

The state variable formulation ww of a system is given as o 2 0 x 1 x1 x1 1 > H=> H> H + > H u , x1 ^0 h = 0 , x2 ^0 h = 0 and y = 61 0@> H 0 - 1 x2 1 x2 xo2 MCQ 3.5

The response y ^ t h to the unit step input is (B) 1 - 1 e-2t - 1 e-t (A) 1 - 1 e-2t 2 2 2 2 (C) e-2t - e-t

MCQ 3.6

The system is (A) controllable but not observable (B) not controllable but observable (C) both controllable and observable (D) both not controllable and not observable YEAR 2012

MCQ 3.7

(D) 1 - e-t

TWO MARKS

The state variable description of an LTI system is given by Jxo1N J 0 a1 0NJx1N J0N K O K OK O K O Kxo2O = K 0 0 a2OKx2O + K0O u Kxo O Ka 0 0OKx 3O K 1O 3 3 L P L PL P L P Jx1N K O y = _1 0 0iKx2O Kx 3O GATE MCQ Electrical L P Engineering (Vol-1, 2 & 3) by RK Kanodia & Ashish Murolia

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Page 253

where y is the output and u is the input. The system is controllable for (A) a1 ! 0, a2 = 0, a 3 ! 0 (B) a1 = 0, a2 ! 0, a 3 ! 0 (C) a1 = 0, a 3 ! 0, a 3 = 0 (D) a1 ! 0, a2 ! 0, a 3 = 0 MCQ 3.8

The feedback system shown below oscillates at 2 rad/s when

(A) K = 2 and a = 0.75 (C) K = 4 and a = 0.5

(B) K = 3 and a = 0.75 (D) K = 2 and a = 0.5

Statement for Linked Answer Questions 9 and 10 :

MCQ 3.9

MCQ 3.10

The transfer function of a compensator is given as Gc (s) = s + a s+b Gc (s) is a lead compensator if (A) a = 1, b = 2 (B) a = 3, b = 2 (C) a =- 3, b =- 1 (D) a = 3, b = 1

(C)

6 rad/s

O N

.no

w

YEAR 2011 MCQ 3.11

ww

co .(D) a 1/ i d

3 rad/s ONE MARK

The frequency response of a linear system G (jw) is provided in the tubular form below G (jw)

1.3

+G (jw) - 130c (A) 6 dB and 30c (C) - 6 dB and 30c MCQ 3.12

A I D

The phase of the above lead compensator is maximum at (A) 2 rad/s (B).in 3 rad/s

1.2

1.0

0.8

0.5

0.3

- 140c

- 150c

- 160c

- 180c

- 200c

(B) 6 dB and - 30c (D) - 6 dB and - 30c

The steady state error of a unity feedback linear system for a unit step input is 0.1. The steady state error of the same system, for a pulse input r (t) having a magnitude of 10 and a duration of one second, as shown in the figure is

(A) 0 (C) 1

(B) 0.1 (D) 10

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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia MCQ 3.13

Page 254

An open loop system represented by the transfer function (s - 1) is G (s) = (s + 2) (s + 3) (A) Stable and of the minimum phase type (B) Stable and of the non–minimum phase type (C) Unstable and of the minimum phase type (D) Unstable and of non–minimum phase type YEAR 2011

TWO MARKS

MCQ 3.14

The open loop transfer function G (s) of a unity feedback control system is given as K bs + 2 l 3 G (s) = 2 s (s + 2) From the root locus, at can be inferred that when K tends to positive infinity, (A) Three roots with nearly equal real parts exist on the left half of the s -plane (B) One real root is found on the right half of the s -plane (C) The root loci cross the jw axis for a finite value of K; K ! 0 (D) Three real roots are found on the right half of the s -plane

MCQ 3.15

A two loop position control system is shown below

A I D

O N

no w.

.in

co ia.

d

ww

The gain K of the Tacho-generator influences mainly the (A) Peak overshoot (B) Natural frequency of oscillation (C) Phase shift of the closed loop transfer function at very low frequencies (w " 0) (D) Phase shift of the closed loop transfer function at very high frequencies (w " 3) YEAR 2010 MCQ 3.16

TWO MARKS

1 plotted in the s (s + 1) (s + 2) complex G (jw) plane (for 0 < w < 3) is The frequency response of G (s) =

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MCQ 3.17

o = AX + Bu with A = >- 1 2H, B = >0H is The system X 0 2 1

A I D

(A) Stable and controllable (C) Unstable but controllable MCQ 3.18

(B) Stable but uncontrollable (D) Unstable and uncontrollable

The characteristic equation of a closed-loop system is s (s + 1) (s + 3) k (s + 2) = 0, k > 0 .Which of the following statements is true ? n (A) Its root are always real o.i c . diniathe range - 1 < Re [s] < 0 (B) It cannot have a breakaway point o .n along the asymptotes Re [s] =- 1 w (C) Two of its roots tend to infinity ww (D) It may have complex roots in the right half plane.

O N

YEAR 2009 MCQ 3.19

MCQ 3.20

Page 255

ONE MARK

The measurement system shown in the figure uses three sub-systems in cascade whose gains are specified as G1, G2, 1/G3 . The relative small errors associated with each respective subsystem G1, G2 and G3 are e1, e2 and e3 . The error associated with the output is :

(A) e1 + e2 + 1 e3

(B) e1 e2 e3

(C) e1 + e2 - e3

(D) e1 + e2 + e3

The polar plot of an open loop stable system is shown below. The closed loop system is

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Page 256

(A) always stable (B) marginally stable (C) un-stable with one pole on the RH s -plane (D) un-stable with two poles on the RH s -plane MCQ 3.21

MCQ 3.22

The first two rows of Routh’s tabulation of a third order equation are as follows. s3 2 2 s2 4 4 This means there are (A) Two roots at s = ! j and one root in right half s -plane (B) Two roots at s = ! j2 and one root in left half s -plane (C) Two roots at s = ! j2 and one root in right half s -plane (D) Two roots at s = ! j and one root in left half s -plane

O N

A I D

The asymptotic approximation of the log-magnitude frequency plot of a inis shown.v/s . o system containing only real poles and zeros Its transfer function is .c

w

ww

(A)

10 (s + 5) s (s + 2) (s + 25)

(C)

100 (s + 5) s (s + 2) (s + 25)

ia

d .no

1000 (s + 5) s (s + 2) (s + 25) 80 (s + 5) (D) 2 s (s + 2) (s + 25)

(B)

2

YEAR 2009 MCQ 3.23

TWO MARKS

The unit-step response of a unity feed back system with open loop transfer function G (s) = K/ ((s + 1) (s + 2)) is shown in the figure. The value of K is

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(A) 0.5 (C) 4 MCQ 3.24

Page 257

(B) 2 (D) 6

The open loop transfer function of a unity feed back system is given by G (s) = (e - 0.1s) /s . The gain margin of the is system is (A) 11.95 dB (B) 17.67 dB (C) 21.33 dB (D) 23.9 dB

Common Data for Question 25 and 26 : A system is described by the following state and output equations dx1 (t) =- 3x1 (t) + x2 (t) + 2u (t) dt dx2 (t) =- 2x2 (t) + u (t) dt y (t) = x1 (t) when u (t) is the input and y (t) is the output.in MCQ 3.25

MCQ 3.26

A I D

O N

.co

The system transfer function is dia o .n (A) 2 s + 2 (B) 2 s + 3 w w s + 5s - 6 s + 5s + 6 w (C) 2 2s + 5 (D) 2 2s - 5 s + 5s + 6 s + 5s - 6 The state-transition matrix of the above system is e - 3t 0 e - 3t e - 2t - e - 3t (B) = (A) = - 2t G - 3t - 2t G e +e e 0 e - 2t e - 3t e - 2t + e - 3t (C) = G 0 e - 2t

e 3t e - 2t - e - 3t (D) = G 0 e - 2t

YEAR 2008 MCQ 3.27

A function y (t) satisfies the following differential equation : dy (t) + y (t) = d (t) dt where d (t) is the delta function. Assuming zero initial condition, and denoting the unit step function by u (t), y (t) can be of the form (B) e - t (A) et (C) et u (t) YEAR 2008

MCQ 3.28

ONE MARK

(D) e - t u (t) TWO MARK

The transfer function of a linear time invariant system is given as

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Page 258

1 s + 3s + 2 The steady state value of the output of the system for a unit impulse input applied at time instant t = 1 will be (A) 0 (B) 0.5 (C) 1 (D) 2 G (s) =

2

MCQ 3.29

The transfer functions of two compensators are given below : 10 (s + 1) C1 = , C2 = s + 10 (s + 10) 10 (s + 1) Which one of the following statements is correct ? (A) C1 is lead compensator and C2 is a lag compensator (B) C1 is a lag compensator and C2 is a lead compensator (C) Both C1 and C2 are lead compensator (D) Both C1 and C2 are lag compensator

MCQ 3.30

The asymptotic Bode magnitude plot of a minimum phase transfer function is shown in the figure :

A I D

O N

no w.

.in

co ia.

d

ww

This transfer function has (A) Three poles and one zero (C) Two poles and two zero MCQ 3.31

(B) Two poles and one zero (D) One pole and two zeros

Figure shows a feedback system where K > 0

The range of K for which the system is stable will be given by (A) 0 < K < 30 (B) 0 < K < 39 (C) 0 < K < 390 (D) K > 390 MCQ 3.32

The transfer function of a system is given as 100 s2 + 20s + 100 The system is (A) An over damped system (B) An under damped system (C) A critically damped system (D) An unstable system

Statement for Linked Answer Question 27 and 28.

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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia

MCQ 3.33

Page 259

The state space equation of a system is described by Xo = AX + Bu,Y = CX where X is state vector, u is input, Y is output and 0 1 0 A == , B = = G, C = [1 0] G 0 -2 1 The transfer function G(s) of this system will be s (A) (B) s + 1 (s + 2) s (s - 2) s 1 (D) (s - 2) s (s + 2) A unity feedback is provided to the above system G (s) to make it a closed loop system as shown in figure. (C)

MCQ 3.34

For a unit step input r (t), the steady state error in the input will be (A) 0 (B) 1 (C) 2 (D) 3

MCQ 3.35

IA

D O

YEAR 2007

The system shown in the figure is

N

no w.

ONE MARK

.in

co ia.

d

ww

(A) Stable (B) Unstable (C) Conditionally stable (D) Stable for input u1 , but unstable for input u2 YEAR 2007 MCQ 3.36

If x = Re [G (jw)], and y = Im [G (jw)] then for w " 0+ , the Nyquist plot for G (s) = 1/s (s + 1) (s + 2) is (A) x = 0 (B) x =- 3/4 (C) x = y - 1/6

MCQ 3.37

TWO MARKS

(D) x = y/ 3

The system 900/s (s + 1) (s + 9) is to be such that its gain-crossover frequency becomes same as its uncompensated phase crossover frequency and provides a 45c phase margin. To achieve this, one may use (A) a lag compensator that provides an attenuation of 20 dB and a phase lag of 45c at the frequency of 3 3 rad/s (B) a lead compensator that provides an amplification of 20 dB and a phase

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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 260

lead of 45c at the frequency of 3 rad/s (C) a lag-lead compensator that provides an amplification of 20 dB and a phase lag of 45c at the frequency of 3 rad/s (D) a lag-lead compensator that provides an attenuation of 20 dB and phase lead of 45c at the frequency of 3 rad/s MCQ 3.38

If the loop gain K of a negative feed back system having a loop transfer function K (s + 3) / (s + 8) 2 is to be adjusted to induce a sustained oscillation then (A) The frequency of this oscillation must be 4 3 rad/s (B) The frequency of this oscillation must be 4 rad/s (C) The frequency of this oscillation must be 4 or 4 3 rad/s (D) Such a K does not exist

MCQ 3.39

The system shown in figure below

O N

A I D

can be reduced to the form

no w.

.in

co ia.

d

ww

with (A) X = c0 s + c1, Y = 1/ (s2 + a0 s + a1), Z = b0 s + b1 (B) X = 1, Y = (c0 s + c1) / (s2 + a0 s + a1), Z = b0 s + b1 (C) X = c1 s + c0, Y = (b1 s + b0) / (s2 + a1 s + a0), Z = 1 (D) X = c1 s + c0, Y = 1/ (s2 + a1 s + a), Z = b1 s + b0 MCQ 3.40

Consider the feedback system shown below which is subjected to a unit step input. The system is stable and has following parameters Kp = 4, Ki = 10, w = 500 and x = 0.7 .The steady state value of Z is

(A) 1 (C) 0.1

(B) 0.25 (D) 0

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Page 261

Data for Q.41 and Q.42 are given below. Solve the problems and choose the correct answers. R-L-C circuit shown in figure

MCQ 3.41

For a step-input ei , the overshoot in the output e0 will be (A) 0, since the system is not under damped (B) 5 % (C) 16 % (D) 48 %

MCQ 3.42

If the above step response is to be observed on a non-storage CRO, then it would be best have the ei as a (A) Step function (B) Square wave of 50 Hz (C) Square wave of 300 Hz (D) Square wave of 2.0 KHz YEAR 2006

MCQ 3.43

O N

A I D .in

.co a i For a system with the transfer function d no 3 (s - 2) . w , H (s) = 2 4s - 2s + 1 ww

ONE MARK

the matrix A in the state space form Xo = AX + Bu is equal to V V R R S1 0 0 W S0 1 0 W (A) S 0 1 0 W (B) S 0 0 1 W SS- 1 2 - 4 WW SS- 1 2 - 4 WW XV TR V X RT 0 1 0 1 0 0 W W S S (C) S3 - 2 1 W (D) S 0 0 1 W SS1 - 2 4 WW SS- 1 2 - 4 WW X X T T YEAR 2006 MCQ 3.44

TWO MARKS

Consider the following Nyquist plots of loop transfer functions over w = 0 to w = 3 . Which of these plots represent a stable closed loop system ?

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(A) (1) only (C) all, except (3) MCQ 3.45

Page 262

(B) all, except (1) (D) (1) and (2) only

The Bode magnitude plot H (jw) =

10 4 (1 + jw) is (10 + jw) (100 + jw) 2

A I D

O N

no w.

.in

co ia.

d

ww

MCQ 3.46

A closed-loop system has the characteristic function (s2 - 4) (s + 1) + K (s - 1) = 0 . Its root locus plot against K is

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YEAR 2005 MCQ 3.47

MCQ 3.48

.in

co ia.

d

ONE MARK

A system with zero initial conditions .nohas the closed loop transfer function. w s2 + 4 ww T (s) = (s + 1) (s + 4) The system output is zero at the frequency (A) 0.5 rad/sec (B) 1 rad/sec (C) 2 rad/sec (D) 4 rad/sec Figure shows the root locus plot (location of poles not given) of a third order system whose open loop transfer function is

(A) K3 s

MCQ 3.49

O N

A I D

Page 263

K s2 (s + 1) K K (C) (D) s (s2 + 1) s (s2 - 1) The gain margin of a unity feed back control system with the open loop transfer (B)

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function G (s) =

(s + 1) is s2

1 2 (D) 3

(A) 0 (C)

(B) 2

YEAR 2005 MCQ 3.50

Page 264

TWO MARKS

A unity feedback system, having an open loop gain K (1 - s) , G (s) H (s) = (1 + s) becomes stable when (A) K > 1 (C) K < 1

MCQ 3.51

When subject to a unit step input, the closed loop control system shown in the figure will have a steady state error of

(A) - 1.0 (C) 0 MCQ 3.52

(B) K > 1 (D) K < - 1

A I D

O N

no w.

.in- 0.5 o(B) c . ia (D) 0.5

d

In the G (s) H (s)-plane, the Nyquist plot of the loop transfer function w w pe G (s) H (s) = s passes through the negative real axis at the point (A) (- 0.25, j0) (B) (- 0.5, j0) (C) 0 (D) 0.5 -0.25s

MCQ 3.53

If the compensated system shown in the figure has a phase margin of 60c at the crossover frequency of 1 rad/sec, then value of the gain K is

(A) 0.366 (C) 1.366

(B) 0.732 (D) 2.738

Data for Q.54 and Q.55 are given below. Solve the problem and choose the correct answer. 0 1 1 X (t) + = Gu (t) with the initial G 0 -3 0 T condition X (0) = [- 1, 3] and the unit step input u (t) has

A state variable system Xo (t) = = MCQ 3.54

The state transition matrix

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MCQ 3.55

1 (A) = 0

1 3

(1 - e- 3t) G e- 3t

1 (B) > 0

1 3

1 (C) > 0

1 3

(e3 - t - e- 3t) H e- 3t

1 (D) > 0

(1 - e- t) H e- t

Page 265

(e- t - e- 3t) H e- t

The state transition equation t - e-t (A) X (t) = = - t G e

1 - e-t (B) X (t) = = - 3t G 3e

t - e 3t (C) X (t) = = - 3t G 3e

t - e - 3t (D) X (t) = = - t G e

YEAR 2004

ONE MARK

MCQ 3.56

The Nyquist plot of loop transfer function G (s) H (s) of a closed loop control system passes through the point (- 1, j 0) in the G (s) H (s)plane. The phase margin of the system is (A) 0c (B) 45c (C) 90c (D) 180c

MCQ 3.57

Consider the function, 5 F (s) = 2 s (s + 3s + 2) where F (s) is the Laplace transform of the of the function f (t). The initial value of f (t) is equal to (A) 5 (B).i25n (C)

MCQ 3.58

5 3

O N

A I D no

. ww

.co a i d (D) 0

For a tachometer, if q (t) is the w rotor displacement in radians, e (t) is the output voltage and Kt is the tachometer constant in V/rad/sec, then the transfer funcE (s) tion, will be Q (s) (A) Kt s2 (C) Kt s

(B) Kt s (D) Kt

YEAR 2004

TWO MARKS

MCQ 3.59

For the equation, s3 - 4s2 + s + 6 = 0 the number of roots in the left half of s -plane will be (A) Zero (B) One (C) Two (D) Three

MCQ 3.60

For the block diagram shown, the transfer function

2 (A) s +2 1 s

C (s) is equal to R (s)

2 (B) s + s2 + 1 s

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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia 2 (C) s + s + 1 s

MCQ 3.61

MCQ 3.62

1 s2 + s + 1 o = AX where The state variable description of a linear autonomous system is, X X is the two dimensional state vector and A is the system matrix given by 0 2 . The roots of the characteristic equation are A == 2 0G (A) - 2 and + 2 (B) - j2 and + j2 (C) - 2 and - 2 (D) + 2 and + 2 (D)

The block diagram of a closed loop control system is given by figure. The values of K and P such that the system has a damping ratio of 0.7 and an undamped natural frequency wn of 5 rad/sec, are respectively equal to

(A) 20 and 0.3 (C) 25 and 0.3 MCQ 3.63

MCQ 3.64

(B) 20 and 0.2 (D) 25 and 0.2

A I D

The unit impulse response of a second order under-damped system starting from rest is given by c (t) = 12.5e - 6t sin 8t, t $ 0 . The steady-state value of the unit step response of the system is equal to (A) 0 (B) 0.25 .in (C) 0.5 .co(D) 1.0

O N

dia o .n input x (t) = sin t . In the steady-state, the In the system shown in figure,wthe response y (t) will be ww

(A)

1 sin (t - 45c) 2

(C) sin (t - 45c) MCQ 3.65

(B)

1 sin (t + 45c) 2

(D) sin (t + 45c)

The open loop transfer function of a unity feedback control system is given as 1. G (s) = as + s2 The value of ‘a ’ to give a phase margin of 45c is equal to (A) 0.141 (B) 0.441 (C) 0.841 (D) 1.141 YEAR 2003

MCQ 3.66

Page 266

ONE MARK

A control system is defined by the following mathematical relationship d2 x + 6 dx + 5x = 12 (1 - e - 2t) dt dt2 The response of the system as t " 3 is (A) x = 6 (B) x = 2 (C) x = 2.4 (D) x =- 2

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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia MCQ 3.67

MCQ 3.68

A lead compensator used for a closed loop controller has the following transfer function K (1 + as ) (1 + bs ) For such a lead compensator (A) a < b (B) b < a (C) a > Kb (D) a < Kb 2 A second order system starts with an initial condition of = G without any exter3 e - 2t 0 nal input. The state transition matrix for the system is given by = G. The 0 e-t state of the system at the end of 1 second is given by 0.271 0.135 (A) = (B) = G 1.100 0.368G 0.271 (C) = 0.736G

0.135 (D) = 1.100 G

YEAR 2003 MCQ 3.69

Page 267

A I D

TWO MARKS

A control system with certain excitation is governed by the following mathematical equation d2 x + 1 dx + 1 x = 10 + 5e- 4t + 2e- 5t 2 dt 18 dt2 The natural time constant of the response of ithe n system are . o (A) 2 sec and 5 sec (B) c 3 sec and 6 sec ia. d (C) 4 sec and 5 sec .no (D) 1/3 sec and 1/6 sec

MCQ 3.70

O N

w w w The block diagram shown in figure gives a unity feedback closed loop control system. The steady state error in the response of the above system to unit step input is

(A) 25% (C) 6% MCQ 3.71

The roots of the closed loop characteristic equation of the system shown above (Q-5.55)

(A) - 1 and - 15 (C) - 4 and - 15 MCQ 3.72

(B) 0.75 % (D) 33%

(B) 6 and 10 (D)- 6 and - 10

The following equation defines a separately excited dc motor in the form of a differential equation

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Page 268

d2 w + B dw + K2 w = K V dt J dt LJ LJ a The above equation may be organized in the state-space form as follows R 2 V Sd w W dw S dt2 W = P dt + QV > H a S dw W w S dt W Where Tthe PX matrix is given by K - B - LJ - K - BJ (B) = LJ (A) = J G G 1 0 0 1 2

2

0 1 (C) =- K - B G LJ J

1 0 (D) =- B - K G J LJ

2

2

MCQ 3.73

The loop gain GH of a closed loop system is given by the following expression K s (s + 2) (s + 4) The value of K for which the system just becomes unstable is (A) K = 6 (B) K = 8 (C) K = 48 (D) K = 96

MCQ 3.74

The asymptotic Bode plot of the transfer function K/ [1 + (s/a)] is given in figure. The error in phase angle and dB gain at a frequency of w = 0.5a are respectively

A I D

O N

no w.

.in

co ia.

d

ww

(A) 4.9c, 0.97 dB (C) 4.9c, 3 dB MCQ 3.75

(B) 5.7c, 3 dB (D) 5.7c, 0.97 dB

The block diagram of a control system is shown in figure. The transfer function G (s) = Y (s) /U (s) of the system is

(A)

1 s 18`1 + j`1 + s j 12 3

(B)

1 s 27`1 + j`1 + s j 6 9

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(C)

1 s 27`1 + j`1 + s j 12 9

(D)

1 s 27`1 + j`1 + s j 9 3

YEAR 2002 MCQ 3.76

Page 269

ONE MARK

The state transition matrix for the system Xo = AX with initial state X (0) is (A) (sI - A) - 1 (B) eAt X (0) (C) Laplace inverse of [(sI - A) - 1] (D) Laplace inverse of [(sI - A) - 1 X (0)] YEAR 2002

MCQ 3.77

MCQ 3.78

TWO MARKS

2 3 1 X + = Gu , which of the following statements is true ? G 0 5 0 (A) The system is controllable but unstable (B) The system is uncontrollable and unstable (C) The system is controllable and stable (D) The system is uncontrollable and stable For the system Xo = =

A I D

A unity feedback system has an open loop transfer function, G (s) = K2 . The s root locus plot is

O N

no w.

.in

co ia.

d

ww

MCQ 3.79

MCQ 3.80

The transfer function of the system described by d2 y dy + = du + 2u 2 dt dt dt with u as input and y as output is (s + 2) (s + 1) (A) 2 (B) 2 (s + s) (s + s) (C) 2 2 (D) 22s (s + s) (s + s) For the system 2 0 1 Xo = = X + = Gu ; Y = 84 0B X, 0 4G 1 with u as unit impulse and with zero initial state, the output y , becomes (A) 2e2t (B) 4e2t (C) 2e 4t

(D) 4e 4t

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Page 270

MCQ 3.81

The eigen values of the system represented by R0 1 0 0 V S W 0 0 1 0W S Xo = S X are 0 0 0 1W W (A) 0, 0, 0, 0SS (B) 1, 1, 1, 1 0 0 0 1W T 1 X (C) 0, 0, 0, (D) 1, 0, 0, 0

MCQ 3.82

*A single input single output system with y as output and u as input, is described by d2 y dy du 2 + 2 dt + 10y = 5 dt - 3u dt for an input u (t) with zero initial conditions the above system produces the same output as with no input and with initial conditions dy (0-) =- 4 , y (0-) = 1 dt input u (t) is (B) 1 d (t) - 7 e- 3t u (t) (A) 1 d (t) - 7 e(3/5)t u (t) 5 25 5 25 (C) - 7 e- (3/5)t u (t) (D) None of these 25

MCQ 3.83

A I D

*A system is described by the following differential equation d2 y dy + - 2y = u (t) e- t dt dt2 dy the state variables are given as x1 = y and .xin - y l et , the state 2 =b dt .co variable representation of the system iis a od 1 1 x1 1 1 xo1 xo1 1 e- t x 1 n . (B) > o H = > H> H + > H u (t) (A) > o H = > - tH> H + > H u (t) w 0 1 x2 0 0 x2 x2 0 e x2 ww

O N

1 xo1 1 e- t x 1 (C) > o H = > >x H + >0H u (t) H x2 0 -1 2

(D) none of these

Common Data Question Q.84-86*.

MCQ 3.84

MCQ 3.85

MCQ 3.86

The open loop transfer function of a unity feedback system is given by 2 (s + a) G (s) = s (s + 2) (s + 10) Angles of asymptotes are (A) 60c, 120c, 300c (B) 60c, 180c, 300c (C) 90c, 270c, 360c (D) 90c, 180c, 270c Intercepts of asymptotes at the real axis is (A) - 6

(B) - 10 3

(C) - 4

(D) - 8

Break away points are (A) - 1.056 , - 3.471 (C) - 1.056, - 6.9433

(B) - 2.112, - 6.9433 (D) 1.056, - 6.9433

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Page 271

YEAR 2001 MCQ 3.87

ONE MARK

The polar plot of a type-1, 3-pole, open-loop system is shown in Figure The closed-loop system is

(A) always stable (B) marginally stable (C) unstable with one pole on the right half s -plane (D) unstable with two poles on the right half s -plane. MCQ 3.88

-3 1 Given the homogeneous state-space equation xo = = x the steady state 0 - 2G T value of xss = lim x (t), given the initial state value of x (0) = 810 - 10B is

A I D

t"3

-3 (B) xss = = G -2

0 (A) xss = = G 0

- 10 (C) xss = = 10 G YEAR 2001 MCQ 3.89

O N

3 (D) xss = = G .in 3

no

. ww

co

. dia

TWO MARKS

The asymptotic approximation w of the log-magnitude versus frequency plot of a minimum phase system with real poles and one zero is shown in Figure. Its transfer functions is

(A)

20 (s + 5) s (s + 2) (s + 25)

(C)

20 (s + 5) s (s + 2) (s + 25) 2

10 (s + 5) (s + 2) 2 (s + 25) 50 (s + 5) (D) 2 s (s + 2) (s + 25) (B)

Common Data Question Q.90-93*. A unity feedback system has an open-loop transfer function of G (s) = 10000 2 s (s + 10)

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Page 272

MCQ 3.90

Determine the magnitude of G (jw) in dB at an angular frequency of w = 20 rad/sec. (A) 1 dB (B) 0 dB (C) - 2 dB (D) 10 dB

MCQ 3.91

The phase margin in degrees is (A) 90c (C) - 36.86c

(B) 36.86c (D) - 90c

The gain margin in dB is (A) 13.97 dB (C) - 13.97 dB

(B) 6.02 dB (D) None of these

The system is (A) Stable (C) Marginally stable

(B) Un-stable (D) can not determined

MCQ 3.92

MCQ 3.93

MCQ 3.94

*For the given characteristic equation s3 + s2 + Ks + K = 0 The root locus of the system as K varies from zero to infinity is

A I D

O N

.in

no w.

co ia.

d

ww

************

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Page 273

SOLUTION

SOL 3.1

Option (B) is correct. From the given plot, we obtain the slope as Slope =

20 log G2 - 20 log G1 log w2 - log w1

From the figure 20 log G2 20 log G1 and w1 w2 So, the slope is

=- 8 dB = 32 dB = 1 rad/s = 10 rad/s

Slope = - 8 - 32 log 10 - log 1 =- 40 dB/decade Therefore, the transfer function can be given as G ^s h = k2 S at w = 1 G ^ jwh = k 2 = k w n o.i In decibel, c . ia 20 log G ^ jwh = 20 log k = 32 o d n w..8 or, k = 10 =w39 Hence, the Transfer functionwis G ^s h = k2 = 392.8 s s

A I D

O N 32

SOL 3.2

20

Option (B) is correct. The Laplace transform of unit step fun n is U ^s h = 1 s So, the O/P of the system is given as Y ^s h = b 1 lb 1 l s s = 12 s For zero initial condition, we check dy ^ t h u^t h = dt U ^s h = SY ^s h - y ^0 h & U ^s h = s c 12 m - y ^0 h & s 1 or, U ^s h = s Hence, the O/P is correct which is Y ^s h = 12 s

^y ^0 h = 0h

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Page 274

its inverse Laplace transform is given by y ^ t h = tu ^ t h SOL 3.3

Option (A) is correct. For the given SFG, we have two forward paths Pk1 = ^1 h^s-1h^s-1h^1 h = s-2 Pk2 = ^1 h^s-1h^1 h^1 h = s-1 since, all the loops are touching to both the paths Pk1 and Pk2 so, Dk 1 = Dk 2 = 1 Now, we have D = 1 - (sum of individual loops) + (sum of product of nontouching loops) Here, the loops are L1 = ^- 4h^1 h =- 4 L2 = ^- 4h^s-1h = 4s-1 L 3 = ^- 2h^s-1h^s-1h =- 2s-2 L 4 = ^- 2h^s-1h^1 h =- 2s-1 As all the loop L1, L2, L 3 and L 4 are touching to each other so,

A I D

D = 1 - ^L1 + L2 + L 3 + L 4h

= 1 - ^- 4 - 4s-1 - 2s-2 - 2s-1h = 5 + 6s-1 + 2s-2 n From Mason’s gain formulae o.i c . ia Y ^s h d S P k Dk o = n D U ^s h w.

O N w

SOL 3.4

s-2 + s-w1 = 5 + 6s-1 + 2s-2 = 2s+1 5s + 6s + 2 Option (C) is correct. Given, open loop transfer function G ^s h = 10Ka = Ka 1 1 + 10s s + 10 By taking inverse Laplace transform, we have g ^ t h = e- t 1 10

Comparing with standard form of transfer function, Ae-t/t , we get the open loop time constant, tol = 10 Now, we obtain the closed loop transfer function for the given system as G ^s h 10Ka H ^s h = = 1 10 s + 10Ka + 1 + G ^s h Ka = s + ^Ka + 101 h By taking inverse laplace transform, we get h ^ t h = ka .e-^k + ht So, the time constant of closed loop system is obtained as tcl = 1 1 ka + 10 a

1 10

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Page 275

(approximately) tcl = 1 ka Now, given that ka reduces open loop time constant by a factor of 100. i.e., tcl = tol 100 1 = 10 or, 100 ka or, ka = 10 or,

SOL 3.5

Option (A) is correct. Given, the state variable formulation, - 2 0 x1 1 xo1 > o H = > 0 - 1H>x H + >1H u x2 2 x1 and y = 61 0@> H x2

....(1) ....(2)

From Eq. (1) we get xo1 = 2x1 + u Taking Laplace transform So,

sX1 - x1 ^0 h =- 2X1 + 1 (Here, X1 denotes Laplace transform of x1 ) s

^s + 2h X1 = s1

D O

1 s ^s + 2h Now, from Eq. (2) we have n o.i y = x1 c . Taking Laplace transform both the sides, dia o .n Y = XL w 1 ww or, Y = s ^s + 2h or, Y = 1 ;1 - 1 E 2 s s+2 Taking inverse Laplace transform y = 1 8u ^ t h - e-2t u ^ t hB 2 1 = - 1 e-2t 2 2 Option (A) is correct. From the given state variable system, we have -2 0 A => 0 1H or,

X1 =

N

SOL 3.6

IA

^x1 ^0 h = 0h

....(3)

(from eq. (3))

^for t > 0h

1 B = > H; C = 61 0@ 1 Now, we obtain the controllability matrix CM = 6B : AB@ and

1 -2 => 2 1H and the observability matrix is obtained as C OM = > H CA

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1 0 H => -2 0 So, we get Rank of the controllability matrix " Rank ^CM h = 2 Rank of the observability matrix " Rank ^OM h = 1 Since, the order of state variable is 2 ^x1 and x2h. Therefore, we have Rank ^CM h = order of state variables but, Rank (OM ) < order of state variables Thus, system is controllable but not observable SOL 3.7

Option (D) is correct. General form of state equations are given as xo = Ax + Bu yo = Cx + Du For the given problem R 0 a 0V R0V 1 W S S W A = S 0 0 a2W, B = S0W SSa SS1WW 0 0WW 3 XVR V R VT X TR 0 a 0 1 WS0W S 0W S AB = S 0 0 a2WS0W = Sa2W SSa 0 0WWSS1WW SS 0WW 3 XT X VTR XV R TR V 0 0 a1 a2WS0W Sa1 a2n S i .W A2 B = Sa2 a 3 0 0WS0W =.Sco 0W ia SS 0 a a 0WWSSo1dWW SS 0WW 3 1 .nXT X T X T w For controllability it is necessary that following matrix has a tank of n = 3 . ww R0 0 a a V 1 2W S 2 U = 6B : AB : A B@ = S0 a2 0W SS1 0 0WW So, a2 ! 0 X T a1 a 2 ! 0 & a1 ! 0 a 3 may be zero or not.

A I D

O N

SOL 3.8

Option (A) is correct. K (s + 1) Y (s) = 3 [R (s) - Y (s)] s + as2 + 2s + 1 K (s + 1) K (s + 1) R (s) = 3 E 2 s + as2 + 2s + 1 s + as + 2s + 1 Y (s) [s3 + as2 + s (2 + k) + (1 + k)] = K (s + 1) R (s) Transfer Function, K (s + 1) Y (s) H (s) = = 3 2 R (s) s + as + s (2 + k) + (1 + k) Routh Table : Y (s) ;1 +

3

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SOL 3.9

Page 277

For oscillation, a (2 + K) - (1 + K) =0 a a = K+1 K+2 Auxiliary equation A (s) = as2 + (k + 1) = 0 s2 =- k + 1 a k + 1 (k + 2) =- (k + 2) 2 s = (k + 1) s = j k+2 jw = j k + 2 (Oscillation frequency) w = k+2 = 2 k =2 and a = 2 + 1 = 3 = 0.75 2+2 4 Option (A) is correct. jw + a GC (s) = s + a = s+b jw + b Phase lead angle, f = tan-1 a w k - tan-1 a w k a b Jw - wN -1 K a bO = tan 2 KK OO w 1+ ab L P n oa.)i c -1 w (b .= tan c ia 2 m d +w oab n . For phase-lead compensation f > ww 0 w b-a > 0 b >a Note: For phase lead compensator zero is nearer to the origin as compared to pole, so option (C) can not be true.

A I D

O N

SOL 3.10

Option (A) is correct. f = tan-1 a w k - tan-1 a w k a b 1/a 1/b df = =0 2 2 dw 1 +awk 1 +awk a b 2 2 1 + w = 1+1w a ab2 b b a2 1 - 1 = w2 1 - 1 a b ab b a b l w = ab = 1 # 2 =

SOL 3.11

2 rad/ sec

Option (A) is correct. Gain margin is simply equal to the gain at phase cross over frequency ( wp ). Phase cross over frequency is the frequency at which phase angle is equal to - 180c. From the table we can see that +G (jwp) =- 180c, at which gain is 0.5. 1 GM = 20 log 10 e = 20 log b 1 l = 6 dB 0.5 G (jwp) o

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Phase Margin is equal to 180c plus the phase angle fg at the gain cross over frequency ( wg ). Gain cross over frequency is the frequency at which gain is unity. From the table it is clear that G (jwg) = 1, at which phase angle is - 150c fPM = 180c + +G (jwg) = 180 - 150 = 30c SOL 3.12

Option (A) is correct. We know that steady state error is given by sR (s) ess = lim s " 0 1 + G (s) where

R (s) " input G (s) " open loop transfer function For unit step input R (s) = 1 s sb 1 l s So ess = lim = 0.1 s " 0 1 + G (s) 1 + G (0) = 10 G (0) = 9 Given inputr (t)

A I D

R (s) = 10 :1 - 1 e-sD = 10 :1 - e D s s s So steady state error n (1 - e-s) o.i 0 c s # 10 . s ia10 (1 - e ) = 0 el = d ss = lim o 1+9 s"0 1 + G (s).n w Option (B) is correct. ww Transfer function having at least one zero or pole in RHS of s -plane is called non-minimum phase transfer function. s-1 G (s) = (s + 2) (s + 3) • In the given transfer function one zero is located at s = 1 (RHS), so this is a non-minimum phase system. • Poles - 2, - 3 , are in left side of the complex plane, So the system is stable or

SOL 3.13

SOL 3.14

= 10 [m (t) - m (t - 1)]

-s

O N

Option (A) is correct. K bs + 2 l 3 G (s) = 2 s (s + 2) Steps for plotting the root-locus (1) Root loci starts at s = 0, s = 0 and s =- 2 (2) n > m , therefore, number of branches of root locus b = 3 (3) Angle of asymptotes is given by (2q + 1) 180c , q = 0, 1 n-m (2 # 0 + 1) 180c (I) = 90c (3 - 1) (2 # 1 + 1) 180c (II) = 270c (3 - 1) (4) The two asymptotes intersect on real axis at centroid

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- 2 - b- 2 l 3 x = =- 2 n-m 3-1 3 (5) Between two open-loop poles s = 0 and s =- 2 there exist a break away point. s2 (s + 2) K =2 bs + 3 l dK = 0 ds

/ Poles - / Zeroes =

s =0 Root locus is shown in the figure

A I D

Three roots with nearly equal parts exist on the left half of s -plane. SOL 3.15

O N

Option (A) is correct. .in The system may be reduced as shown below .co

w

ww

ia

d .no

1 s (s + 1 + K ) Y (s) 1 = 2 = 1 R (s) 1 + s + s (1 + K ) + 1 s (s + 1 + K ) This is a second order system transfer function, characteristic equation is s2 + s (1 + K) + 1 = 0 Comparing with standard form s2 + 2xwn s + wn2 = 0 We get x = 1+K 2 Peak overshoot M p = e- px/

1 - x2

So the Peak overshoot is effected by K . SOL 3.16

Option (A) is correct. Given

G (s) =

1 s (s + 1) (s + 2)

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1 jw (jw + 1) (jw + 2) 1 G (jw) = 2 w w + 1 w2+ 4 +G (jw) =- 90c - tan- 1 (w) - tan- 1 (w/2) In nyquist plot For w = 0, G (jw) = 3 G (jw) =

+G (jw) =- 90c For w = 3, G (jw) = 0 +G (jw) =- 90c - 90c - 90c =- 270c Intersection at real axis 1 1 G (jw) = = jw (jw + 1) (jw + 2) jw (- w2 + j3w + 2) - 3w2 - jw (2 - w2) 1 = # - 3w2 + jw (2 - w2) - 3w2 - jw (2 - w2) - 3w2 - jw (2 - w2) = 9w4 + w2 (2 - w2) 2 2 jw (2 - w2) = 4 -23w 9w + w (2 - w2) 2 9w4 + w2 (2 - w2) 2 At real axis

A I D

Im [G (jw)] = 0 w (2 - w2) So, =0 n 9w4 + w2 (2 - w2) o.i c . 2 - w2 = 0 & w = 2 rad/sec dia o n .response At w = 2 rad/sec, magnitude is w w w 1 G (jw) at w = 2 = =1 => A - lI = > H H 0 2 0 l 0 2 - lH A - lI = (- 1 - l) (2 - l) - 2 # 0 = 0 & l1, l2 =- 1, 2 Since eigen values of the system are of opposite signs, so it is unstable Controllability : 0 -1 2 , B=> H A => H 1 0 2 2 AB = > H 2 0 2 [B: AB] = > H 1 2 Y 0 6B: AB@ = So it is controllable.

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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia SOL 3.18

Page 281

Option (C) is correct. Given characteristic equation s (s + 1) (s + 3) + K (s + 2) = 0 ; s (s2 + 4s + 3) + K (s + 2) = 0 s3 + 4s2 + (3 + K) s + 2K = 0 From Routh’s tabulation method

K>0

s3

1

3+K

s2

4

2K

s1

4 (3 + K) - 2K (1) 12 + 2K = >0 4 4

s0

2K

There is no sign change in the first column of routh table, so no root is lying in right half of s -plane. For plotting root locus, the equation can be written as K (s + 2) =0 1+ s (s + 1) (s + 3)

A I D

Open loop transfer function G (s) =

K (s + 2) s (s + 1) (s + 3)

O N

Root locus is obtained in following steps: n o.si =- 3 1. No. of poles n = 3 , at s = 0, s =- 1 and c . ia 2. No. of Zeroes m = 1, at s =- 2no d w. between s = 0 and s =- 1, between s =- 3 w 3. The root locus on real axis lies w and s =- 2 . 4. Breakaway point lies between open loop poles of the system. Here breakaway point lies in the range - 1 < Re [s] < 0 . 5. Asymptotes meet on real axis at a point C , given by C =

/ real part of poles - / real parts of zeroes

n-m (0 - 1 - 3) - (- 2) = 3-1

=- 1 As no. of poles is 3, so two root loci branches terminates at infinity along asymptotes Re (s) =- 1 SOL 3.19

SOL 3.20

Option (D) is correct. Overall gain of the system is written as G = G1 G 2 1 G3 We know that for a quantity that is product of two or more quantities total percentage error is some of the percentage error in each quantity. so error in overall gain G is 3 G = e1 + e2 + 1 e3 Option (D) is correct.

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Page 282

From Nyquist stability criteria, no. of closed loop poles in right half of s -plane is given as Z = P-N P " No. of open loop poles in right half s -plane N " No. of encirclement of (- 1, j0)

Here N =- 2 (` encirclement is in clockwise direction) P = 0 (` system is stable) So, Z = 0 - (- 2) Z = 2 , System is unstable with 2-poles on RH of s -plane. SOL 3.21

A I D

Option (D) is correct. Given Routh’s tabulation. s3

2

2

s2

4

4

s1

0

0

O N

.in

co ia.

So the auxiliary equation is given by, od

.n

4s2 + 4 = 0 ww w s2 =- 1 s =! j From table we have characteristic equation as 2s3 + 2s + 4s2 + 4 = 0 s3 + s + 2s2 + 2 = 0 s (s2 + 1) + 2 (s2 + 1) = 0 (s + 2) (s2 + 1) = 0 s =- 2 , s = ! j SOL 3.22

Option (B) is correct. Since initial slope of the bode plot is - 40 dB/decade, so no. of poles at origin is 2. Transfer function can be written in following steps: 1. Slope changes from - 40 dB/dec. to - 60 dB/dec. at w1 = 2 rad/sec., so at w1 there is a pole in the transfer function. 2. Slope changes from - 60 dB/dec to - 40 dB/dec at w2 = 5 rad/sec., so at this frequency there is a zero lying in the system function. 3. The slope changes from - 40 dB/dec to - 60 dB/dec at w3 = 25 rad/sec, so there is a pole in the system at this frequency. Transfer function T (s) =

K (s + 5) s (s + 2) (s + 25) 2

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Page 283

Constant term can be obtained as. T (jw) at w = 0.1 = 80 So,

80 = 20 log

K (5) (0.1) 2 # 50

K = 1000 therefore, the transfer function is 1000 (s + 5) T (s) = 2 s (s + 2) (s + 25) SOL 3.23

Option (D) is correct. From the figure we can see that steady state error for given system is ess = 1 - 0.75 = 0.25 Steady state error for unity feed back system is given by sR (s) ess = lim = G s " 0 1 + G (s) s ^ 1s h ; R (s) = 1 (unit step input) = lim s s"0> K H 1+ (s + 1) (s + 2) = 1K = 2 2+K 1+ 2 So, ess = 2 = 0.25 2+K

A I D

2 = 0.5 + 0.25K K = 1. 5 = 6 0.25

SOL 3.24

O N

.in

co ia.

d Option (D) is correct. no . w figure is given by, Open loop transfer function ofwthe w - 0.1s G (s) = e s - j0.1w G (jw) = e jw Phase cross over frequency can be calculated as,

+G (jwp) =- 180c 180 b- 0.1wp # p l - 90c =- 180c 0.1wp # 180c = 90c p 0.1wp = 90c # p 180c wp = 15.7 rad/sec So the gain margin (dB) 1 = 20 log e = 20 log G (jwp) o >

1 1 b 15.7 l H

= 20 log 15.7 = 23.9 dB SOL 3.25

Option (C) is correct. Given system equations dx1 (t) =- 3x1 (t) + x2 (t) + 2u (t) dt

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dx2 (t) =- 2x2 (t) + u (t) dt y (t) = x1 (t) Taking Laplace transform on both sides of equations. sX1 (s) =- 3X1 (s) + X2 (s) + 2U (s) (s + 3) X1 (s) = X2 (s) + 2U (s) Similarly sX2 (s) =- 2X2 (s) + U (s) (s + 2) X2 (s) = U (s) From equation (1) & (2) U (s) (s + 3) X1 (s) = + 2U (s) s+2 U (s) 1 + 2 (s + 2) X1 (s) = E s + 3; s + 2 (2s + 5) = U (s) (s + 2) (s + 3) From output equation, Y (s) = X1 (s) So,

SOL 3.26

Y (s) = U (s)

...(1) ...(2)

A I D

(2s + 5) (s + 2) (s + 3)

System transfer function (2s + 5) Y (s) (2s + 5) = 2 T.F = = U (s) (s + 2) (s + 3) s + 5s + 6 Option (B) is correct. .in o c . Given state equations in matrix formia can be written as, d o - 3 1 .n x1 2 xo1 + > H u (t) >o H = > 0 w H > H w x2 w - 2 x2 1 dX (t) = AX (t) + Bu (t) dt State transition matrix is given by

O N

f (t) = L- 1 6F (s)@ F (s) = (sI - A) - 1 s 0 -3 1 (sI - A) = > H - > 0 s 0 - 2H s + 3 -1 (sI - A) = > 0 s + 2H s+2 1 1 > 0 s + 3H (s + 3) (s + 2) R V 1 S 1 W (s + 3) (s + 3) (s + 2)W S -1 So F (s) = (sI - A) = S W 1 S 0 (s + 2) W T - 3t - 2t X e e - e- 3t -1 f (t) = L [F (s)] = > H 0 e- 2t (sI - A) - 1 =

SOL 3.27

Option (D) is correct. Given differential equation for the function dy (t) + y (t) = d (t) dt

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Taking Laplace on both the sides we have, sY (s) + Y (s) = 1 (s + 1) Y (s) = 1 Y (s) =

1 s+1

Taking inverse Laplace of Y (s) y (t) = e- t u (t), t > 0 SOL 3.28

Option (A) is correct. Given transfer function G (s) =

1 s + 3s + 2 2

r (t) = d (t - 1) R (s) = L [d (t - 1)] = e- s Output is given by -s Y (s) = R (s) G (s) = 2 e s + 3s + 2 Steady state value of output -s lim y (t) = lim sY (s) = lim 2 se =0 t"3 s"0 s " 0 s + 3s + 2 Input

SOL 3.29

Option (A) is correct. For C1 Phase is given by

A I D

O N

qC = tan- 1 (w) - tan- 1 a wok.in c .10 a i d woN Jw - n . -1 K 10 O w = tan w wKK 1 + w2 OO 10 P L = tan- 1 c 9w 2 m > 0 (Phase lead) 10 + w Similarly for C2 , phase is qC2 = tan- 1 a w k - tan- 1 (w) 10 J w - wN - 1 K 10 O = tan 2 KK O w 1+ O 10 P L = tan- 1 c - 9w 2 m < 0 (Phase lag) 10 + w Option (C) is correct. From the given bode plot we can analyze that: 1. Slope - 40 dB/decade"2 poles 2. Slope - 20 dB/decade (Slope changes by + 20 dB/decade)"1 Zero 3. Slope 0 dB/decade (Slope changes by + 20 dB/decade)"1 Zero 1

SOL 3.30

So there are 2 poles and 2 zeroes in the transfer function. SOL 3.31

Option (C) is correct. Characteristic equation for the system K =0 1+ s (s + 3) (s + 10)

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s (s + 3) (s + 10) + K = 0 s3 + 13s2 + 30s + K = 0 Applying Routh’s stability criteria s3

1

30

s2

13

K

s1

(13 # 30) - K 13 K

s0

For stability there should be no sign change in first column So, 390 - K > 0 & K < 390 K >0 0 < K < 90 SOL 3.32

Option (C) is correct. Given transfer function is 100 s2 + 20s + 100 Characteristic equation of the system is given by H (s)) =

or

A I D

s2 + 20s + 100 = 0 wn2 = 100 & wn = 10 rad/sec. 2xwn = 20 n x = 20 = 1 o.i c 2 # 10 . a

O N

di

(x = 1) so system is critically damped. .no SOL 3.33

w

Option (D) is correct. ww State space equation of the system is given by, o = AX + Bu X Y = CX Taking Laplace transform on both sides of the equations.

So

sX (s) (sI - A) X (s) X (s) ` Y (s) Y (s) T.F =

= AX (s) + BU (s) = BU (s) = (sI - A) - 1 BU (s) = CX (s) = C (sI - A) - 1 BU (s)

Y (s) = C (sI - A) - 1 B U (s)

s 0 0 1 s -1 => (sI - A) = > H - > H 0 s 0 -2 0 s + 2H R V 1 W S1 1 >s + 2 1H = Ss s (s + 2)W (sI - A) - 1 = S0 1 W s (s + 2) 0 s S (s + 2) W T X Transfer function

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V V R R 1 W S1 S 1 W s s (s + 2)W 0 Ss (s + 2)W = 1 0 G (s) = C [sI - A] - 1 B = 81 0BSS > H 8 B S 1 W 1 W1 S0 (s + 2) W S (s + 2) W X X T T 1 = s (s + 2) SOL 3.34

Option (A) is correct. Steady state error is given by,

Here

SOL 3.35

sR (s) ess = lim = G s " 0 1 + G (s) H (s) R (s) = L [r (t)] = 1 (Unit step input) s 1 G (s) = s (s + 2)

H (s) = 1 (Unity feed back) V R sb 1 l W S s W So, ess = lim S 1 s"0S W + 1 S s (s + 2) W X T s (s + 2) = lim = G =0 s " 0 s (s + 2) + 1 Option (D) is correct. For input u1 , the system is (u2 = 0)

A I D

O N

no w.

.in

co ia.

d

ww

System response is (s - 1) (s - 1) (s + 2) H1 (s) = = (s - 1) 1 (s + 3) 1+ (s + 2) (s - 1) Poles of the system is lying at s =- 3 (negative s -plane) so this is stable. For input u2 the system is (u1 = 0)

System response is 1 (s - 1) (s + 2) H2 (s) = = 1 s ( ) ( s 1) (s + 3) 1+ 1 (s - 1) (s + 2) One pole of the system is lying in right half of s -plane, so the system is unstable.

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Page 288

Option (B) is correct. Given function is. 1 s (s + 1) (s + 2) 1 G (jw) = jw (1 + jw) (2 + jw) G (s) =

By simplifying 1 - jw 2 - jw - jw 1 1 G (jw) = c 1 # jw - jw mc 1 + jw # 1 - jw mc 2 + jw # 2 - jw m = c-

- jw (2 - w2 - j3w) jw 1 - j w 2 - j w = w2 mc 1 + w2 mc 4 + w2 m w2 (1 + w2) (4 + w2)

jw (w2 - 2) - 3w2 + w2 (1 + w2) (4 + w2) w2 (1 + w2) (4 + w2) G (jw) = x + iy x = Re [G (jw)] w " 0 = - 3 =- 3 1#4 4 Option (D) is correct. Let response of the un-compensated system is 900 H UC (s) = s (s + 1) (s + 9) Response of compensated system. 900 HC (s) = G (s) s (s + 1) (s + 9) C =

+

SOL 3.37

A I D

O N

.in

Where GC (s) " Response of compensator .co a i d Given that gain-crossover frequency no of compensated system is same as phase . crossover frequency of un-compensated system ww w So, (wg) compensated = (wp) uncompensated - 180c = +H UC (jwp) - 180c =- 90c - tan- 1 (wp) - tan- 1 a

wp 9k

J w + wp N p 9 O 90c = tan KK 2 O K 1 - wp O 9 P L 2 w 1- p = 0 9 -1

wp = 3 rad/sec. So, (wg) compensated = 3 rad/sec. At this frequency phase margin of compensated system is fPM = 180c + +HC (jwg) 45c = 180c - 90c - tan- 1 (wg) - tan- 1 (wg /9) + +GC (jwg) 45c = 180c - 90c - tan- 1 (3) - tan- 1 (1/3) + +GC (jwg) R 1 V 3 + S 3 WW + +GC (jwg) 45c = 90c - tan- 1 S SS1 - 3 b 1 lWW 3 X T

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45c = 90c - 90c + +GC (jwg) +GC (jwg) = 45c The gain cross over frequency of compensated system is lower than un-compensated system, so we may use lag-lead compensator. At gain cross over frequency gain of compensated system is unity so. HC (jwg) = 1 900 GC (jwg) wg

wg2 + 1 wg2 + 81

=1

GC (jwg) = 3 9 + 1 9 + 81 = 3 # 30 = 1 900 900 10 in dB GC (wg) = 20 log b 1 l 10 =- 20 dB (attenuation) SOL 3.38

Option (B) is correct. Characteristic equation for the given system, K (s + 3) =0 1+ (s + 8) 2 (s + 8) 2 + K (s + 3) = 0 s2 + (16 + K) s + (64 + 3K) = 0 By applying Routh’s criteria. 64 + 3K s2 1 s

16 + K

s0

64 + 3K

1

O N

A I D

0

For system to be oscillatory

no w.

.in

co ia.

d

ww

16 + K = 0 & K =- 16 Auxiliary equation A (s) = s2 + (64 + 3K) = 0 &

SOL 3.39

s2 + 64 + 3 # (- 16) = 0 s2 + 64 - 48 = 0 s2 =- 16 & jw = 4j w = 4 rad/sec

Option (D) is correct. From the given block diagram we can obtain signal flow graph of the system. Transfer function from the signal flow graph is written as c 0 P + c1 P s s2 T.F = a Pb a 0 1 + 1 + 2 - 2 0 - Pb1 s s s s (c 0 + c1 s) P (s + a1 s + a 0) - P (b 0 + sb1) (c 0 + c1 s) P 2 s ^ + a1 s + a 0 h = P (b + sb1) 1- 2 0 s + a1 s + a 0 from the given reduced form transfer function is given by =

2

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Page 290

XYP 1 - YPZ by comparing above two we have T.F =

X = (c 0 + c1 s) 1 Y = 2 s + a1 s + a 0 Z = (b 0 + sb1) SOL 3.40

Option (A) is correct. For the given system Z is given by Z = E (s) Ki s Where E (s) " steady state error of the system Here sR (s) E (s) = lim s " 0 1 + G (s) H (s) Input

R (s) = 1 (Unit step) s

A I D

w2 G (s) = b Ki + K p le 2 s s + 2xws + w2 o H (s) = 1 (Unity feed back) So,

SOL 3.41

O N

R V n sb 1 l i S W . s co Wb Ki l . Z = lim S a 2 i s"0S Ki W s d w p lo 2 2 W S1 + b s + K.n w (s + 2xws + w ) X T w w Ki = lim = Ki = 1 2 s"0 >s + (Ki + K p s) 2 w H Ki 2 (s + 2xws + w )

Option (C) is correct. System response of the given circuit can be obtained as. 1 bCs l e 0 (s) H (s) = = 1 ei (s) bR + Ls + Cs l 1 b LC l 1 H (s) = = LCs2 + RCs + 1 s2 + R s + 1 L LC Characteristic equation is given by, s2 + R s + 1 = 0 L LC Here natural frequency wn = 1 LC 2xwn = R L Damping ratio x = R LC = R C 2 L 2L Here

x = 10 2

1 # 10- 3 = 0.5 (under damped) 10 # 10- 6

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Page 291

So peak overshoot is given by SOL 3.42 SOL 3.43

% peak overshoot = e Option ( ) is correct.

- px 1 - x2

# 100 = e

- p # 0.5 1 - (0.5) 2

# 100 = 16%

Option (B) is correct. In standard form for a characteristic equation give as sn + an - 1 sn - 1 + ... + a1 s + a 0 = 0 in its state variable representation matrix A is given as R V 1 0 g 0 W S 0 S 0 0 1 g 0 W A =S W Sh h h h h W S- a 0 - a1 - a2 g - an - 1W T X Characteristic equation of the system is

SOL 3.44

SOL 3.45

4s2 - 2s + 1 = 0 So, a2 = 4, a1 =- 2, a 0 = 1 R 0 1 0 VW RS 0 S 0 1 W=S 0 A =S 0 SS- a - a - a WW SS- 1 0 1 2 T X T Option (A) is correct. In the given options only in option (A) circle (- 1, j0), So this is stable.

1 0 VW 0 1W 2 - 4WW X

A I D

O N

the nyquist plot does not enclose the unit

in

Option (A) is correct. co. . a di Given function is, o n . 10 4 (1 + jw )w w H (jw) = w + jw ) 2 (10 + jw) (100 Function can be rewritten as, 10 4 (1 + jw) H (jw) = 2 10 91 + j w C 10 4 91 + j w C 10 100 =

0.1 (1 + jw) w w 2 a1 + j 10 ka1 + j 100 k

The system is type 0, So, initial slope of the bode plot is 0 dB/decade. Corner frequencies are w1 = 1 rad/sec w 2 = 10 rad/sec w 3 = 100 rad/sec As the initial slope of bode plot is 0 dB/decade and corner frequency w1 = 1 rad/ sec, the Slope after w = 1 rad/sec or log w = 0 is(0 + 20) =+ 20 dB/dec. After corner frequency w2 = 10 rad/sec or log w2 = 1, the Slope is (+ 20 - 20) = 0 dB/dec. Similarly after w3 = 100 rad/sec or log w = 2 , the slope of plot is (0 - 20 # 2) =- 40 dB/dec. Hence (A) is correct option. SOL 3.46

Option (B) is correct.

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Page 292

Given characteristic equation. (s2 - 4) (s + 1) + K (s - 1) = 0 K (s - 1) or =0 1+ 2 (s - 4) (s + 1) So, the open loop transfer function for the system. K (s - 1) , no. of poles n = 3 G (s) = (s - 2) (s + 2) (s + 1) no of zeroes m = 1 Steps for plotting the root-locus (1) Root loci starts at s = 2, s =- 1, s =- 2 (2) n > m , therefore, number of branches of root locus b = 3 (3) Angle of asymptotes is given by (2q + 1) 180c , q = 0, 1 n-m (I)

(2 # 0 + 1) 180c = 90c (3 - 1)

(II)

(2 # 1 + 1) 180c = 270c (3 - 1)

A I D

(4) The two asymptotes intersect on real axis at / Poles - / Zeroes = (- 1 - 2 + 2) - (1) =- 1 x = 3-1 n-m (5) Between two open-loop poles s =- 1 and s =- 2 there exist a break away point. .in (s2 - 4) (s + i1a).co K =(s -n1o) d

O N dK = 0 ds

.

w ww

s =- 1.5

SOL 3.47

Option (C) is correct. Closed loop transfer function of the given system is, s2 + 4 T (s) = (s + 1) (s + 4) T (jw) =

(jw) 2 + 4 (jw + 1) (jw + 4)

If system output is zero 4 - w2 T (jw) = =0 ^ jw + 1h (jw + 4) 4 - w2 = 0 w2 = 4 & w = 2 rad/sec SOL 3.48

Option (A) is correct. From the given plot we can see that centroid C (point of intersection) where asymptotes intersect on real axis) is 0 So for option (a) G (s) = K3 s

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Centroid = SOL 3.49

Page 293

/ Poles - / Zeros = 0 - 0 = 0 n-m

3-0

Option (A) is correct. Open loop transfer function is. (s + 1) G (s) = s2 jw + 1 G (jw) = - w2 Phase crossover frequency can be calculated as. +G (jwp) =- 180c tan- 1 (wp) =- 180c wp = 0 Gain margin of the system is. G.M =

SOL 3.50

1 G (jwp)

=

1 = 2 wp + 1 w2p

w2p =0 w2p + 1

Option (C) is correct. Characteristic equation for the given system

A I D

1 + G (s) H (s) = 0 (1 - s) =0 1+K (1 + s)

O N

(1 + s) + K (1 - s) = 0 n s (1 - K) + (1 + K) = 0 o.i c . For the system to be stable, coefficient dia of characteristic equation should be of o .n same sign. w w 1 - K > 0 , K + 1 >w0 K < 1, K > - 1 -1 < K < 1 K H - > H = > = s2 - 4 = 0 0 s 2 0 -2 s H

A I D

s1, s2 = ! 2 SOL 3.62

Option (D) is correct. For the given system, characteristic equation ican n be written as, . o c 1 + K (1 + sP) = 0 ia. d s (s + 2) o

O N

.n

s (s + 2) + K (1 + sP) = w 0w w s2 + s (2 + KP) + K = 0 From the equation. wn = K = 5 rad/sec (given) So, K = 25 and 2xwn = 2 + KP 2 # 0.7 # 5 = 2 + 25P or P = 0.2 so K = 25 , P = 0.2 SOL 3.63

Option (D) is correct. Unit - impulse response of the system is given as, c (t) = 12.5e- 6t sin 8t , t $ 0 So transfer function of the system. H (s) = L [c (t)]

=

12.5 # 8 (s + 6) 2 + (8) 2

100 s2 + 12s + 100 Steady state value of output for unit step input, H (s) =

lim y (t) = lim sY (s) = lim sH (s) R (s)

t"3

s"0

s"0

100 1 = 1.0 = lim s ; 2 E s s"0 s + 12s + 100

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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia SOL 3.64

Page 298

Option (A) is correct. System response is. H (s) = s s+1 jw H (jw) = jw + 1 Amplitude response w w+1 Given input frequency w = 1 rad/sec. 1 So H (jw) w = 1 rad/sec = = 1 1+1 2 Phase response H (jw) =

SOL 3.65

qh (w) = 90c - tan- 1 (w) qh (w) w = 1 = 90c - tan- 1 (1) = 45c So the output of the system is y (t) = H (jw) x (t - qh) = 1 sin (t - 45c) 2 Option (C) is correct. Given open loop transfer function jaw + 1 G (jw) = (jw) 2 Gain crossover frequency (wg) for the system.

A I D

O N

G (jwg) = 1 a2 wg2 + 1 =1 - wg2

no w.

.in

co ia.

d

ww

a2 wg2 + 1 = wg4 wg4 - a2 wg2 - 1 = 0 Phase margin of the system is

SOL 3.66

fPM = 45c = 180c + +G (jwg) 45c = 180c + tan- 1 (wg a) - 180c tan- 1 (wg a) = 45c wg a = 1 From equation (1) and (2) 1 -1-1 = 0 a4 a 4 = 1 & a = 0.841 2 Option (C) is correct. Given system equation is. d 2 x + 6 dx + 5x = 12 (1 - e- 2t) dt dt 2 Taking Laplace transform on both side. s2 X (s) + 6sX (s) + 5X (s) = 12 :1 - 1 D s s+2

...(1)

(2)

(s2 + 6s + 5) X (s) = 12 ; 2 E s (s + 2)

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Page 299

System transfer function is X (s) =

24 s (s + 2) (s + 5) (s + 1)

Response of the system as t " 3 is given by lim f (t) = lim sF (s) (final value theorem) t"3

s"0

24 = lim s ; s"0 s (s + 2) (s + 5) (s + 1)E = SOL 3.67

24 = 2.4 2#5

Option (A) is correct. Transfer function of lead compensator is given by. K a1 + s k a H (s) = s a1 + b k R w V S1 + j a a kW H (jw) = K S W SS1 + j a w kWW b T X So, phase response of the compensator is. qh (w) = tan- 1 a w k - tan- 1 a w k a b

A I D

O N

Jw - wN -1 K a b O = tan-i1nw (b - a) = tan 2 KK O o. ; ab + w2 E 1 + w O ia.c ab Pd L o qh should be positive for phase lead .ncompensation w w(b - a) w So, qh (w) = tan- 1w; >0 ab + w2 E b >a

SOL 3.68

Option (A) is correct. Since there is no external input, so state is given by X (t) = f (t) X (0) f (t) "state transition matrix X [0] "initial condition e- 2t 0 2 So x (t) = > H> H 0 e- t 3 2e- 2t x (t) = > - t H 3e At t = 1, state of the system 0.271 2e- 2 x (t) t = 1 = > - 1H = > 1.100H 2e

SOL 3.69

Option (B) is correct. Given equation d2 x + 1 dx + 1 x = 10 + 5e- 4t + 2e- 5t dt2 2 dt 18 Taking Laplace on both sides we have

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Page 300

s2 X (s) + 1 sX (s) + 1 X (s) = 10 + 5 + 2 2 18 s s+4 s+5 10 (s + 4) (s + 5) + 5s (s + 5) + 2s (s + 4) (s2 + 1 s + 1 ) X (s) = 2 18 s (s + 4) (s + 5) System response is, 10 (s + 4) (s + 5) + 5s (s + 5) + 2s (s + 4) X (s) = s (s + 4) (s + 5) bs2 + 1 s + 1 l 2 18 10 (s + 4) (s + 5) + 5s (s + 5) + 2s (s + 4) s (s + 4) (s + 5) bs + 1 lbs + 1 l 3 6 We know that for a system having many poles, nearness of the poles towards imaginary axis in s -plane dominates the nature of time response. So here time constant given by two poles which are nearest to imaginary axis. Poles nearest to imaginary axis s1 =- 1 , s2 =- 1 6 3 =

So, time constants ) SOL 3.70

A I D

t1 = 3 sec t2 = 6 sec

Option (A) is correct. Steady state error for a system is given by sR (s) n ess = lim s " 0 1 + G (s) H (s) o.i c . dia Where input R (s) = 1 (unit step) o s .n w w G (s) = b 3 lb w15 l s + 15 s + 1

O N

H (s) = 1 So

SOL 3.71

(unity feedback) sb 1 l s ess = lim = 15 = 15 60 15 + 45 45 s"0 1+ (s + 15) (s + 1)

%ess = 15 # 100 = 25% 60 Option (C) is correct. Characteristic equation is given by Here

So,

1 + G (s) H (s) = 0 H (s) = 1

(unity feedback) G (s) = b 3 lb 15 l s + 15 s + 1 1 + b 3 lb 15 l = 0 s + 15 s + 1

(s + 15) (s + 1) + 45 = 0 s2 + 16s + 60 = 0 (s + 6) (s + 10) = 0 s =- 6, - 10 SOL 3.72

Option (A) is correct.

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Page 301

Given equation can be written as, d 2 w =- b dw - K 2 w + K V J dt LJ LJ a dt 2 Here state variables are defined as, dw = x 1 dt w = x2 So state equation is 2 xo1 =- B x1 - K x2 + K Va J LJ LJ xo2 = dw = x1 dt In matrix form K/LJ xo1 - B/J - K 2 /LJ x1 >o H = > >x H + > 0 H Va H x2 1 0 2 R 2 V Sd w W S dt2 W = P >dwH + QVa dt S dw W S dt W T X So matrix P is

A I D

- B/J - K 2 /LJ > 1 H 0 SOL 3.73

O N

Option (C) is correct. in Characteristic equation of the system is given co. by

a.

di 1 + GH = 0 o n w. K w = 0 1+ s (s + 2) (s + 4) w

s (s + 2) (s + 4) + K = 0 s3 + 6s2 + 8s + K = 0 Applying routh’s criteria for stability s3

1

8

s2

6 K - 48 6

K

s

1

s0

SOL 3.74

K

System becomes unstable if K - 48 = 0 & K = 48 6 Option (A) is correct. The maximum error between the exact and asymptotic plot occurs at corner frequency. Here exact gain(dB) at w = 0.5a is given by 2

1 + w2 a (0.5a) 2 1/2 = 20 log K - 20 log ;1 + E a2 = 20 log K - 0.96 Gain(dB) calculated from asymptotic plot at w = 0.5a is gain(dB) w = 0.5a = 20 log K - 20 log

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Page 302

= 20 log K Error in gain (dB) = 20 log K - (20 log K - 0.96) dB = 0.96 dB Similarly exact phase angle at w = 0.5a is. qh (w) w = 0.5a =- tan- 1 a w k =- tan- 1 b 0.5a l =- 26.56c a a Phase angle calculated from asymptotic plot at (w = 0.5a) is - 22.5c Error in phase angle =- 22.5 - (- 26.56c) = 4.9c SOL 3.75

Option (B) is correct. Given block diagram

Given block diagram can be reduced as

Where

O N

A I D

G1 =

1 bs l

.in

co = 1ia. osd+ 3

1 + b 1 l .n sw w w1 bs l G2 = = 1 s + 12 1 1 + b l 12 s Further reducing the block diagram. 3

2G1 G2 1 + (2G1 G2) 9 (2) b 1 lb 1 l s + 3 s + 12 = 1 + (2) b 1 lb 1 l (9) s + 3 s + 12

Y (s) =

2 2 = 2 (s + 3) (s + 12) + 18 s + 15s + 54 1 2 = = s (s + 9) (s + 6) 27 a1 + ka1 + s k 9 6 Option (C) is correct. Given state equation is, o = AX X =

SOL 3.76

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Page 303

Taking Laplace transform on both sides of the equation, sX (s) - X (0) = AX (s) (sI - A) X (s) = X (0) X (s) = (sI - A) - 1 X (0) = F (s) X (0) Where f (t) = L- 1 [F (s)] = L- 1 [(sI - A) - 1] is defined as state transition matrix SOL 3.77

Option (B) is correct. State equation of the system is given as, o = >2 3H X + >1H u X 0 5 0 Here

1 2 3 A = > H, B = > H 0 0 5

Check for controllability: 2 3 1 2 AB = > H> H = > H 0 5 0 0 1 2 U = [B : AB] = > H 0 0

A I D

U = (1 # 0 - 2 # 0) = 0 Matrix U is singular, so the system is uncontrollable. Check for Stability: Characteristic equation of the system is obtained as,

O N

sI - A = 0 n o.i s 0 2 3 c . (sI - A) = > H - > H ia 0 s 0 5 od .n s-2 w -w3 => 0 ws - 5H

sI - A = (s - 2) (s - 5) = 0 s = 2, s = 5 There are two R.H.S Poles in the system so it is unstable. SOL 3.78

Option (B) is correct. Given open loop transfer function, no of poles = 2 G (s) = K2 , s no of zeroes = 0 For plotting root locus: (1) Poles lie at s1, s2 = 0 (2) So the root loci starts (K = 0) from s = 0 and s = 0 (3) As there is no open-loop zero, root loci terminates (K = 3) at infinity. (4) Angle of asymptotes is given by (2q + 1) 180c , q = 0, 1 n-m So the two asymptotes are at an angle of (i)

(2 # 0 + 1) 180c = 90c 2

(2 # 1 + 1) 180c (ii) = 270c 2 (5) The asymptotes intersect on real axis at a point given by

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x=

/ Poles - / zeros n-m

Page 304

= 0-0 = 0 2

(6) Break away points 1 + K2 = 0 s K =- s2 dK =- 2s = 0 & s = 0 ds So the root locus plot is.

SOL 3.79

Option (A) is correct. System is described as. d2 y dy = du + 2u + dt dt2 dt

A I D

Taking Laplace transform on both sides.

s2 Y (s) + sY (s) = sU (s) + 2U (s) (s2 + s) Y (s) = (s + 2) U (s) n So, the transfer function is o.i c . Y (s) (s + 2) dia T.F = = 2 o U (s) (s + s)w.n SOL 3.80

O N w

w Option (A) is correct. Here, we have 1 2 0 A = > H, B = > H, C = [4, 0] 1 0 4

We know that transfer function of the system is given by. Y (s) G (s) = = C (sI - A) - 1 B U (s) s 0 2 0 s-2 0 [sI - A ] = > H - > H = > 0 s 0 4 0 s - 4H (s - 4) 0 1 > 0 (s - 2)H (s - 2) (s - 4) R V S 1 0 W (s - 2) W = SS 1 W S 0 (s - 4)W T X R V R V S 1 W S 1 0 W1 (s - 2) S(s - 2)W W = [4 0] SS 1 W>1H = [4 0] S 1 W S(s - 4)W S 0 (s - 4)W T X X T 4 = (s - 2)

(sI - A) - 1 =

So,

Y (s) U (s) Y (s) U (s)

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Page 305

Here input is unit impulse so U (s) = 1 and output Y (s) = 4 (s - 2) Taking inverse Laplace transfer we get output y (t) = 4e2t SOL 3.81

Option (D) is correct. Given state equation R V S0 1 0 0W S W o = S0 0 1 0W X X S0 0 0 1W S0 0 0 1W RT VX S0 1 0 0W S0 0 1 0W Here A =S W S0 0 0 1W S0 0 0 1W T obtained X as Eigen value can be A - lI = 0 R V R V S0 1 0 0W Sl 0 0 0W S0 0 1 0W S0 l 0 0W (A - lI) = S W-S W S0 0 0 1W S0 0 l 0W S0 0 0 1W S0 0 0 lW T X T X n R V o.i c l 1 0 0 . S W ia S 0 -l 1 0 nWo d =S w1. WW w S 0 0 -l w S 0 0 0 1 - lW T X A - lI = l3 (1 - l) = 0 l1, l2, l3 = 0 , l4 = 1

A I D

O N

or SOL 3.82

Option (A) is correct. Input-output relationship is given as d 2y dy du 2 + 2 dt + 10y = 5 dt - 3u dt Taking Laplace transform on both sides with zero initial condition. s 2 Y (s) + 2sY (s) + 10Y (s) = 5sU (s) - 3U (s) (s2 + 2s + 10) Y (s) = (5s - 3) U (s) (5s - 3) Output Y (s) = 2 U (s) (s + 2s + 10) With no input and with given initial conditions, output is obtained as d 2y dy 2 + 2 dt + 10y = 0 dt Taking Laplace transform (with initial conditions) [s2 Y (s) - sy (0) - y' (0)] + 2 [sY (s) - y (0)] + 10Y (s) = 0 Given that y' (0) =- 4 , y (0) = 1 [s2 Y (s) - s - (- 4)] + 2 (s - 1) + 10Y (s) = 0

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Page 306

Y (s) [s2 + 2s + 10] = (s - 2) (s - 2) Y (s) = 2 (s + 2s + 10) Output in both cases are same so (5s - 3) (s - 2) U (s) = 2 (s2 + 2s + 10) (s + 2s + 10) U (s) =

(s - 2) (5s - 10) =1 5 (5s - 3) (5s - 3)

(5s - 3) 7 = 1= 5 5s - 3 (5s - 3)G 7 U (s) = 1 ;1 5 (5s - 3)E Taking inverse Laplace transform, input is u (t) = 1 :d (t) - 5 e3/5t u (t)D 5 5 = 1 d (t) - 7 e3/5t u (t) 5 25 SOL 3.83

A I D

Option (C) is correct. d 2 y dy + - 2y = u (t) e- t dt 2 dt State variable representation is given as o = AX + Bu X Or Here

O N

...(1)

n

o.i x1 xo1 c . > o H = A >x H + Bu ia x2 2 od

x1 = y , x 2 = b

n

. ww

dy t - yw le dt

dx1 = dy = x e- t + y = x e- t + x 2 2 1 dt dt dx1 = x + x e- t + (0) u (t) or 1 2 dt Similarly 2 dx2 = d y et + dy et - et dy - yet dt dt dt dt 2 d 2y Put from equation (1) dt 2 dx2 = u (t) e- t - dy + 2y et - yet So, : D dt dt = u (t) -

...(2)

dy t e + 2yet - yet dt

= u (t) - [x2 e- t + y] et + yet = u (t) - x2 dx2 = 0 - x + u (t) 2 dt

...(3)

From equation (2) and (3) state variable representation is 0 xo1 1 e- t x 1 >o H = > >x H + >1H u (t) H x2 0 -1 2

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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia SOL 3.84

Page 307

Option (B) is correct. Characteristic equation of the system 1 + G (s) = 0 2 (s + a) =0 1+ s (s + 2) (s + 10) s (s + 2) (s + 10) + 2 (s + a) = 0 s3 + 12s2 + 20s + 2s + 2a = 0 s3 + 12s2 + 22s + 2a = 0 2a =0 1+ 3 s + 12s2 + 22s No of poles n = 3 No. of zeros m = 0 Angle of asymptotes (2q + 1) 180c fA = , q = 0, 1, 2 n-m fA =

(2q + 1) 180c = (2q + 1) 60c 3

A I D

fA = 60c, 180c, 300c SOL 3.85

Option (A) is correct. Asymptotes intercepts at real axis at the point / real Parts of Poles - / real Parts of zeros C = n-m in Poles at

no

. ww

w

C = 0 - 2 - 10 - 0 =- 4 3-0 Option (C) is correct. Break away points da = 0 ds So

SOL 3.86

O N

s1 = 0 s2 =- 2 s 3 =- 10

o.

.c dia

a =- 1 [s3 + 12s2 + 22s] 2 da =- 1 [3s2 + 24s + 22] = 0 2 ds s1, s2 =- 1.056, - 6.9433 SOL 3.87

Option ( ) is correct.

SOL 3.88

Option (A) is correct. Given state equation o = >- 3 1 H X X 0 -2 Or

o = AX , where A = >- 3 1 H X 0 -2

Taking Laplace transform on both sides. sX (s) - X (0) = AX (s)

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Page 308

X (s) (sI - A) = X (0) X (s) = (sI - A) - 1 X (0) Steady state value of X is given by xss = lim sX (s) = lim s (sI - A) - 1 X (0) s"0

s"0

s 0 -3 1 s + 3 -1 => (sI - A) = > H - > H 0 s 0 -2 0 s + 2H s+2 1 1 > 0 s + 3H (s + 3) (s + 2) R V 1 S 1 W (s + 3) (s + 2) (s + 3)W S =S W 1 S 0 (s + 2) W T value X So the steady state R V 1 S 1 W (s + 3) (s + 2) (s + 3)W 10 S xss = lim s S W>- 10H 1 s"0 S 0 (s + 2) W TR XV 10 10 S W 0 (s + 3) (s + 2) (s + 3)W S => H = lim s S W 0 s"0 - 10 S W (s + 2) T X Option (D) is correct. n So no. of poles at origin is 2. Initial slope of the bode plot is - 40 cdB/dec. o.i . Then slope increased by - 20 dB/dec. diaat w = 2 rad/sec, so one poles lies at this o .n changes by + 20 dB/dec, so there is one zero frequency. At w = 5 rad/sec slope w ww slope decrease by - 20 dB/dec at w = 25 so one lying at this frequency. Further pole of the system is lying at this frequency. Transfer function K (s + 5) H (s) = 2 s (s + 2) (s + 25) At w = 0.1, gain is 54 dB, so 5K 54 = 20 log (0.1) 2 (2) (25) (sI - A- 1) =

SOL 3.89

A I D

O N

K = 50 H (s) = SOL 3.90

50 (s + 5) s (s + 2) (s + 25) 2

Option (B) is correct. Open loop transfer function of the system is 10 4 G (s) = s (s + 10) 2 10 4 10 4 = jw (jw + 10) 2 jw (100 - w2 + j20w) 10 4 G (jw) = w (100 - w2) 2 + 400w2

G (jw) = Magnitude

At w = 20 rad/sec

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Page 309

10 4 20 9 # 10 4 + 16 # 10 4 10 4 =1 = 20 # 5 # 102 Magnitude in dB = 20 log 10 G (j20) = 20 log 10 1 = 0 dB G (j20) =

SOL 3.91

Option (C) is correct. Since G (j w) = 1 at w = 20 rad/sec, So this is the gain cross-over frequency wg = 20 rad/sec Phase margin fPM = 180c + +G (jwg) 20 wg +G (jwg) =- 90c - tan- 1 = 100 - wg2 G fPM = 180 - 90c - tan- 1 ; 20 # 20 2 E 100 - (20) =- 36.86c

SOL 3.92

Option (C) is correct. To calculate the gain margin, first we have to obtain phase cross over frequency (wp). At phase cross over frequency

A I D

+G (jwp) =- 180c 20wp =- 180c - 90c - tan- 1 = 100 - w2p G 20wp n = 90c tan- 1 = o.i 100 - w2p G c . dia rad/sec. 100 - w2p = 0 & wp = 10 o .n w 1 Gain margin in dB = 20 log 10 e ww G (jwp) o 10 4 G (jwp) = G (j10) = 10 (100 - 100) 2 + 400 (10) 2

O N

10 4 =5 10 # 2 # 102 G.M. = 20 log 10 b 1 l =- 13.97 dB 5 Option (B) is correct. Since gain margin and phase margin are negative, so the system is unstable. =

SOL 3.93

SOL 3.94

Option (C) is correct. Given characteristic equation s3 + s2 + Ks + K = 0 K (s + 1) =0 1+ 3 s + s2 K (s + 1) =0 1+ 2 s (s + 2) so open loop transfer function is K (s + 1) G (s) = 2 s (s + 1) root-locus is obtained in following steps: 1. Root-loci starts(K = 0 ) at s = 0 , s = 0 and s =- 2

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2. 3. 4.

Page 310

There is one zero at s =- 1, so one of root-loci terminates at s =- 1 and other two terminates at infinity No. of poles n = 3 , no of zeros ,m = 1 Break - Away points

dK = 0 ds Asymptotes meets on real axis at a point C / poles - / zeros C = n-m (0 + 0 - 2) - (- 1) =- 0.5 = 3-1 ***********

A I D

O N

no w.

.in

co ia.

d

ww

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3

ELECTRICAL & ELECTRONIC MEASUREMENTS

YEAR 2013 MCQ 3.1

ONE MARK

Three moving iron type voltmeters are connected as shown below. Voltmeter readings are V , V1 and V2 as indicated. The correct relation among the voltmeter readings is

(A) V = V1 + V2 2 2 (C) V = V1 V2 MCQ 3.2

A I D

(B) V = V1 + V2 (D) V = V2 - V1

The input impedance of the permanent magnet moving coil (PMMC) voltmeter is infinite. Assuming that the diode shown in the figure below is ideal, the read.in o ing of the voltmeter in Volts is c .

O N w

ww

(A) 4.46 (C) 2.23 YEAR 2013

ia

d .no

(B) 3.15 (D) 0 TWO MARKS

MCQ 3.3

Two magnetically uncoupled inductive coils have Q factors q1 and q2 at the chosen operating frequency. Their respective resistances are R1 and R2 . When connected in series, their effective Q factor at the same operating frequency is (A) q1 + q2 (B) ^1/q1h + ^1/q2h (C) ^q1 R1 + q2 R2h / ^R1 + R2h (D) ^q1 R2 + q2 R1h / ^R1 + R2h

MCQ 3.4

A strain gauge forms one arm of the bridge shown in the figure below and has a nominal resistance without any load as Rs = 300 W . Other bridge resistances are R1 = R2 = R 3 = 300 W . The maximum permissible current through the strain gauge is 20 mA. During certain measurement when the bridge is excited by

GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 312

maximum permissible voltage and the strain gauge resistance is increased by 1% over the nominal value, the output voltage V0 in mV is

(A) 56.02 (C) 29.85

(B) 40.83 (D) 10.02

YEAR 2012 MCQ 3.5

ONE MARK

A periodic voltage waveform observed on an oscilloscope across a load is shown. A permanent magnet moving coil (PMMC) meter connected across the same load reads

A I D

O N

no w.

.in

co ia.

d

ww

(A) 4 V (C) 8 V

(B) 5 V (D) 10 V

MCQ 3.6

The bridge method commonly used for finding mutual inductance is (A) Heaviside Campbell bridge (B) Schering bridge (C) De Sauty bridge (D) Wien bridge

MCQ 3.7

For the circuit shown in the figure, the voltage and current expressions are v (t) = E1 sin (wt) + E 3 sin (3wt) and i (t) = I1 sin (wt - f1) + I 3 sin (3wt - f3) + I5 sin (5wt) The average power measured by the wattmeter is

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Page 313

(A) 1 E1 I1 cos f1 2 (B) 1 [E1 I1 cos f1 + E1 I 3 cos f3 + E1 I5] 2 (C) 1 [E1 I1 cos f1 + E 3 I 3 cos f3] 2 (D) 1 [E1 I1 cos f1 + E 3 I1 cos f1] 2 YEAR 2012 MCQ 3.8

TWO MARKS

An analog voltmeter uses external multiplier settings. With a multiplier setting of 20 kW, it reads 440 V and with a multiplier setting of 80 kW, it reads 352 V. For a multiplier setting of 40 kW, the voltmeter reads (A) 371 V (B) 383 V (C) 394 V (D) 406 V YEAR 2011

MCQ 3.9

ONE MARK

O N

(C) both (1) and (2) are true MCQ 3.10

ww

w

ia

d .no

(D) both (1) and (2) are false

The bridge circuit shown in the figure below is used for the measurement of an unknown element ZX . The bridge circuit is best suited when ZX is a

(A) low resistance (C) low Q inductor MCQ 3.11

A I D

Consider the following statement (1) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the current coil. (2) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the voltage coil circuit. in (A) (1) is true but (2) is false co. (1) is false but (2) is true .(B)

(B) high resistance (D) lossy capacitor

A dual trace oscilloscope is set to operate in the ALTernate mode. The control input of the multiplexer used in the y -circuit is fed with a signal having a frequency equal to (A) the highest frequency that the multiplexer can operate properly (B) twice the frequency of the time base (sweep) oscillator (C) the frequency of the time base (sweep) oscillator (D) haif the frequency of the time base (sweep) oscillator

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YEAR 2011 MCQ 3.12

TWO MARKS

A 4 12 digit DMM has the error specification as: 0.2% of reading + 10 counts. If a dc voltage of 100 V is read on its 200 V full scale, the maximum error that can be expected in the reading is (A) ! 0.1% (B) ! 0.2% (C) ! 0.3% (D) ! 0.4% YEAR 2010

MCQ 3.13

Page 314

ONE MARK

A wattmeter is connected as shown in figure. The wattmeter reads.

A I D

(A) Zero always (B) Total power consumed by Z1 and Z 2 (C) Power consumed by Z1 (D) Power consumed by Z2

MCQ 3.14

O N

.in o c . An ammeter has a current range of 0-5 ia A, and its internal resistance is 0.2 W. dA, o In order to change the range to .0-25 we need to add a resistance of n w (A) 0.8 W in series with the w meter w (B) 1.0 W in series with the meter (C) 0.04 W in parallel with the meter (D) 0.05 W in parallel with the meter

MCQ 3.15

As shown in the figure, a negative feedback system has an amplifier of gain 100 with ! 10% tolerance in the forward path, and an attenuator of value 9/100 in the feedback path. The overall system gain is approximately :

(A) 10 ! 1% (C) 10 ! 5%

(B) 10 ! 2% (D) 10 ! 10%

YEAR 2010 MCQ 3.16

TWO MARKS

The Maxwell’s bridge shown in the figure is at balance. The parameters of the inductive coil are.

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(A) (B) (C) (D)

Page 315

R = R2 R 3 /R 4, L = C 4 R2 R 3 L = R2 R 3 /R 4, R = C 4 R2 R 3 R = R 4 /R2 R 3, L = 1/ (C 4 R2 R 3) L = R 4 /R2 R 3, R = 1/ (C 4 R2 R 3)

YEAR 2009

ONE MARK

MCQ 3.17

The pressure coil of a dynamometer type wattmeter is (A) Highly inductive (B) Highly resistive (C) Purely resistive (D) Purely inductive

MCQ 3.18

The two inputs of a CRO are fed with two stationary periodic signals. In the X-Y mode, the screen shows a figure which changes from ellipse to circle and back to ellipse with its major axis changing orientation slowly and repeatedly. The following inference can be made from this. (A) The signals are not sinusoidal .in o c . (B) The amplitudes of the signals are very close but not equal dia o (C) The signals are sinusoidal with.ntheir frequencies very close but not equal ww phase difference between the signals (D) There is a constant but w small

A I D

YEAR 2009 MCQ 3.19

TWO MARKS

The figure shows a three-phase delta connected load supplied from a 400V, 50 Hz, 3-phase balanced source. The pressure coil (PC) and current coil (CC) of a wattmeter are connected to the load as shown, with the coil polarities suitably selected to ensure a positive deflection. The wattmeter reading will be

(A) 0 (C) 800 Watt MCQ 3.20

O N

(B) 1600 Watt (D) 400 Watt

An average-reading digital multi-meter reads 10 V when fed with a triangular wave, symmetric about the time-axis. For the same input an rms-reading meter will read (B) 10 (A) 20 3 3

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(C) 20 3

(D) 10 3

YEAR 2008 MCQ 3.21

ONE MARK

Two 8-bit ADCs, one of single slope integrating type and other of successive approximate type, take TA and TB times to convert 5 V analog input signal to equivalent digital output. If the input analog signal is reduced to 2.5 V, the approximate time taken by the two ADCs will respectively, be (A) TA, TB (B) TA /2, TB (C) TA, TB /2 (D) TA /2, TB /2 YEAR 2008

MCQ 3.22

Page 316

TWO MARKS

Two sinusoidal signals p (w1, t) = A sin w1 t and q (w2 t) are applied to X and Y inputs of a dual channel CRO. The Lissajous figure displayed on the screen shown below : The signal q (w2 t) will be represented as

A I D

O N

no (A) q (w2 t) = A sin w2 t, w2 = 2w1w. ww (C) q (w2 t) = A cos w2 t, w2 = 2w1 MCQ 3.23

.in

co ia.

d

(B) q (w2 t) = A sin w2 t, w2 = w1 /2 (D) q (w2 t) = A cos w2 t, w2 = w1 /2

The ac bridge shown in the figure is used to measure the impedance Z .

If the bridge is balanced for oscillator frequency f = 2 kHz, then the impedance Z will be (A) (260 + j0) W (B) (0 + j200) W (C) (260 - j200) W (D) (260 + j200) W YEAR 2007 MCQ 3.24

ONE MARK

The probes of a non-isolated, two channel oscillocope are clipped to points A, B and C in the circuit of the adjacent figure. Vin is a square wave of a suitable low frequency. The display on Ch1 and Ch2 are as shown on the right. Then the “Signal” and “Ground” probes S1, G1 and S2, G2 of Ch1 and Ch2 respectively are connected to points :

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(A) A, B, C, A (C) C, B, A, B

(B) A, B, C, B (D) B, A, B, C

YEAR 2007 MCQ 3.25

Page 317

TWO MARKS

A bridge circuit is shown in the figure below. Which one of the sequence given below is most suitable for balancing the bridge ?

A I D

O N

no w.

(A) First adjust (B) First adjust (C) First adjust (D) First adjust YEAR 2006 MCQ 3.26

.in

co ia.

d

w

w adjust R1 R4 , and then R2 , and then adjust R3 R2 , and then adjust R4 R4 , and then adjust R2 ONE MARK

The time/div and voltage/div axes of an oscilloscope have been erased. A student connects a 1 kHz, 5 V p-p square wave calibration pulse to channel-1 of the scope and observes the screen to be as shown in the upper trace of the figure. An unknown signal is connected to channel-2(lower trace) of the scope. It the time/div and V/div on both channels are the same, the amplitude (p-p) and period of the unknown signal are respectively

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(A) 5 V, 1 ms (C) 7.5 V, 2 ms MCQ 3.27

(B) 5 V, 2 ms (D) 10 V, 1 ms

A sampling wattmeter (that computes power from simultaneously sampled values of voltage and current) is used to measure the average power of a load. The peak to peak voltage of the square wave is 10 V and the current is a triangular wave of 5 A p-p as shown in the figure. The period is 20 ms. The reading in W will be

(A) 0 W (C) 50 W

(B) 25 W (D) 100 W

A I D

YEAR 2006 MCQ 3.28

MCQ 3.29

Page 318

TWO MARKS

A current of - 8 + 6 2 (sin wt + 30%) A is passed through three meters. They are a centre zero PMMC meter, a true rms meter and a moving iron instrument. The respective reading (in A) will be (A) 8, 6, 10 (B)n8, 6, 8 o.i c . (C) - 8 ,10,10 ia (D) - 8 ,2,2

O N

d no . A variable w is related to three ww other variables x ,y ,z as w = xy/z . The variw ables are measured with meters of accuracy ! 0.5% reading, ! 1% of full scale value and ! 1.5% reading. The actual readings of the three meters are 80, 20 and 50 with 100 being the full scale value for all three. The maximum uncertainty in the measurement of w will be (A) ! 0.5% rdg (B) ! 5.5% rdg (C) ! 6.7 rdg (D) ! 7.0 rdg

MCQ 3.30

A 200/1 Current transformer (CT) is wound with 200 turns on the secondary on a toroidal core. When it carries a current of 160 A on the primary, the ratio and phase errors of the CT are found to be - 0.5% and 30 minutes respectively. If the number of secondary turns is reduced by 1 new ratio-error(%) and phaseerror(min) will be respectively (A) 0.0, 30 (B) - 0.5, 35 (C) - 1.0, 30 (D) - 1.0, 25

MCQ 3.31

R1 and R4 are the opposite arms of a Wheatstone bridge as are R3 and R2 . The source voltage is applied across R1 and R3 . Under balanced conditions which one of the following is true (A) R1 = R3 R4 /R2 (B) R1 = R2 R3 /R4 (C) R1 = R2 R4 /R3 (D) R1 = R2 + R3 + R4

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Page 319

YEAR 2005 MCQ 3.32

MCQ 3.33

The Q-meter works on the principle of (A) mutual inductance (C) series resonance

(B) self inductance (D) parallel resonance

A PMMC voltmeter is connected across a series combination of DC voltage source V1 = 2 V and AC voltage source V2 (t) = 3 sin (4t) V. The meter reads (A) 2 V (B) 5 V (C) (2 +

MCQ 3.34

ONE MARK

3 /2) V

(D) ( 17 /2) V

A digital-to-analog converter with a full-scale output voltage of 3.5 V has a resolution close to 14 mV. Its bit size is (A) 4 (B) 8 (C) 16 (D) 32 YEAR 2005

MCQ 3.35

MCQ 3.36

MCQ 3.37

The simultaneous application of signals x (t) and y (t) to the horizontal and vertical plates, respectively, of an oscilloscope, produces a vertical figure-of-8 display. If P and Q are constants and x (t) = P sin (4t + 30c), then y (t) is equal to (B) Q sin (2t + 15c) (A) Q sin (4t - 30c) (C) Q sin (8t + 60c) (D) Q sin (4t + 30c)

A I D

O N

n A DC ammeter has a resistance of 0.1 W and o.i its current range is 0-100 A. If c . the range is to be extended to 0-500 A, diathen meter required the following shunt o resistance .n w (A) 0.010 W (B) 0.011 W ww (C) 0.025 W (D) 1.0 W The set-up in the figure is used to measure resistance R .The ammeter and voltmeter resistances are 0.01W and 2000 W, respectively. Their readings are 2 A and 180 V, respectively, giving a measured resistances of 90 W The percentage error in the measurement is

(A) 2.25% (C) 4.5% MCQ 3.38

TWO MARKS

(B) 2.35% (D) 4.71%

A 1000 V DC supply has two 1-core cables as its positive and negative leads : their insulation resistances to earth are 4 MW and 6 MW, respectively, as shown in the figure. A voltmeter with resistance 50 kW is used to measure the insulation of the cable. When connected between the positive core and earth, then voltmeter reads

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(A) 8 V (C) 24 V MCQ 3.39

Page 320

(B) 16 V (D) 40 V

Two wattmeters, which are connected to measure the total power on a threephase system supplying a balanced load, read 10.5 kW and - 2.5 kW, respectively. The total power and the power factor, respectively, are (A) 13.0 kW, 0.334 (B) 13.0 kW, 0.684 (C) 8.0 kW, 0.52 (D) 8.0 kW, 0.334 YEAR 2004

ONE MARK

MCQ 3.40

A dc potentiometer is designed to measure up to about 2 V with a slide wire of 800 mm. A standard cell of emf 1.18 V obtains balance at 600 mm. A test cell is seen to obtain balance at 680 mm. The emf of the test cell is (A) 1.00 V (B) 1.34 V (C) 1.50 V (D) 1.70 V

MCQ 3.41

The circuit in figure is used to measure the.in power consumed by the load. The co current coil and the voltage coil of theawattmeter have 0.02 W and 1000W resist. i d o ances respectively. The measurednpower compared to the load power will be

A I D

O N

.

w ww

(A) 0.4 % less (C) 0.2% more MCQ 3.42

(B) 0.2% less (D) 0.4% more

A galvanometer with a full scale current of 10 mA has a resistance of 1000 W. The multiplying power (the ratio of measured current to galvanometer current) of 100 W shunt with this galvanometer is (A) 110 (B) 100 (C) 11 (D) 10 YEAR 2004

MCQ 3.43

TWO MARKS

A CRO probe has an impedance of 500 kW in parallel with a capacitance of 10 pF. The probe is used to measure the voltage between P and Q as shown in figure. The measured voltage will be

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(A) 3.53 V (C) 4.54 V

Page 321

(B) 4.37 V (D) 5.00 V

MCQ 3.44

A moving coil of a meter has 100 turns, and a length and depth of 10 mm and 20 mm respectively. It is positioned in a uniform radial flux density of 200 mT. The coil carries a current of 50 mA. The torque on the coil is (A) 200 mNm (B) 100 mNm (C) 2 mNm (D) 1 mNm

MCQ 3.45

A dc A-h meter is rated for 15 A, 250 V. The meter constant is 14.4 A-sec/rev. The meter constant at rated voltage may be expressed as (A) 3750 rev/kWh (B) 3600 rev/kWh (C) 1000 rev/kWh (D) 960 rev/kWh

MCQ 3.46

A moving iron ammeter produces a full scale torque of 240 mNm with a deflection of 120c at a current of 10 A . The rate of change of self induction (mH/ radian) of the instrument at full scale is (A) 2.0 mH/radian (B) 4.8 mH/radian n (C) 12.0 mH/radian (D) o.i114.6 mH/radian

MCQ 3.47

A I D

O N

a.c i d A single-phase load is connected between no R and Y terminals of a 415 V, sym. w metrical, 3-phase, 4-wire system with phase sequence RYB. A wattmeter is ww in figure. The power factor of the load is 0.8 connected in the system as shown lagging. The wattmeter will read

(A) - 795 W (C) + 597 W

(B) - 597 W (D) + 795 W

MCQ 3.48

A 50 Hz, bar primary CT has a secondary with 500 turns. The secondary supplies 5 A current into a purely resistive burden of 1 W. The magnetizing ampere-turns is 200. The phase angle between the primary and second current is (A) 4.6c (B) 85.4c (C) 94.6c (D) 175.4c

MCQ 3.49

The core flux in the CT of Prob Q.44, under the given operating conditions is (A) 0 (B) 45.0 mWb (C) 22.5 mWb (D) 100.0 mWb

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Page 322

YEAR 2003

ONE MARK

MCQ 3.50

A Manganin swap resistance is connected in series with a moving coil ammeter consisting of a milli-ammeter and a suitable shunt in order to (A) minimise the effect of temperature variation (B) obtain large deflecting torque (C) reduce the size of the meter (D) minimise the effect of stray magnetic fields

MCQ 3.51

The effect of stray magnetic field on the actuating torque of a portable instrument is maximum when the operating field of the instrument and the stray fields are (A) perpendicular (B) parallel (C) inclined at 60%

MCQ 3.52

A reading of 120 is obtained when standard inductor was connected in the circuit of a Q-meter and the variable capacitor is adjusted to value of 300 pF. A lossless capacitor of unknown value Cx is then connected in parallel with the variable capacitor and the same reading was obtained when the variable capacitor is readjusted to a value of 200 pF. The value of Cx in pF is (A) 100 (B) 200 (C) 300 (D) 500 YEAR 2003

MCQ 3.53

(D) inclined at 30%

A I D

O N

n

.i .co

The simplified block diagram of a 10-bit dia A/D converter of dual slope integrator o type is shown in figure. The 10-bit .n counter at the output is clocked by a 1 MHz w w overhead for the control logic, the maximum clock. Assuming negligible w timing frequency of the analog signal that can be converted using this A/D converter is approximately

(A) 2 kHz (C) 500 Hz MCQ 3.54

TWO MARKS

(B) 1 kHz (D) 250 Hz

The items in Group-I represent the various types of measurements to be made with a reasonable accuracy using a suitable bridge. The items in Group-II represent the various bridges available for this purpose. Select the correct choice of the item in Group-II for the corresponding item in Group-I from the following List-I

List-II

P.

Resistance in the milli-ohm 1. range

Wheatstone Bridge

Q.

Low values of Capacitance

R.

Comparison of resistance which 3. are nearly equal

2.

Kelvin Double Bridge Schering Bridge

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S.

Inductance of a coil with a 4. large time-constant

Codes : (A) P=2, Q=3, R=6, S=5 (C) P=2, Q= 3, R=5, S=4 MCQ 3.55

Wien’s Bridge

5.

Hay’s Bridge

6.

Carey-Foster Bridge (B) P=2, Q=6, R=4, S=5 (D) P=1, Q=3, R=2, S=6

A rectifier type ac voltmeter of a series resistance Rs , an ideal full-wave rectifier bridge and a PMMC instrument as shown in figure. The internal. resistance of the instrument is 100 W and a full scale deflection is produced by a dc current of 1 mA. The value of Rs required to obtain full scale deflection with an ac voltage of 100 V (rms) applied to the input terminals is

(A) 63.56 W (C) 89.93 W MCQ 3.56

Page 323

O N

A I D

(B) 69.93 W (D) 141.3 kW

n o.i c . A wattmeter reads 400 W when its current dia coil is connected in the R-phase and o its pressure coil is connected between .n this phase and the neutral of a symmetriw cal 3-phase system supplyingwawbalanced star connected 0.8 p.f. inductive load.

This phase sequence is RYB. What will be the reading of this wattmeter if its pressure coil alone is reconnected between the B and Y phases, all other connections remaining as before ? (A) 400.0 (B) 519.6 (C) 300.0 (D) 692.8 MCQ 3.57

The inductance of a certain moving-iron ammeter is expressed as L = 10 + 3q - (q2 /4) mH , where q is the deflection in radians from the zero position. The control spring torque is 25 # 10 - 6 Nm/radian. The deflection of the pointer in radian when the meter carries a current of 5 A, is (A) 2.4 (B) 2.0 (C) 1.2 (D) 1.0

MCQ 3.58

A 500A/5A, 50 Hz transformer has a bar primary. The secondary burden is a pure resistance of 1 W and it draws a current of 5 A. If the magnetic core requires 250 AT for magnetization, the percentage ratio error is (A) 10.56 (B) - 10.56 (C) 11.80 (D) - 11.80

MCQ 3.59

The voltage-flux adjustment of a certain 1-phase 220 V induction watt-hour meter is altered so that the phase angle between the applied voltage and the flux due to it is 85c(instead of 90c). The errors introduced in the reading of this meter when the current is 5 A at power factor of unity and 0.5 lagging are

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respectively (A) 3.8 mW, 77.4 mW (C) - 4.2 W, - 85.1 W MCQ 3.60

Page 324

(B) - 3.8 mW, - 77.4 mW (D) 4.2 W, 85.1 W

Group-II represents the figures obtained on a CRO screen when the voltage signals Vx = Vxm sin wt and Vy = Vym sin (wt + F) are given to its X and Y plates respectively and F is changed. Choose the correct value of F from Group-I to match with the corresponding figure of Group-II. Group-I Group-II P. F = 0

Q. F = p/2

R. p < F < 3p/2

A I D

S. F = 3p/2

O N

Codes : .no w (A) P=1, Q= 3, R=6, S=5ww (C) P=2, Q= 3, R=5, S=4

.in

co ia.

d

(B) P=2, Q= 6, R=4, S=5 (D) P=1, Q=5, R=6, S=4

YEAR 2002

ONE MARK

MCQ 3.61

Two in-phase, 50 Hz sinusoidal waveforms of unit amplitude are fed into channel-1 and channel-2 respectively of an oscilloscope. Assuming that the voltage scale, time scale and other settings are exactly the same for both the channels, what would be observed if the oscilloscope is operated in X-Y mode ? (A) A circle of unit radius (B) An ellipse (C) A parabola (D) A straight line inclined at 45c with respect to the x-axis.

MCQ 3.62

The line-to-line input voltage to the 3-phase, 50 Hz, ac circuit shown in Figure is 100 V rms. Assuming that the phase sequence is RYB, the wattmeters would read.

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Page 325

(B) W1 =500 W and W2 =500 W (D) W1 =250 W and W2 =750 W

(A) W1 =886 W and W2 =886 W (C) W1 =0 W and W2 =1000 W YEAR 2001

ONE MARK

MCQ 3.63

If an energy meter disc makes 10 revolutions in 100 seconds when a load of 450 W is connected to it, the meter constant (in rev/kWh) is (A) 1000 (B) 500 (C) 1600 (D) 800

MCQ 3.64

The minimum number of wattmeter(s) required to measure 3-phase, 2-wire balanced or unbalanced power is (A) 1 (B) 2 in (C) 3 co. 4 .(D)

MCQ 3.65

A I D

(A) 20.0 (C) 25.0

MCQ 3.66

O N

dia o A 100 mA ammeter has an internal.nresistance of 100 W. For extending its range w to measure 500 mA , the shunt required is of resistance (in W) ww (B) 22.22 (D) 50.0

Resistance R1 and R2 have, respectively, nominal values of 10 W and 5 W, and tolerance of ! 5% and ! 10% . The range of values for the parallel combination of R1 and R2 is (A) 3.077 W to 3.636 W (B) 2.805 W to 3.371 W (C) 3.237 W to 3.678 W (D) 3.192 W to 3.435 W ***********

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SOLUTIONS

SOL 3.1

Option (B) is correct. For an ideal voltmeter interval resistance is always zero. So we can apply the KVL along the two voltmeters as or

SOL 3.2

Page 326

V - V1 - V2 = 0 V = V1 + V2

Option (A) is correct. For the + ve half cycle of I/p voltage, diode will be forward biased (Vg = 0 , ideal diode) Therefore, the voltmeter will be short circuited and reads (for + ve half cycle) V1 = 0 volt Now, for - ve half cycle, diode will be reverse biased and treated as open cirn cuit. So, the voltmeter reads the voltage across o.i 100 kW. Which is given by c . 14.14 di0ac V2 = 100 # 100 .no+ 1 w w So, V2,rms = 14wvolt 2 Therefore, the average voltage for the whole time period is obtained as V + V2, rms 0 + ^14/ 2 h Vave = 1 = = 14 2 2 2 2 = 4.94 . 4.46 volt

A I D

O N

SOL 3.3

Option (C) is correct. The quality factor of the inductances are given by q 1 = wL 1 R1 and q 2 = wL 2 R2 So, in series circuit, the effective quality factor is given by XLeq = wL 1 + wL 2 Q = Req R1 + R 2 w L 1 + wL 2 R = 1 R 2 R1 R 2 1 + 1 R 2 R1 q1 q + 2 R R 2 2 = 1 + 1 R 2 R1

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=

Page 327

q1 R1 + q 2 R 2 R1 + R 2

SOL 3.4

Option (C) is correct.

SOL 3.5

Option (A) is correct. PMMC instrument reads average (dc) value. 20 T 1 v (t) dt Vavg = 1 v (t) dt = 3 T 0 20 # 10 0

#

#

10 = 1 ; tdt + 20 0

#

20

12

# (- 5) dt + # 5dtE 10

12

2 10

20 = 1 c :t D - 5 6t @12 + 5 6t @12 m 10 20 2 0

SOL 3.6

SOL 3.7

= 1 [50 - 5 (2) + 5 (8)] = 80 = 4 V 20 20 Option (A) is correct. Heaviside mutual inductance bridge measures mutual inductance is terms of a known self-inductance. Option (C) is correct. Let wt = q , we have instaneous voltage and current as follows. v (t) = E1 sin q + E 3 sin 3q i (t) = I1 sin (q - f1) + I 3 sin (3q - f3) + I5 sin (5q) We know that wattmeter reads the average power, which is gives as 2p ...(i) P = 1 v (t) i (t) d.qin 2p 0 o c . We can solve this integration using following dia results. o n 2p w.b) dq = 1 AB cos (a - b) (i) 1 A sin (q + a):B sin (w q+ 2p 0 2 w

A I D

O N

#

#

SOL 3.8

2p

(ii) 1 2p

#

(iii) 1 2p

# A sin (mq + a):B cos (nq + b) dq = 0

(iv) 1 2p

# A sin (mq + a):B cos (nq + b) dq = 0

0

A sin (q + a):B cos (q + a) dq = 1 AB sin (a - b) 2

2p

0

2p

0

Result (iii) and (iv) implies that power is transferred between same harmonics of voltages and currents. Thus integration of equation (i) gives. P = 1 E1 I1 cos f + 1 E 3 I 3 cos f3 2 2 Option (D) is correct. A voltmeter with a multiplier is shown in figure below.

Here

Im = Fully scale deflection current of meter.

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Rm Rs V Vm V V Vm Here when, So,

Page 328

= Internal resistance of meter = Voltage across the meter = Full range voltage of instrument = Im Rm = Im (Rm + Rs) = R m + Rs = 1 + Rs Rm Rm

Rs1 = 20 kW , Vm1 = 440 V V = 1 + 20k 440 Rm

...(i)

Rs2 = 80 kW , Vm2 = 352 V V = 1 + 80 k So, ...(ii) 352 Rm Solving equation (i) and (ii), we get V = 480 V , Rm = 220 kW So when Rs3 = 40 kW , Vm3 = ? 480 = 1 + 40 k & V - 406 V m2 Vm3 220 k Option (A) is correct. The compensating coil compensation the effect of impedance of current coil. When,

SOL 3.9

SOL 3.10

A I D

Option (C) is correct. Let Z1 = R1 || jwC1 .in so admittance Y1 = 1 = 1 + jwCa1 .co Z1 R 1 di

O N

o

.Zn4 = R 4 Z2 = R2 and w w Let ZX = RXw + jwLX (Unknown impedance) For current balance condition of the Bridge Z 2 Z 4 = Z X Z1 = Z X Y1 Let

ZX = Z2 Z 4 Y1

R X + jw L X = R 2 R 4 b 1 + jw C 1 l R1 Equating imaginary and real parts RX = R2 R 4 and LX = R2 R 4 C1 R1 Quality factor of inductance which is being measured Q = wL X = wR 1 C 1 RX From above equation we can see that for measuring high values of Q we need a large value of resistance R 4 which is not suitable. This bridge is used for measuring low Q coils. Note: We can observe directly that this is a maxwell’s bridge which is suitable for low values of Q (i.e. Q < 10 ) SOL 3.11

Option (C) is correct. In the alternate mode it switches between channel A and channel B, letting each through for one cycle of horizontal sweep as shown in the figure.

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SOL 3.12

Page 329

A I D

O N

.in

co ia.

Option (C) is correct. od n 1 . to 199.99 So error of 10 counts is equal to 4 digit display will read from 000.00 ww 2 w = ! 0.10 V For 100 V, the maximum error is e = ! (100 # 0.002 + 0.1) = ! 0.3 V

Percentage error SOL 3.13

= ! 0.3 # 100 % = ! 0.3 % of reading 100 Option (D) is correct. Since potential coil is applied across Z2 as shown below

Wattmeter read power consumed by Z2 SOL 3.14

Option (D) is correct. Given that full scale current is 5 A

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SOL 3.15

Page 330

Current in shunt Il = IR - I fs = 25 - 5 = 20 A 20 # Rsh = 5 # 0.2 Rsh = 1 = .05 W 20 Option (A) is correct. Overall gain of the system is 100 g = = 10 (zero error) 1 + 100 b 9 l 100 Gain with error 110 100 + 10% g = = 10.091 = 110 9 #9 1+ 1 + (100 + 10%) b l 100 100

A I D

error 3 g = 10.091 - 10 - 0.1 Similarly 90 100 - 10% g = = 9 1 + (100 - 10%) 1 + 90 # 9 100 100 = 9.89 n o.i c error 3 g = 9.89 - 10 -- 0.1 . dia o So gain g = 10 ! 0.1 = 10 ! 1 % .n SOL 3.16

O N

Option (A) is correct. At balance condition

w

ww

-j = R2 R 3 wC 4 m - jR 4 wC 4 (R + jwL) = R2 R 3 j c R 4 - wC 4 m - jRR 4 wLR 4 + = R2 R 3 R 4 wC 4 wC 4 - jRR 4 LR 4 + = R2 R 3 R 4 wC 4 C4 Comparing real & imaginary parts. RR 4 = R2 R 3 wC 4 wC 4 R = R2 R 3 R4 Similarly, LR 4 = R R R 2 3 4 C4 (R + jwL) c R 4
0 i.e., VY is a higher voltage than VZ So, the diode will be in cutoff region. Therefore, there will no voltage difference between X and W node. i.e., VWX = 0 Now, for - ve half cycle all the four diodes will active and so, X and W terminal is short circuited i.e., VWX = 0 Hence, VWX = 0 for all t SOL 3.8

Option (B) is correct.

A I D

From the circuit, we have

Is = I Z + I L or, (1) I Z = Is - I L n i Since, voltage across zener diode is 5 V so, ocurrent through 100 W resistor is . c . a obtained as di o 10 5 n . A Is = = 0.05 100 ww w Therefore, the load current is given by IL = 5 RL Since, for proper operation, we must have IZ $ Iknes So, from Eq. (1), we write 0.05 A - 5 $ 10 mA RL 50 mA - 5 $ 10 mA RL 40 mA $ 5 RL 40 # 10-3 $ 5 RL RL 1 -3 # 5 40 # 10 5 # RL 40 # 10-3 or, 125 W # RL Therefore, minimum value of RL = 125 W Now, we know that power rating of Zener diode is given by PR = VZ IZ^maxh IZ^maxh is maximum current through zener diode in reverse bias. Maximum currrent through zener diode flows when load current is zero. i.e.,

O N

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IZ^maxh = Is = 10 - 5 = 0.05 100 PR = 5 # 0.05 W = 250 mW

Therefore, SOL 3.9

SOL 3.10

SOL 3.11

Page 394

Option (A) is correct. Prime implicants are the terms that we get by solving K-map

F = XY + XY 1prime 44 2 44 3 implicants

Option (D) is correct. Let v > 0.7 V and diode is forward biased. Applying Kirchoff’s voltage law 10 - i # 1k - v = 0 10 - :v - 0.7 D (1000) - v = 0 500 10 - (v - 0.7) # 2 - v = 0 v = 11.4 = 3.8 V > 0.7 3 So, i = v - 0.7 = 3.8 - 0.7 = 6.2 mA 500 500 Option (B) is correct.

A I D

O N

no w.

ww

(Assumption is true)

.in

co ia.

d

Y = 1, when A > B

A = a1 a 0, B = b1 b 0

a1

a0

b1

b0

Y

0

1

0

0

1

1

0

0

0

1

1

0

0

1

1

1

1

0

0

1

1

1

0

1

1

1

1

1

0

1

Total combination = 6 SOL 3.12

Option (A) is correct. The given circuit is

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Page 395

Condition for the race-around It occurs when the output of the circuit (Y1, Y2) oscillates between ‘0’ and ‘1’ checking it from the options. 1. Option (A): When CLK = 0 Output of the NAND gate will be A1 = B1 = 0 = 1. Due to these input to the next NAND gate, Y2 = Y1 : 1 = Y1 and Y1 = Y2 : 1 = Y2 . If Y1 = 0 , Y2 = Y1 = 1 and it will remain the same and doesn’t oscillate. If Y2 = 0 , Y1 = Y2 = 1 and it will also remain the same for the clock period. So, it won’t oscillate for CLK = 0 . So, here race around doesn’t occur for the condition CLK = 0 . 2. Option (C): When CLK = 1, A = B = 1 A1 = B1 = 0 and so Y1 = Y2 = 1 And it will remain same for the clock period. So race around doesn’t occur for the condition. 3. Option (D): When CLK = 1, A = B = 0 So, A1 = B1 = 1 And again as described for Option (B) race around doesn’t occur for the condition. So, Option (A) will be correct. SOL 3.13

A I D

Option (D) is correct. DC Analysis :

O N

no w.

.in

co ia.

d

ww

Using KVL in input loop, VC - 100IB - 0.7 = 0 VC = 100IB + 0.7 IC - IE = 13.7 - VC = (b + 1) IB 12k 13.7 - VC = 100I B 12 # 103 Solving equation (i) and (ii),

...(i)

...(ii)

IB = 0.01 mA Small Signal Analysis : Transforming given input voltage source into equivalent current source.

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This is a shunt-shunt feedback amplifier. Given parameters, rp = VT = 25 mV = 2.5 kW IB 0.01 mA b 100 = 0.04 s gm = = rp 2.5 # 1000 Writing KCL at output node v0 + g v + v0 - vp = 0 m p RC RF v 0 : 1 + 1 D + v p :gm - 1 D = 0 RC RF RF Substituting RC = 12 kW, RF = 100 kW, gm = 0.04 s v 0 (9.33 # 10-5) + v p (0.04) = 0 v 0 =- 428.72Vp Writing KCL at input node vi = v p + v p + v p - vo = v 1 + 1 + 1 - v 0 p: Rs Rs rp RF Rs rp RF D RF = v p (5.1 # 10-4) - v 0 RF Substituting Vp from equation (i)

Page 396

...(i)

A I D

vi = - 5.1 # 10-4 v - v 0 0 428.72 Rs RF

SOL 3.14

O N

vi -6 -5 (source resistance) R = 10 kW 3 =- 1.16 # 10 v 0 - 1 # 10 .ivn 10 # 10 o 0 s c . ia vi -5 d 1 . 116 10 =o # .n 10 # 103 w 1 ww Av = v 0 = - 8.96 3 vi 10 # 10 # 1.116 # 10-5 Option (D) is correct. Let Qn + 1 is next state and Qn is the present state. From the given below figure.

If A = 0, If A = 1, So state diagram is

SOL 3.15

D = Y = AX 0 + AX1 Qn + 1 = D = AX 0 + AX1 Qn + 1 = A Qn + AQn Qn + 1 = Qn Qn + 1 = Qn

X 0 = Q , X1 = Q (toggle of previous state)

Option (B) is correct. First we obtain the transfer function.

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Page 397

0 - Vi (jw) 0 - Vo (jw) =0 + 1 +R R2 1 jw C Vo (jw) - Vi (jw) = 1 +R R2 1 jw C Vo (jw) =-

Vi (jw) R2 R1 - j 1 wC

1 " 3, so V = 0 o wC 1 " 0, so V (jw) =- R2 V (jw) At w " 3 (higher frequencies), o R1 i wC The filter passes high frequencies so it is a high pass filter. H (jw) = Vo = - R2 Vi R1 - j 1 wC At w " 0 (Low frequencies),

A I D

O N

in

. 2co . H (3) = - R2 =iR a R1 d R1

no 2. times of maximum gain 6H (3)@ w w

At 3 dB frequency, gain will be

So,

w H ^ jw0h = 1 H (3) 2 R2 = 1 b R2 l 2 R1 R 12 + 21 2 w0 C 2R 12 = R 12 + w0 =

SOL 3.16

Option (D) is correct.

1 & R2 = 1 1 w C2 w 2C 2 2 0

1 R1 C

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A I D

O N

So, it will act as a Band pass filter. SOL 3.17

Option (D) is correct.

Page 398

no w.

.in

co ia.

d

ww

The first half of the circuit is a differential amplifier (negative feedback) Va =- (Vi) Second op-amp has a positive feedback, so it acts as an schmitt trigger. Since Va =- Vi this is a non-inverting schmitt trigger. Threshold value VTH = 12 = 6 V 2 VTL =- 6 V SOL 3.18

Option (A) is correct.

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Y = X5X = X X + XX = XX + X X = X+X = 1 SOL 3.19

Option (C) is correct. LXI D, DISP LP : CALL SUB LP + 3 When CALL SUB is executed LP+3 value is pushed(inserted) in the stack. POP H & HL = LP + 3 DAD D

& HL = HL + DE = LP + 3 + DE

PUSH H SOL 3.20

& The last two value of the stack will be HL value i.e, LP + DISP + 3

Option (D) is correct. Zener Diode is used as stabilizer. The circuit is assumed to be as

A I D

O N

no w.

.in

co ia.

d

ww

We can see that both BE and BC Junction are forwarded biased. So the BJT is operating in saturation. Collector current IC = 12 - 0.2 = 5.36 mA 2.2k Y bIB Note:- In saturation mode IC SOL 3.21

Option (C) is correct. The characteristics equation of the JK flip-flop is Q n + 1 = JQ n + KQn From figure it is clear that

Qn + 1 is the next state

J = QB ; K = QB The output of JK flip flop QA (n + 1) = QB QA + QB QA = QB (QA + QA) = QB Output of T flip-flop QB (n + 1) = Q A

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SOL 3.22

Page 400

Clock pulse

QA

QB

QA (n + 1)

QB (n + 1)

Initially(tn )

1

0

1

0

tn + 1

1

0

1

0

tn + 2

1

0

1

0

tn + 3

1

0

1

0

Option (C) is correct. We can obtain three operating regions depending on whether the Zener and PN diodes are forward biased or reversed biased. 1. vi #- 0.7 V , zener diode becomes forward biased and diode D will be off so the equivalent circuit looks like

A I D

The output vo =- 0.7 V 2. When - 0.7 1 vi # 5.7 , both zener and diode D will be off. The circuit is

O N

no w.

.in

co ia.

d

ww

Output follows input i.e vo = vi Note that zener goes in reverse breakdown(i.e acts as a constant battery) only when difference between its p-n junction voltages exceeds 10 V. 3. When vi > 5.7 V , the diode D will be forward biased and zener remains off, the equivalent circuit is

vo = 5 + 0.7 = 5.7 V SOL 3.23

Option (B) is correct. Since the op-amp is ideal v+ = v- =+ 2 volt By writing node equation v- - 0 + v- - vo = 0 R 2R

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Page 401

2 + (2 - vo) = 0 R 2R 4 + 2 - vo = 0 vo = 6 volt SOL 3.24

Option (B) is correct. Given circuit is,

We can observe that diode D2 is always off, whether D1 ,is on or off. So equivalent circuit is.

O N

A I D

D1 is ON in this condition and

no w10.

.in

co ia.

d

V0 = ww10 + 10 # 10 = 5 volt

SOL 3.25

Option (A) is correct. By writing KVL equation for input loop (Base emitter loop) 10 - (10 kW) IB - VBE - V0 = 0 Emitter current IE = V0 100

...(1)

IC - IE = bIB V0 = 100I B 100 V0 IB = 10 # 103

So,

Put IB into equation (1) 10 - (10 # 103)

& SOL 3.26

V0 - 0.7 - V0 = 0 10 # 103 9.3 - 2V0 = 0 V0 = 9.3 = 4.65 A 2

Option (A) is correct. The circuit is

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Page 402

Output Y is written as Y = X5B Since each gate has a propagation delay of 10 ns.

SOL 3.27

A I D

O N

.in

Option (D) is correct. .co a i d CALL, Address performs two operations no . w (1) PUSH PC & Save the ww contents of PC (Program Counter) into stack. SP = SP - 2 (decrement) ((SP)) ! (PC) (2) Addr stored in PC.

(PC) ! Addr SOL 3.28

Option (B) is correct. Function F can be minimized by grouping of all 1’s in K-map as following.

F = X Y + YZ SOL 3.29

Option (D) is correct. Since F = X Y + YZ In option (D)

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Page 403

SOL 3.30

Option (A) is correct. Figure shows current characteristic of diode during switching.

SOL 3.31

Option (B) is correct. The increasing order of speed is as following Magnetic tape> CD-ROM> Dynamic RAM>Cache Memory>Processor register

SOL 3.32

Option (B) is correct. Equivalent circuit of given amplifier

A I D

O N

.in o c . Feedback samples output voltage and iadds a a negative d o input. .n w So, it is a voltage-voltage feedback. ww

feedback voltage (vfb) to

SOL 3.33

Option () is correct. NOR and NAND gates considered as universal gates.

SOL 3.34

Option (A) is correct. Let voltages at positive and negative terminals of op-amp are V+ and V- respectively, then V+ = V- = Vs (ideal op-amp) In the circuit we have, V- - 0 + V- - V0 (s) = 0 1 R ` Cs j (RCs) V- + V- - V0 (s) = 0 (1 + RCs) Vs = V0 (s) Similarly current Is is, Is = Vs - V0 R Is = RCs Vs R Is = jwCVs Is = wCVs + + 90% Is = 2pf # 10 # 10 - 6 # 10 Is = 2 # p # 50 # 10 # 10 - 6 # 10

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Page 404

Is = 10p mA, leading by 90% SOL 3.35

Option (D) is correct. Input and output power of a transformer is same Pin = Pout for emitter follower, voltage gain (A v) = 1 current gain (Ai) > 1 Power (Pout) = Av Ai Pin Since emitter follower has a high current gain so Pout > Pin

SOL 3.36

Option (D) is correct. For the given instruction set, XRA A & XOR A with A & A = 0 MVI B, F0 H&B = F0 H SUB B &A = A - B A B 2’s complement of (- B) A + (- B) = A - B

SOL 3.37

= 00000000 = 1111 0 0 0 0 = 0 0 010 0 0 0 = 0 0 010 0 0 0 = 10 H

A I D

Option (D) is correct. This is a schmitt trigger circuit, output can takes two states only. n VOH =+ 6 volt o.i c . VOL =- 3 volt dia o .n terminals of op-amp is given as Threshold voltages at non-inverting w w VTH - 6 + VTH - 0w= 0 2 1

O N

3VTH - 6 = 0 VTH = 2 V (Upper threshold)

Similarly VTL - (- 3) VTL =0 + 2 1 3VTL + 3 = 0 VTL =- 1 V (Lower threshold) For Vin < 2 Volt, V0 =+ 6 Volt Vin > 2 Volt, V0 =- 3 Volt Vin < - 1 Volt V0 =+ 6 Volt Vin > - 1 Volt V0 =- 3 Volt Output waveform

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SOL 3.38

Page 405

Option (A) is correct. Assume the diode is in reverse bias so equivalent circuit is

A I D

V0 = 10 sin wt # 10 = 5 sin wt 10 + 10 Due to resistor divider, voltage across diode VD < 0 (always). So it in reverse bias for given input. Output, V0 = 5 sin wt Output voltage

SOL 3.39

O N

Option (C) is correct.

no w.

.in

co ia.

d

ww

This is a current mirror circuit. Since b is high so IC1 = IC2, IB1 = IB2 VB = (- 5 + 0.7) =- 4.3 volt Diode D1 is forward biased. So, current I is, I = IC2 = IC1 0 - (- 4.3) = = 4.3 mA 1 SOL 3.40

Option (D) is correct. In positive half cycle of input, diode D1 is in forward bias and D2 is off, the equivalent circuit is

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Page 406

Capacitor C1 will charge upto + 5 volt. VC1 =+ 5 volt In negative halt cycle diode D1 is off and D2 is on.

Now capacitor VC2 will charge upto - 10 volt in opposite direction. SOL 3.41

Option () is correct. Let input Vin is a sine wave shown below

A I D

O N

.in

co ia.

d

According to given transfer characteristics of rectifiers output of rectifier P is. .no

w

ww

Similarly output of rectifier Q is

Output of a full wave rectifier is given as

To get output V0

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Page 407

V0 = K (- VP + VQ) K - gain of op-amp So, P should connected at inverting terminal of op-amp and Q with non-inverting terminal. SOL 3.42

Option () is correct.

SOL 3.43

Option (C) is correct. For low frequencies, w " 0 , so 1 " 3 wC Equivalent circuit is,

A I D

Applying node equation at positive and negative input terminals of op-amp. vA - vi + vA - vo = 0 R1 R2

O N

2vA = vi + vo ,

Similarly,

.in

a R1 = R 2 = R A

co ia.

vA - vi + vA - 0 = 0 d R3 R4 no . 2vA = w vinw,

w

a R 3 = R 4 = RB

So, vo = 0 It will stop low frequency signals. For high frequencies, w " 3 , then 1 " 0 wC Equivalent circuit is,

Output, vo = vi So it will pass high frequency signal. This is a high pass filter. SOL 3.44

Option (D) is correct. In Q.7.21 cutoff frequency of high pass filter is given by, 1 wh = 2pRA C

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Page 408

Here given circuit is a low pass filter with cutoff frequency, 1 2 wL = = R 2 R p A AC 2p C 2 wL = 2wh When both the circuits are connected together, equivalent circuit is,

So this is is Band pass filter, amplitude response is given by.

SOL 3.45

Option (B) is correct. In SOP form, F is written as F = Sm (1, 3, 5, 6) = X Y Z + X YZ + XY Z + XYZ Solving from K- map

A I D

O N

.in

co ia.

od

F = XZ + Y Z.n+ XYZ w w In POS form F = (Yw + Z) (X + Z) (X + Y + Z ) Since all outputs are active low so each input in above expression is complemented F = (Y + Z ) (X + Z ) (X + Y + Z)

SOL 3.46

Option (B) is correct. Given that SP = 2700 H PC = 2100 H HL = 0000 H Executing given instruction set in following steps, DAD SP & Add register pair (SP) to HL register HL = HL + SP HL = 0000 H + 2700 H HL = 2700 H PCHL & Load program counter with HL contents PC = HL = 2700 H So after execution contents are, PC = 2700 H, HL = 2700 H

SOL 3.47

Option (D) is correct. If transistor is in normal active region, base current can be calculated as following, By applying KVL for input loop, 10 - IC (1 # 103) - 0.7 - 270 # 103 IB = 0

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Page 409

bIB + 270 IB = 9.3 mA, ` IC = bIB IB (b + 270) = 9.3 mA IB = 9.3 mA = 0.025 mA 270 + 100 In saturation, base current is given by, 10 - IC (1) - VCE - IE (1) = 0 10 = I C (sat) 2

IC - IE VCE - 0

IC (sat) = 5 mA IC (sat) = 5 = .050 mA b 100 IB 1 IB(sat), so transistor is in forward active region. IB(sat) =

SOL 3.48

Option (B) is correct. In the circuit

A I D

O N

We can analyze that the transistor is operating in active region. .in VBE(ON) = 0.6 volt .co

ia

d VB - VE = 0.6 no . w 6.6 - VE = 0.6 ww VE = 6.6 - 0.6 = 6 volt At emitter (by applying KCL),

IE = IB + IL IE = 6 - 6.6 + 6 - 0.6 amp 1 kW 10 W VCE = VC - VE = 10 - 6 = 4 volt Power dissipated in transistor is given by. PT = VCE # IC = 4 # 0.6 = 2.4 W SOL 3.49

` IC - IE = 0.6 amp

Option (D) is correct. This is a voltage-to-current converter circuit. Output current depends on input voltage.

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Page 410

Since op-amp is ideal v+ = v- = v1 Writing node equation. v1 - v + v1 - 0 = 0 R1 R2 v1 c R1 + R2 m = V R1 R1 R2

R2 R1 + R2 m Since the op-amp is ideal therefore iL = i1 = v1 = V c R2 m r R1 + R2 r Option (D) is correct. In the circuit output Y is given as v1 = V c

SOL 3.50

Y = [A 5 B] 5 [C 5 D] Output Y will be 1 if no. of 1’s in the input is odd. SOL 3.51

Option () is correct. This is a class-B amplifier whose efficiency is given as h = p VP 4 VCC

A I D

where VP " peak value of input signal VCC " supply voltage here VP = 7 volt, VCC = 10 volt so, SOL 3.52

O N

in

. co100 . h = p # 7ia# = 54.95% - 55% 10 4 d .no

Option (B) is correct. ww w In the circuit the capacitor starts charging from 0 V (as switch was initially closed) towards a steady state value of 20 V. for t < 0 (initial) for t " 3 (steady state)

So at any time t , voltage across capacitor (i.e. at inverting terminal of op-amp) is given by vc (t) = vc (3) + [vc (0) - vc (3)] e

-t RC

-t RC

vc (t) = 20 (1 - e ) Voltage at positive terminal of op-amp v+ - vout v+ - 0 =0 + 10 100 v+ = 10 vout 11 Due to zener diodes, - 5 # vout # + 5 So, v+ = 10 (5) V 11

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Page 411

Transistor form - 5 V to + 5 V occurs when capacitor charges upto v+ . So 20 (1 - e - t/RC ) = 10 # 5 11 1 - e - t/RC = 5 22 17 = e - t/RC 22 t = RC ln ` 22 j = 1 # 103 # .01 # 10 - 6 # 0.257 = 2.57 msec 17 Voltage waveforms in the circuit is shown below

SOL 3.53

A I D

Option (B) is correct. First convert the given number from hexadecimal to its binary equivalent, then binary to octal. Hexadecimal no. AB. CD 1 0 10 S 1 0 1 1 $ 1A BB 1 0B0C S 1 1 .0in 1 Binary equivalent S C A B .coD

O N

ia d o To convert in octal group three binary .n digits together as shown w 0 1 0 1 0 1 0 1 1 $ 1 1 0 0 1 1 SSSSSS ww 0 1 0 2

So, SOL 3.54

5

3

6

3

2

(AB.CD) H = (253.632) 8

Option (B) is correct. In a 555 astable multi vibrator circuit, charging of capacitor occurs through resistor (RA + RB) and discharging through resistor RB only. Time for charging and discharging is given as. TC = 0.693 (RA + RB) C = 0.693 RB C But in the given circuit the diode will go in the forward bias during charging, so the capacitor will charge through resistor RA only and discharge through RB only. a So

SOL 3.55

RA = RB TC = TD

Option (A) is correct. First we can check for diode D2 . Let diode D2 is OFF then the circuit is

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Page 412

In the above circuit diode D1 must be ON, as it is connected with 10 V battery now the circuit is

Because we assumed diode D2 OFF so voltage across it VD2 # 0 and it is possible only when D3 is off.

A I D

So, all assumptions are true. SOL 3.56

Option (D) is correct. In the positive half cycle of input, Diode D1 will be reverse biased and equivalent circuit is. .in

O N

no

. ww

co

. dia

w

Since there is no feed back to the op-amp and op-amp has a high open loop gain so it goes in saturation. Input is applied at inverting terminal so. VP =- VCC =- 12 V In negative half cycle of input, diode D1 is in forward bias and equivalent circuit is shown below.

Output VP = Vg + VOp-amp is at virtual ground so V+ = V- = 0 and VP = Vg = 0.7 V Voltage wave form at point P is

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SOL 3.57

Page 413

Option (A) is correct. In the circuit when Vi < 10 V, both D1 and D2 are off. So equivalent circuit is,

Output,

Vo = 10 volt

A I D

When Vi > 10 V (D1 is in forward bias and D2 is off So the equivalent circuit is,

O N

no w.

.in

co ia.

d

ww

Output, Vo = Vi Transfer characteristic of the circuit is

SOL 3.58

Option (B) is correct. Assume that BJT is in active region, thevenin equivalent of input circuit is obtained as

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Page 414

Vth - Vi + Vth - (- 12) = 0 15 100 20Vth - 20Vi + 3Vth + 36 23Vth Vth Thevenin resistance Rth

=0 = 20 # 5 - 36 , Vi = 5 V = 2.78 V = 15 KW || 100 KW = 13.04 KW

So the circuit is

Writing KVL for input loop

A I D

2.78 - Rth IB - 0.7 = 0 IB = 0.157 mA Current in saturation is given as, I IB(sat) = C(sat) b n o.i 12 . 2 c . IC(sat) = = 5.4 mA 2.2 o dia n mA .45 5 . w So, IB(sat) = ww 30 = 0.181 mA

O N

Since IB (sat) > IB , therefore assumption is true. SOL 3.59

Option (C) is correct. Here output of the multi vibrator is V0 = ! 12 volt Threshold voltage at positive terminal of op-amp can be obtained as following When output V0 =+ 12 V, equivalent circuit is,

writing node equation at positive terminal of op-amp Vth - 12 + Vth - 0 = 0 10 10

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Page 415

Vth = 6 volt (Positive threshold) So, the capacitor will charge upto 6 volt. When output V0 =- 12 V, the equivalent circuit is.

node equation Vth + 12 + Vth - 0 = 0 2 10 5 Vth + 60 + Vth = 0 Vth =- 10 volt (negative threshold) So the capacitor will discharge upto - 10 volt. At terminal P voltage waveform is.

A I D

O N

no w.

.in

co ia.

d

ww

SOL 3.60

Option () is correct.

SOL 3.61

Option () is correct.

SOL 3.62

Option (A) is correct. Function F can be obtain as, F = I0 S1 S0 + I1 S1 S0 + I2 S1 S0 + I3 S1 S0 = AB C + A B C + 1 $ BC + 0 $ BC = AB C + A BC + BC = AB C + A BC + BC (A + A) = AB C + A BC + ABC + A BC = S (1, 2, 4, 6)

SOL 3.63

SOL 3.64

Option (A) is correct. MVI H and MVI L stores the value 255 in H and L registers. DCR L decrements L by 1 and JNZ checks whether the value of L is zero or not. So DCR L executed 255 times till value of L becomes ‘0’. Then DCR H will be executed and it goes to ‘Loop’ again, since L is of 8 bit so no more decrement possible and it terminates. Option (A) is correct. XCHG & Exchange the contain of DE register pair with HL pair So now addresses of memory locations are stored in HL pair.

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INR M & pair. SOL 3.65

Page 416

Increment the contents of memory whose address is stored in HL

Option (A) is correct. From the circuit we can observe that Diode D1 must be in forward bias (since current is flowing through diode). Let assume that D2 is in reverse bias, so equivalent circuit is.

Voltage Vn is given by Vn = 1 # 2 = 2 Volt Vp = 0 Vn > Vp (so diode is in reverse bias, assumption is true) Current through D2 is ID2 = 0 SOL 3.66

SOL 3.67

SOL 3.68

A I D

Option (C) is correct. SHLD transfers contain of HL pair to memory location. SHLD 2050 & L " M [2050H] n H " M [2051H] o.i c Option (D) is correct. . ia = 2 V dVGS This is a N-channel MOSFET with o .n VTH =+ 1 Vww w VDS(sat) = VGS - VTH VDS(sat) = 2 - 1 = 1 V Due to 10 V source VDS > VDS(sat) so the NMOS goes in saturation, channel conductivity is high and a high current flows through drain to source and it acts as a short circuit. So, Vab = 0

O N

Option (C) is correct. Let the present state is Q(t), so input to D-flip flop is given by, D = Q (t) 5 X Next state can be obtained as,

and

Q (t + 1) = D = Q (t) 5 X = Q (t) X + Q (t) X = Q (t), if X = 1 Q (t + 1) = Q (t), if X = 0

So the circuit behaves as a T flip flop. SOL 3.69

Option (B) is correct. Since the transistor is operating in active region. IE . bIB

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Page 417

IB = IE = 1 mA = 10 mA b 100 SOL 3.70

Option (C) is correct. Gain of the inverting amplifier is given by, 6 Av =- RF =- 1 # 10 , RF = 1 MW R1 R1 6

R1 =- 1 # 10 Av Av =- 10 to - 25 so value of R1 6 R1 = 10 = 100 kW 10

for Av =- 10

6 R1' = 10 = 40 kW 25

for Av =- 25

R1 should be as large as possible so R1 = 100 kW SOL 3.71

Option (B) is correct. Direct coupled amplifiers or DC-coupled amplifiers provides gain at dc or very low frequency also.

SOL 3.72

Option (C) is correct. Since there is no feedback in the circuit and ideally op-amp has a very high value of open loop gain, so it goes into saturation (ouput is either + V or - V ) for small values of input. The input is applied to negative terminal of op-amp, so in positive half cycle it goes to + V . saturates to - V and in negative half cycleoit.in

SOL 3.73 CHECK

SOL 3.74

A I D

O N

a.c

Option (B) is correct. di o n From the given input output waveforms truth table for the circuit is drawn as w. w w X1 X2 Q 1 0 1 0 0 1 0 1 0 In option (A), for X1 = 1, Q = 0 so it is eliminated. In option (C), for X1 = 0, Q = 0 (always), so it is also eliminated. In option (D), for X1 = 0, Q = 1, which does not match the truth table. Only option (B) satisfies the truth table.

Option (D) is correct. In the given circuit NMOS Q1 and Q3 makes an inverter circuit. Q4 and Q5 are in parallel works as an OR circuit and Q2 is an output inverter. So output is Q = X1 + X2 = X1 .X2

SOL 3.75

Option (D) is correct. Let Q (t) is the present state then from the circuit,

So, the next state is given by

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Page 418

Q (t + 1) = Q (t) (unstable) SOL 3.76

Option (B) is correct. Trans-conductance of MOSFET is given by gm = 2iD 2VGS (2 - 1) mA = = 1 mS (2 - 1) V

SOL 3.77

Option (D) is correct. Voltage gain can be obtain by small signal equivalent circuit of given amplifier.

So, Voltage gain

vo =- gm vgs RD vgs = vin vo =- gm RD vin Av = vo =- gm RD vi

A I D

=- (1 mS) (10 kW) =- 10

SOL 3.78

O N

Option (C) is correct. Given circuit,

no w.

.in

co ia.

d

ww

In the circuit

SOL 3.79

V1 = 3.5 V (given) Current in zener is. IZ = V1 - VZ = 3.5 - 3.33 = 2 mA RZ 0.1 # 10 Option (C) is correct. This is a current mirror circuit. Since VBE is the same in both devices, and transistors are perfectly matched, then IB1 = IB2 and IC1 = IC2 From the circuit we have,

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IR = IC1 + IB1 + IB2 = IC1 + 2IB2 = IC2 + 2IC2 b IR = IC2 c1 + 2 m b IR IC2 = I = 2 c1 + b m IR can be calculate as

Page 419

a IB1 = IB2 a IC1 = IC2, IC2 = bIB2

A I D

IR = - 5 + 03.7 =- 4.3 mA 1 # 10 So, SOL 3.80

I =

4. 3 - 4.3 mA 2 1 + ` 100 j .in

O N

o

c Option (B) is correct. ia. d The small signal equivalent circuit.nofo given amplifier w

ww

Here the feedback circuit samples the output voltage and produces a feed back current Ifb which is in shunt with input signal. So this is a shunt-shunt feedback configuration. SOL 3.81

Option (A) is correct. In the given circuit output is stable for both 1 or 0. So it is a bistable multivibrator.

SOL 3.82

Option (A) is correct. Since there are two levels (+ VCC or - VCC ) of output in the given comparator circuit. For an n -bit Quantizer 2n = No. of levels 2n = 2 n =1

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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia SOL 3.83

Page 420

Option (C) is correct. From the circuit, we can see the that diode D2 must be in forward Bias. For D1 let assume it is in reverse bias. Voltages at p and n terminal of D1 is given by Vp and Vn Vp < Vn (D1 is reverse biased)

Applying node equation Vp - 5 Vp + 8 =0 + 1 1 2Vp =- 3 Vp =- 1.5 Vn = 0 Vp < Vn (so the assumption is true and D1 is in reverse bias) and current in D1 ID1 = 0 mA SOL 3.84

A I D

Option (D) is correct. The small signal ac equivalent circuit of given amplifier is as following.

O N

no w.

.in

co ia.

d

ww

RB = (10 kW < 10 kW) = 5 kW gm = 10 ms 50 a gm rp = b & rp = = 5 kW 10 # 10 - 3 Input resistance Here

Rin = RB < rp = 5 kW < 5 kW = 2.5 kW SOL 3.85

Option (D) is correct. For PMOS to be biased in non-saturation region. VSD < VSD(sat) and

So,

VSD(sat) = VSG + VT VSD(sat) = 4 - 1 = 3 Volt VSD < 3 VS - VD < 3 4 - ID R < 3 1 < ID R

"a VSG = 4 - 0 = 4 volt

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ID R > 1, R > 1000 W

Page 421

ID = 1 mA

SOL 3.86

Option () is correct.

SOL 3.87

Option (B is correct. If op-amp is ideal, no current will enter in op-amp. So current ix is v - vy ...(1) ix = x 1 # 106 (ideal op-amp) v+ = v- = vx vx - vy + vx - 0 3 = 0 3 100 # 10 10 # 10 vx - vy + 10vx = 0 11vx = vy For equation (1) & (2) ix = vx - 11v6 x =- 10v6x 1 # 10 10 Input impedance of the circuit. 6 Rin = vx =- 10 =- 100 kW 10 ix

SOL 3.88

Option (A) is correct. Given Boolean expression,

...(2)

A I D

O N

n Y = (A $ BC + D) (A $ D +.B o i $ C) c . = (A $ BCD) + (ABC dia $ B $ C ) + (AD) + B C D o .n + B C D = A BCD + AD w ww+ 1) + B C D = AD + B C D = AD (BC

SOL 3.89

Option (D) is correct. In the given circuit, output is given as. Y = (A0 5 B0) 9 (A1 5 B1) 9 (A2 5 B2) 9 (A3 5 B3) For option (A) Y = (1 5 1) 9 (0 5 0) 9 (1 5 1) 9 (0 5 0) = 0909090 = 1 For option (B) Y = (0 5 0) 9 (1 5 1) 9 (0 5 0) 9 (1 5 1) = 0909090 = 1 For option (C) Y = (0 5 0) 9 (0 5 0) 9 (1 5 1) 9 (0 5 0) = 0909090 = 1 For option (D) Y = (1 5 1) 9 (0 5 0) 9 (1 5 1) 9 (0 5 1) = 0909091 = 0

SOL 3.90

Option (B) is correct. In the given circuit, waveforms are given as,

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SOL 3.91

Page 422

Option (C) is correct. The program is executed in following steps. START MVI A, 14H " one instruction cycle. RLC & rotate accumulator left without carry RLC is executed 6 times till value of accumulator becomes zero. JNZ, JNZ checks whether accumulator value is zero or not, it is executed 5 times. HALT " 1-instruction cycle.

A I D

O N

no w.

.in

co ia.

d

ww

So total no. of instruction cycles are n = 1+6+5+1 = 13 SOL 3.92

Option (B) is correct. In the given circuit Vi = 0 V So, transistor Q1 is in cut-off region and Q2 is in saturation. 5 - IC RC - VCE(sat) - 1.25 = 0

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5 - IC RC - 0.1 - 1.25 = 0 5 - IC RC = 1.35 V0 = 1.35

Page 423

"a V0 = 5 - IC RC

A I D

SOL 3.93

Option (C) is correct. Since there exists a drain current for zero gate voltage (VGS = 0), so it is a depletion mode device. ID increases for negative values of gate voltages so it is a p-type depletion mode device.

SOL 3.94

Option (B) is correct. Applying KVL in input loop,

O N

d no . 4 - (33 # 10 ) IB - VBE - (3.3 #w 10w) IE = 0 w 3

3

.in

co ia.

4 - (33 # 103) IB - 0.7 - (3.3 # 103) (hfe + 1) IB = 0

a IE = (hfe + 1) IB

3.3 = 6(33 # 103) + (3.3 # 103) (99 + 1)@ IB 3.3 IB = 33 # 103 + 3.3 # 103 # 100 IC = hfe IB 99 # 3.3 3.3 = mA = mA [0.33 + 3.3] # 100 0.33 + 3.3

SOL 3.95

Option (D) is correct. Let the voltages at positive and negative terminals of op-amp are v+ and vrespectively. Then by applying nodal equations. v- - vin + v- - vout = 0 R1 R1 2 v-- = vin + vout

..(1)

Similarly, v+ - vin v -0 =0 + + R 1 c jwC m v+ - vin + v+ (jwCR) = 0 v+ (1 + jwCR) = Vin By equation (1) & (2)

..(2)

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2vin = vin + vout 1 + jwCR 2 - 1E = vout vin ; 1 + jwCR (1 - jwCR) vout = vin 1 + jwCR

Page 424

"a v+ = v- (ideal op-amp)

Phase shift in output is given by q = tan - 1 (- wCR) - tan - 1 (wCR) = p - tan - 1 (wCR) - tan - 1 (wCR) Maximum phase shift SOL 3.96

= p - 2 tan - 1 (wCR) q =p

Option (C) is correct. In given circuit MUX implements a 1-bit full adder, so output of MUX is given by. F = Sum = A 5 Q 5 Cin Truth table can be obtain as. Cin P Q Sum 0

0

0

0

0

1

0

1

0

0

1

1

1

0

0

1

0

1

1

1

1

1

0ww

A I D 0 1

O N

1 0

.in

co ia.

1

d no 0 . w

1

0 1

Sum = P Q Cin + PQ Cin + P Q Cin + P Q Cin Output of MUX can be written as F = P Q $ I0 + PQ $ I1 + PQ $ I2 + PQ $ I3 Inputs are, I0 = Cin, I1 = Cin, I2 = Cin, I3 = Cin SOL 3.97

Option (D) is correct. Program counter contains address of the instruction that is to be executed next.

SOL 3.98

Option (A) is correct. For a n -channel enhancement mode MOSFET transition point is given by, a VTH = 2 volt VDS (sat) = VGS - VTH VDS (sat) = VGS - 2 From the circuit, VDS = VGS So VDS (sat) = VDS - 2 & VDS = VDS (sat) + 2 VDS > VDS (sat) Therefore transistor is in saturation region and current equation is given by. ID = K (VGS - VTH ) 2 4 = K (VGS - 2) 2

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Page 425

VGS is given by VGS = VDS = 10 - ID RD = 10 - 4 # 1 = 6 Volt 4 = K (6 - 2) 2 K =1 4 ' ' Now RD is increased to 4 kW, Let current is ID' and voltages are VDS = VGS Applying current equation. So,

' ID' = K (VGS - VTH ) 2 ' ID' = 1 (VGS - 2) 2 4 ' ' VGS = VDS = 10 - ID' # RD' = 10 - 4ID'

So, 4ID' = (10 - 4ID' - 2) 2 = (8 - 4ID' ) 2 = 16 (2 - ID' ) 2 ID' = 4 (4 + I'D2 - 4ID' ) 4I'D2 - 17 + 16 = 0 I'D2 = 2.84 mA SOL 3.99

A I D

Option (D) is correct. Let the voltages at input terminals of op-amp are v- and v+ respectively. So, v+ = v- = 0 (ideal op-amp)

O N

no w.

.in

co ia.

d

ww

Applying node equation at negative terminal of op-amp, 0 - vin + 0 - vx = 0 1 10 At node x vx - 0 + vx - vout + vx - 0 = 0 10 10 1

From equation (1),

SOL 3.100

...(1)

vx + vx - vout + 10vx = 0 12 vx = vout vx = vout 12 vin + vx = 0 1 10 vin =- vout 120 vout =- 120 vin

Option (D) is correct. In the positive half cycle (when Vin > 4 V ) diode D2 conducts and D1 will be off so the equivalent circuit is,

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Page 426

Vout = + 4 Volt In the negative half cycle diode D1 conducts and D2 will be off so the circuit is,

A I D

Applying KVL

Vin - 10I + 4 - 10I = 0 Vin + 4 = I 20 Vin =- 10 V (Maximum value in negative half in cycle) . o c 3 mA So, I = - 10 +i4a.=20 10 d o n . Vin - Vout = w I 10 ww - 10 - Vout =- 3 10 10

O N

Vout =- (10 - 3) Vout =- 7 volt SOL 3.101

Option (C) is correct. In the circuit, the capacitor charges through resistor (RA + RB) and discharges through RB . Charging and discharging time is given as. TC = 0.693 (RA + RB) C TD = 0.693 RB C 1 1 f= 1 = = T TD + TC 0.693 (RA + 2RB) C 1 = 10 # 103 -9 0.693 (RA + 2RB) # 10 # 10

Frequency

14.4 # 103 = RA + 2RB duty cycle = TC = 0.75 T 0.693 (RA + RB) C =3 0.693 (RA + 2RB) C 4

...(1)

4RA + 4RB = 3RA + 6RB

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RA = 2RB

Page 427

...(2)

From (1) and (2) 2RA = 14.4 # 103 RA = 7.21 kW RB = 3.60 kW

and SOL 3.102

Option (B) is correct. Given boolean expression can be written as, F = XYZ + X Y Z + XY Z + XYZ + XYZ = X YZ + Y Z (X + X ) + XY (Z + Z) = XYZ + Y Z + XY = Y Z + Y (X + X Z ) a A + BC = (A + B) (A + C) = Y Z + Y (X + X ) (X + Z ) = Y Z + Y (X + Z ) = Y Z + YX + YZ

SOL 3.103

Option (B) is correct.

A I D

O N

.in

X = X1 5 X 0 , Y = X2 a.co di Serial Input Z = X 5 Y = [X1 5 X0].n 5oX2 w Truth table for the circuit can wwbe obtain as. Clock pulse

Serial Input

Shift register

Initially

1

1010

1

0

1101

2

0

0110

3

0

0011

4

1

0001

5

0

1000

6

1

0100

7 1 1010 So after 7 clock pulses contents of the shift register is 1010 again. SOL 3.104

Option (D) is correct. Characteristic table of the X-Y flip flop is obtained as. X

Y

Qn

Qn+1

0

0

0

1

0

0

1

1

0

1

0

0

0

1

1

1

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1

0

0

1

1

0

1

0

1

1

0

0

1

1

0

0

Page 428

Solving from k-map

Characteristic equation of X-Y flip flop is Qn + 1 = Y Qn + XQn Characteristic equation of a J-K flip-flop is given by Qn + 1 = KQn + J Qn by comparing above two characteristic equations J =Y, K=X SOL 3.105

A I D

Option (A) is correct. Total size of the memory system is given by.

O N

= (212 # 4) # 8 bits = 214 # 8 bits

.in

co ia.

= 214 Bytes o d n w. = 16 Kwbytes

SOL 3.106

w

Option (C) is correct. Executing all the instructions one by one. LXI H, 1FFE & H = (1F) H, L = (FE) H MOV B, M & B = Memory [HL] = Memory [1FFE] INR L & L = L + (1) H = (FF) H MOV A, M & A = Memory [HL] = Memory [1FFF] ADD B & A = A + B INR L & L = L + (1) H = (FF) H + (1) H = 00 MOV M, A & Memory [HL] = A Memory [1F00] = A XOR A & A = A XOR A = 0 So the result of addition is stored at memory address 1F00.

SOL 3.107

Option (D) is correct. Let the initial state Q(t) = 0, So D = Q = 1, the output waveform is.

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SOL 3.108

Page 429

So frequency of the output is, f f out = in = 10 = 5 kHz 2 2 Option (A) is correct. This is a half-wave rectifier circuit, so the DC voltage is given by Vdc = Vm p Equivalent circuit with forward resistance is

DC current in the circuit Idc Idc SOL 3.109

Vm (Vm /p) = p = rf + R (5 + 45) = Vm 50p

A I D

Option (B) is correct. In the positive half cycle zener diode (Dz ) will be in reverse bias (behaves as a constant voltage source) and diode (D) is in forward bias. So equivalent circuit for positive half cycle is. in

O N

o.

no

. ww

.c dia

w

Vo = VD + Vz = 0.7 + 3.3 = 4 Volt In the negative halt cycle, zener diode (Dz ) is in forward bias and diode (D) is in reverse bias mode. So equivalent circuit is. Output

So the peak output is, Vo =

10 # 1 (1 + 1)

Vo = 5 Volt SOL 3.110

Option (A) is correct.

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Page 430

For active low chip select CS = 0 , so the address range can be obtain as, A15 A14 A13 A12 A11 A10 A9 A8 A7 A6 A5 A4 A3 A2 A1 A0 1110 0000 0000 0000 h h h h 1110 1111 1111 1111 So address range is E000-EFFF SOL 3.111

Option (C) is correct. A first order low pass filter is shown in following figure.

Transfer function V0 (jw) 1 1 = # 1 = j w cR +1 V1 (jw) 1 j w C R+ jw C H (jw1) = 0.25 H (jw) =

Given that

1

4p2 f12 (50) 2 (5 SOL 3.112

SOL 3.113

A I D

=1 4 1 16 = w12 R2 C2 + 1

w12 C2 R2 +

O N

w12 R2 C2 -6 2

= 15

.in

co ia.

od

# 10 ) = 15 .n w ww f 1 = 2.46 kHz

Option (A) is correct. In the circuit, voltage at positive terminal of op-amp is given by v+ - vo v+ - 2 =0 + 10 3 3 (v+ - vo) + 10 (v+ - 2) = 0 13v+ = 20 + 3vo Output changes from + 15 V to - 15 V ,when v- > v+ 20 + (3 # 15) v+ = = 5 Volt (for positive half cycle) 13 Option (B) is correct. Output for each stage can be obtain as,

So final output Y is.

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Y = P Q $ R S = (P + Q) $ (R + S) = P+Q+R+S SOL 3.114

Page 431

a AB = A + B

Option (B) is correct. We can analyze that the transistor is in active region. b I IC = (b + 1) E 99 (1 mA) = 0.99 mA = (99 + 1) In the circuit

A I D

In the circuit VBE = 0.7 V VE = IE # 1 kW

O N

.in

co ia.

=1V

d VB - VE = 0.7 no . VB = 0.7 + 1w=w1.7 volt w Current throughR1 IR = VB = 1.7 = 100 mA 17 kW 17 kW IB = IE = 1 mA = 10 mA b+1 (99 + 1) Current through RF , by writing KCL at Base 1

IRF = IB + IR1 = 10 + 100 = 110 mA Current through RC I1 = IC + IRF = 0.99 mA + 110 mA = 1.1 mA SOL 3.115

Option (D) is correct. Output voltage V0 = 15 - I1 RC = 15 - (1.1 mA) (1 kW) = 13.9 V

SOL 3.116

Option (A) is correct. Current in RF IRF = V0 - VB RF 0.11 mA = 13.9 - 1.7 kW RF

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Page 432

RF = 110.9 kW SOL 3.117

Option (A) is correct. By writing node equations in the circuit

Va - Vx + V Cs + (V - V ) Cs = 0 a a y R Va (1 + 2RCs) - Vx - sCRVy = 0

or

(Vy - Va) Cs +

or

Vy =0 R

or Vy (1 + sCR) - Va sCR = 0 From equation (1) & (2) 1 + sCR c sCR m (1 + 2sCR) Vy - Vx - sCRVy = 0 (1 + sCR) (1 + 2sCR) Vy ; - sCR E = Vx sCR Vy

...(1)

...(2)

A I D

O N

(1 + 3sCR + 2s2 C2 R2 - s2 C2 R2) .in co= Vx sCR ia.

no w.

Transfer function

d

w Vy = w sCR 2 2 2 Vx 1 + 3sCR + s C R jwCR jwCR T (jw) = = 2 2 2 2 2 2 1 + j3wCR - C R w (1 - C R w ) + 3jwCR

T (s) =

SOL 3.118

Option (A) is correct. Applying Barkhausen criterion of oscillation phase shift will be zero. +T (jw0) = 0 w0 " frequency of oscillation. 1 - C2 R2 w20 = 0 1 R C2 w0 = 1 RC w20 =

SOL 3.119

2

Option (C) is correct. In figure V0 R RF + R V jw0 CR = y = 2 V0 1 - w0 C2 R2 + j3w0 CR = 1 RC j = =1 3j 3

Vy = T (jw) w So,

Vy V0

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Page 433

R =1 RF + R 3 RF = 2R = 2 # 1 = 2 kW SOL 3.120

Option (C) is correct. By writing truth table for the circuit Q2 Q1 CLK

Q0

Initially

0

0

0

1

0

0

1

2

0

1

0

3

0

1

1

4

1

0

0

1 0 1 All flip flops are reset. When it goes to state 101, output of NAND gate becomes 0 or CLR = 0, so all FFs are reset. Thus it is modulo 4 counter. SOL 3.121

Option (A) is correct. When the switch is closed (i.e. during TON ) the equivalent circuit is

A I D

O N

n

.i Diode is off during TON .writing KVL in the ocircuit. c . ia 100 - (100 # 10- 6) di = 0 no d dt w. w di = w 106 dt

i = # 106 dt = 106 t + i (0) Since initial current is zero i (0) = 0 So, i = 106 t After a duration of TON the current will be maximum given as i Peak = 106 TON When the switch is opened (i.e. during Toff ) the equivalent circuit is

Diode is ON during Toff , writing KVL again 500 =- (100 # 10- 6) di dt i =- 5 # 106 t + i (0) i (0) = i p = 106 TON So, i =- 5 # 106 t + 106 TON After a duration of Toff , current i = 0

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So, & Given that

Page 434

0 =- 5 # 106 t Toff + 106 TON TON = 5 Toff

TON + Toff = 100 m sec TON + TON = 100 m sec 5 TON = 100 = 63.33 m sec 1.2 Peak current SOL 3.122

i p = 106 # TON = 63.33 # 10- 6 # 106 = 63.33 A

Option (C) is correct. When the switch is opened, current flows through capacitor and diode is ON in this condition. so the equivalent circuit during TOFF is

& Initially

At

Duty cycle

A I D

I = C dVc dt Vc = I t + Vc (0) C

O N

.in

co ia.

od

.n Vc (0) = 0 w w Vc = I tw C

t = Toff Vc = I Toff C TON D = = TON TON + TOFF T

TON = DT TOFF = T - TON = T - DT So, Vc = I (T - DT) C = I (1 - D) T C During TOFF , output voltage V0 = 0 volt . SOL 3.123

Option (B) is correct. When the switch is closed, diode is off and the circuit is

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Page 435

In steady state condition C dVc = I2 dt a dVc = I dt C

I2 = C I C V0 =- Vc = - I t C Average output voltage DT = T I V0 = 1 ; # b-C t l dt + T 0 ON

TOFF

#0

0 dtE

2 DT 2 2 2 =- 1 . I :t D =- 1 . I . D T =- I D .T 2 T C 2 0 T C C 2

SOL 3.124

Option (B) is correct. Equivalent hybrid circuit of given transistor amplifier when RE is by passed is shown below.

In the circuit

A I D

O N

ib = vs hie

.in

co ia.

...(1)

d vo = hfe ib .RC = hfe . vs ..n Ro C hw ie w w h R Voltage gain Av = vo = fe C vi hie Equivalent hybrid circuit when RE is not bypassed by the capacitor. 1

In the circuit vs = ib hie + (ib + hfe ib) RE vs = ib [hie + (1 + hfe) RE ] v0 = hfe ib .RC from equation (2) and (3)

...(2) ...(3)

vs hie + (1 + hfe) RE hfe RC Voltage gain, Av2 = v0 = vs hie + (1 + hfe) RE Av1 = hie + (1 + hfe) RE = 1 + (1 + hfe) RE So hie hie Av2 v0 = hfe .RC

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Av < Av 2

SOL 3.125

Page 436

1

Option (C) is correct. Conversion time for different type of ADC is given as Counting type TT " Conversion time TT = 2n TC TC " Clock period Integrating type TT = 2n + 1 TC Successive Approximation type TT = nTC Parallel (flash) type " fastest Conversion time is highest for integrating type ADC. So it is slowest.

SOL 3.126

Option (D) is correct. F = A + B (NOR) Output is 1 when A = B = 0 OR, F = A 9 B (Ex-NOR) Output is 1 when A = B = 0

SOL 3.127

So,

SOL 3.128

f = I0 S1 S0 + I1 S1 S0 + I2 S1 S0 + I3 S1 S0 I0 = 0 , I1 = I2 = I 3 = 1 n f = 0 + xy + xy + xy =.ixy o + xy + xy c . a xy + x = xy + x (y + yd) i= a y+y = 1 o n = (x + x)w (x. + y) A + BC = (A + B) (A + C) w = x +wy a x+x = 1

O N

Option (C) is correct. Since gain-bandwidth product remains constant Therefore

SOL 3.129

A I D

Option (B) is correct. Output of the multiplexer is written as

105 # 10 = 100 # fCL fCL = 10 kHz

Option (B) is correct. Given circuit is an astable multi vibrator circuit, time period is given as 1+b , t = RC T = 2t ln c 1 - bm b " feedback factor

b=

v+ =1 vo 2

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SOL 3.130

Page 437

J1 + 1 N 2 O = 2t ln 3 So, T = 2t ln KK 1 K1 - OO 2P L Option (C) is correct. MVI A, 10 H & MOV (10) H in accumulator A =(10)H MVI B, 10 H & MOV (10) H in register B B = (10) H BACK : NOP ADDB & Adds contents of register B to accumulator and result stores in accumulator A = A + B = (10) H + (10) H 000 10000 ADD 0 0 0 1 0 0 0 0 A=001 00000 = (20) H RLC & Rotate accumulator left without carry

A I D

O N

n

JNC BACK & JUMP TO Back if CY = 0co.i ia. NOP d .no w ADD B &A = A + B w

w

= (40) H + (10) H

0100 0000 ADD 0 0 0 1 0 0 0 0 A=0101 0000 = (60) H

A = (A0) H JNC BACK NOP ADDB & A = A + B = (A0) H + (10) H 1010 0000 ADD 0 0 0 1 0 0 0 0 A=1011 0 0 0 0 A = (B0) H

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Page 438

CY = 1 So it goes to HLT. therefore NOP will be executed 3 times. SOL 3.131

SOL 3.132

Option (D) is correct. Leakage current is given by 1 0.5 # 1 # CV Ql 0.5 # 100 # Q 100 I Leakage = = = t t t -2 10- 9 # 5 = 0.5 # 10 # 0.1-# 6 1 # 10 - 13 = 25 # -10 = 2.5 # 10- 6 = 2.5 mA 10 6 Option (A) is correct. Slew rate is defined as the maximum rate of change in output voltage per unit time. Slew rate = dv0 dt

A I D

v0 = vin Slew rate = dvin , vin = 10 sin wt dt n 10w cos wt = d (10 sin wot).i= c dt . a

For voltage follower, So,

O N

i

= 10.n wo=d 62.8 volt/msec (given)

w

w 62.8 # 106 10 # 2pfw=

6 f = 62.8 # 10 = 1 MHz 62.8

SOL 3.133

Option (C) is correct. Trans conductance of an n-channel JFET, is given by. gm = 2IDS = - 2IDSS c1 - VGS m 2VGS VP VP Trans conductance (gm) is maximum when gate - to - source voltage VGS = 0 (gm) max = - 2IDSS VP gm = (gm) max c1 - VGS m VP (- 3) Here 1 = (gm) max ;1 = (gm) max # 2 (- 5) E 5 (gm) max = 5 = 2.5 2 Option () is correct. The circuit is a synchronous counter. Where input to the flip flops are So,

SOL 3.134

D3 = Q3 + Q2 + Q1 D2 = Q3 , D1 = Q2 , D 0 = Q1 Truth table of the circuit can be drawn as

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Page 439

CLK

Q3

Q2

Q1

Q0

Initial state

1

1

1

0

1

0

1

1

1

2

0

0

1

1

3

0

0

0

1

4

1

0

0

0

5

0

1

0

0

6

0

0

1

0

7

0

0

0

1

8 1 0 0 0 From the truth table we can see that counter states at N = 4 and N = 8 are same. So mod number is 4. SOL 3.135

Option (B) is correct. In the circuit

A I D

O N

no w.

.in

co ia.

d

w

Writing node equation in thewcircuit at the negative terminal of op amp-1 v1 - 1 + v1 - v2 = 0 1 2 3v1 - v2 = 2 Similarly, at the positive terminal of op amp-1 v1 - vo + v1 - 0 = 0 3 1

...(1)

4v1 - vo = 0 At the negative terminals of op-amp-2 - 1 - v2 + - 1 - vo = 0 m c m c 4 8

...(2)

- 2 - 2v2 - 1 - vo = 0 vo + 2v2 =- 3 From equation (1) and (2) 3 vo - 2v2 = 1 4 From equation (3) 3 v - 2 (- 3 - v ) = 1 o 4 o 3 v + v =- 5 o 4 o 7 v =- 5 4 o

...(3)

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SOL 3.136

Page 440

vo =- 20 volt 7 Option (C) is correct. Small signal circuit is (mid-band frequency range)

CE " 0 , for mid-band frequencies vo =- gm vp RC In the input loop vi rp RB + rp - gm RC rp vi So, vo = RB + rp - gm rp RC Gain Av = vo = vi RB + rp Trans-conductance (1 mA) gm = IC = = 1 A/V VT (26 mV) 26 in b gm rp = b0 & rp = 0 .c=o.200 # 26 = 5.2 kX gima d o - 200 #.n (1 kW) So gain Av = =- 6.62 w (25 kw W + 5.2 kW) vp =

A I D

O N w

SOL 3.137

Option (B) is correct. Cut off frequency due to CE is obtained as

f0 =

1 2pReq CE

Req " Equivalent resistance seen through capacitor CE

Req = RE < RB + rp =

RE (RB + rp) RE + RB + rp

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SOL 3.138

So

f0 =

1 (RE + RB + rp) = 10 Hz (given) 2pRE (RB + rp) CE

So,

CE =

(0.1 + 25 + 5.2) # 103 = 1.59 mF 2p # 0.1 (25 + 5.2) # 106

Page 441

Option (D) is correct. We can approximately analyze the circuit at low and high frequencies as following. For low frequencies w " 0 & 1 " 3 (i.e. capacitor is open) wc Equivalent circuit is

So, it does not pass the low frequencies. For high frequencies w " 3 & 1 " 0 (i.e. capacitor is short) wc Equivalent circuit is

A I D

O N

no w.

in R2 oo .=c . v v ia R1 i

d

So it does pass the high frequencies. This is a high pass filter. w

w

SOL 3.139

At high frequency w " 3 & 1 " 0 , capacitor behaves as short circuit wc and gain of the filter is given as Av = - R2 = 10 R1 R2 = 10 R1 Input resistance of the circuit Rin = R1 = 100 kW So, R2 = 10 # 100 kW = 1 MW Transfer function of the circuit Vo (jw) - jwR2 C = 1 + jw R1 C Vi (jw) High frequency gain Av3 = 10 At cutoff frequency gain is - jwc R2 C Av = 10 = 1 + jwc R1 C 2 10 = 2

wc R2 C 1 + wc2 R 12 C2

100 + 100wc2 R 12 C2 = 2wc2 R 22 C2 100 + 100 # wc2 # 1010 # C2 = 2 # wc2 # 1012 # C2 100 = wc2 C2 # 1012

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Page 442

100 wc2 # 1012 1 C = 2pfc # 10 4

C2 =

1 2 # 3.14 # 10 3 # 10 4

=

= 15.92 nF ***********

A I D

O N

no w.

.in

co ia.

d

ww

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3

POWER ELECTRONICS

YEAR 2013 MCQ 3.1

TWO MARKS

Thyristor T in the figure below is initially off and is triggered with a single pulse of width 10 ms . It is given that L = b 100 l mH and C = b 100 l mF . Assump p ing latching and holding currents of the thyristor are both zero and the initial charge on C is zero, T conducts for

A I D

(A) 10 ms (C) 100 ms MCQ 3.2

(B) 50 ms (D) 200 ms

The separately excited dc motor in the figure below has a rated armature current of 20 A and a rated armature voltage of 150 n V. An ideal chopper switchi . o ing at 5 kHz is used to control the armature .c voltage. If La = 0.1 mH , Ra = 1 W , a i d ratio of the chopper to obtain 50% of the neglecting armature reaction, the duty .nothe rated field current is rated torque at the rated speed w and

O N ww

(A) 0.4 (C) 0.6

(B) 0.5 (D) 0.7

Common Data Questions: 3 & 4 In the figure shown below, the chopper feeds a resistive load from a battery source. MOSFET Q is switched at 250 kHz, with duty ratio of 0.4. All elements of the circuit are assumed to be ideal

GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia MCQ 3.3

The Peak to Peak source current ripple in amps is (A) 0.96 (B) 0.144 (C) 0.192 (D) 0.228

MCQ 3.4

The average source current in Amps in steady-state is (A) 3/2 (B) 5/3 (C) 5/2 (D) 15/4

Page 444

Statement for Linked Answer Questions: 5 & 6 The Voltage Source Inverter (VSI) shown in the figure below is switched to provide a 50 Hz, square wave ac output voltage Vo across an RL load. Reference polarity of Vo and reference direction of the output current io are indicated in the figure. It is given that R = 3 ohms, L = 9.55 mH .

MCQ 3.5

MCQ 3.6

A I D

O N

.in

co ia.

d

In the interval when V0 < 0 and .in0 o> 0 the pair of devices which conducts the w load current is ww (A) Q1, Q2 (B) Q 3, Q 4 (C) D1, D2 (D) D 3, D 4 Appropriate transition i.e., Zero Voltage Switching ^ZVS h/Zero Current Switching ^ZCS h of the IGBTs during turn-on/turn-off is (A) ZVS during turn off (B) ZVS during turn-on (C) ZCS during turn off (D) ZCS during turn-on YEAR 2012

ONE MARK

MCQ 3.7

A half-controlled single-phase bridge rectifier is supplying an R-L load. It is operated at a firing angle a and the load current is continuous. The fraction of cycle that the freewheeling diode conducts is (B) (1 - a/p) (A) 1/2 (C) a/2p (D) a/p

MCQ 3.8

The typical ratio of latching current to holding current in a 20 A thyristor is (A) 5.0 (B) 2.0 (C) 1.0 (D) 0.5

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YEAR 2012 MCQ 3.9

Page 445

TWO MARKS

In the circuit shown, an ideal switch S is operated at 100 kHz with a duty ratio of 50%. Given that Dic is 1.6 A peak-to-peak and I 0 is 5 A dc, the peak current in S , is

(A) 6.6 A (C) 5.8 A

(B) 5.0 A (D) 4.2 A

Common Data for Questions 10 and 11 In the 3-phase inverter circuit shown, the load is balanced and the gating scheme is 180c conduction mode. All the switching devices are ideal.

A I D

O N

no w.

.in

co ia.

d

ww

MCQ 3.10

MCQ 3.11

The rms value of load phase voltage is (A) 106.1 V (C) 212.2 V

If the dc bus voltage Vd = 300 V, the power consumed by 3-phase load is (A) 1.5 kW (B) 2.0 kW (C) 2.5 kW (D) 3.0 kW YEAR 2011

MCQ 3.12

(B) 141.4 V (D) 282.8 V

ONE MARK

A three phase current source inverter used for the speed control of an induction motor is to be realized using MOSFET switches as shown below. Switches S1 to S6 are identical switches.

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Page 446

The proper configuration for realizing switches S1 to S6 is

MCQ 3.13

A I D

Circuit turn-off time of an SCR is defined as the time (A) taken by the SCR turn to be off (B) required for the SCR current to become zero (C) for which the SCR is reverse biased by the commutation circuit n (D) for which the SCR is reverse biased cto o.ireduce its current below the holding . ia current od

O N

n

. ww

YEAR 2011 MCQ 3.14

w

TWO MARKS

A voltage commutated chopper circuit, operated at 500 Hz, is shown below.

If the maximum value of load current is 10 A, then the maximum current through the main (M) and auxiliary (A) thyristors will be (A) iM max = 12 A and iA max = 10 A (B) iM max = 12 A and iA max = 2 A (C) iM max = 10 A and iA max = 12 A (D) iM max = 10 A and iA max = 8 A

Statement for Linked Answer Questions: 9 & 10 A solar energy installation utilize a three – phase bridge converter to feed en-

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Page 447

ergy into power system through a transformer of 400 V/400 V, as shown below.

The energy is collected in a bank of 400 V battery and is connected to converter through a large filter choke of resistance 10 W. MCQ 3.15

The maximum current through the battery will be (A) 14 A (B) 40 A (C) 80 A (D) 94 A

MCQ 3.16

The kVA rating of the input transformer is (A) 53.2 kVA (B) 46.0 kVA (C) 22.6 kVA (D) 7.5 kVA YEAR 2010

MCQ 3.17

ONE MARK

The power electronic converter shown in the figure has a single-pole doublethrow switch. The pole P of the switch is connected alternately to throws A and B. The converter shown is a

A I D

O N

no w.

.in

co ia.

d

(A) step down chopper (buck converter) ww (B) half-wave rectifier (C) step-up chopper (boost converter) (D) full-wave rectifier MCQ 3.18

Figure shows a composite switch consisting of a power transistor (BJT) in series with a diode. Assuming that the transistor switch and the diode are ideal, the I -V characteristic of the composite switch is

MCQ 3.19

The fully controlled thyristor converter in the figure is fed from a single-phase

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Page 448

source. When the firing angle is 0c, the dc output voltage of the converter is 300 V. What will be the output voltage for a firing angle of 60c, assuming continuous conduction

(A) 150 V (C) 300 V

(B) 210 V (D) 100p V

YEAR 2009 MCQ 3.20

An SCR is considered to be a semi-controlled device because (A) It can be turned OFF but not ON with a gate pulse. (B) It conducts only during one half-cycle of an alternating current wave. (C) It can be turned ON but not OFF with a gate pulse. (D) It can be turned ON only during one half-cycle of an alternating voltage wave. YEAR 2009

MCQ 3.21

ONE MARK

A I D

O N

.in

co ia.

TWO MARKS

od The circuit shows an ideal diode.n connected to a pure inductor and is connected w to a purely sinusoidal 50 Hz w voltage source. Under ideal conditions the current w waveform through the inductor will look like.

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MCQ 3.22

Page 449

The Current Source Inverter shown in figure is operated by alternately turning on thyristor pairs (T1, T2) and (T3, T4). If the load is purely resistive, the theoretical maximum output frequency obtainable will be

A I D

(A) 125 kHz (C) 500 kHz MCQ 3.23

ww

no w.

d

(B) 250 kHz (D) 50 kHz

In the chopper circuit shown, the main thyristor (TM) is operated at a duty ratio of 0.8 which is much larger the commutation interval. If the maximum allowable reapplied dv/dt on TM is 50 V/ ms , what should be the theoretical minimum value of C1 ? Assume current ripple through L 0 to be negligible.

(A) 0.2 mF (C) 2 mF MCQ 3.24

O N

.in

co ia.

(B) 0.02 mF (D) 20 mF

Match the switch arrangements on the top row to the steady-state V -I characteristics on the lower row. The steady state operating points are shown by large black dots.

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A I D

O N

(A) P-I, Q-II, R-III, S-IV (C) P-IV, Q-III, R-I, S-II YEAR 2008 MCQ 3.25

no w.

d

.inP-II, Q-IV, R-I, S-III o(B) c . ia

(D) P-IV, Q-III, R-II, S-I

ww

ONE MARK

In the single phase voltage controller circuit shown in the figure, for what range of triggering angle (a), the input voltage (V0) is not controllable ?

(A) 0c < a < 45c (C) 90c < a < 180c MCQ 3.26

Page 450

(B) 45c < a < 135c (D) 135c < a < 180c

A 3-phase voltage source inverter is operated in 180c conduction mode. Which one of the following statements is true ? (A) Both pole-voltage and line-voltage will have 3rd harmonic components (B) Pole-voltage will have 3rd harmonic component but line-voltage will be free from 3rd harmonic (C) Line-voltage will have 3rd harmonic component but pole-voltage will be free from 3rd harmonic (D) Both pole-voltage and line-voltage will be free from 3rd harmonic components

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Page 451

YEAR 2008 MCQ 3.27

TWO MARKS

The truth table of monoshot shown in the figure is given in the table below :

Two monoshots, one positive edge triggered and other negative edge triggered, are connected shown in the figure, The pulse widths of the two monoshot outputs Q1 and Q2 are TON and TON respectively. 1

2

1

D O

2

1

1

MCQ 3.28

MCQ 3.29

IA

The frequency and the duty cycle of the signal at Q1 will respectively be TON 1 1 (A) f = (B) f = , D= , D= 1 TON + TON TON + TON TON + TON 5 TON TON (C) f = 1 , D = (D) f = 1 , D = n TON TON TON + TON TON + TON o.i c . A single phase fully controlled bridge converter supplies a load drawing constant dia o and ripple free load current, if the.n triggering angle is 30c, the input power factor w will be ww (A) 0.65 (B) 0.78 (C) 0.85 (D) 0.866

N 1

2

1

2

2

2

1

2

1

1

2

A single-phase half controlled converter shown in the figure feeding power to highly inductive load. The converter is operating at a firing angle of 60c.

If the firing pulses are suddenly removed, the steady state voltage (V0) waveform of the converter will become

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MCQ 3.30

A single phase source inverter is feeding a purely inductive load as shown in the figure. The inverter is operated at 50 Hz in 180c square wave mode. Assume that the load current does not have any dc component. The peak value of the inductor current i0 will be

A I D

(A) 6.37 A (C) 20 A MCQ 3.31

Page 452

(B) 10 A (D) 40 A

A single phase fully controlled converter bridge is used for electrical braking of a separately excited dc motor. The dc motor load is represented by an equivalent circuit as shown in the figure. .in

O N

no

. ww

co

. dia

w

Assume that the load inductance is sufficient to ensure continuous and ripple free load current. The firing angle of the bridge for a load current of I0 = 10 A will be (A) 44c (B) 51c (C) 129c (D) 136c MCQ 3.32

A three phase fully controlled bridge converter is feeding a load drawing a constant and ripple free load current of 10 A at a firing angle of 30c. The approximate Total harmonic Distortion (%THD) and the rms value of fundamental component of input current will respectively be (A) 31% and 6.8 A (B) 31% and 7.8 A (C) 66% and 6.8 A (D) 66% and 7.8 A

MCQ 3.33

In the circuit shown in the figure, the switch is operated at a duty cycle of 0.5. A large capacitor is connected across the load. The inductor current is assumed to be continuous.

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Page 453

The average voltage across the load and the average current through the diode will respectively be (A) 10 V, 2 A (B) 10 V, 8 A (C) 40 V 2 A (D) 40 V, 8 A YEAR 2007

ONE MARK

MCQ 3.34

A single-phase fully controlled thyristor bridge ac-dc converter is operating at a firing angle of 25c and an overlap angle of 10c with constant dc output current of 20 A. The fundamental power factor (displacement factor) at input ac mains is (A) 0.78 (B) 0.827 (C) 0.866 (D) 0.9

MCQ 3.35

A three-phase, fully controlled thyristor bridge converter is used as line commutated inverter to feed 50 kW power 420 V dc to a three-phase, 415 V(line), 50 Hz ac mains. Consider dc link current to be constant. The rms current of the thyristor is (A) 119.05 A (B) 79.37 A in (C) 68.73 A co. 39.68 A .(D)

MCQ 3.36

A I D

O N

dia o .n bridge converter feeds an inductive load. A single phase full-wave half-controlled w The two SCRs in the converter wware connected to a common DC bus. The converter has to have a freewheeling diode. (A) because the converter inherently does not provide for free-wheeling (B) because the converter does not provide for free-wheeling for high values of triggering angles (C) or else the free-wheeling action of the converter will cause shorting of the AC supply (D) or else if a gate pulse to one of the SCRs is missed, it will subsequently cause a high load current in the other SCR.

MCQ 3.37

“Six MOSFETs connected in a bridge configuration (having no other power device) must be operated as a Voltage Source Inverter (VSI)”. This statement is (A) True, because being majority carrier devices MOSFETs are voltage driven. (B) True, because MOSFETs hav inherently anti-parallel diodes (C) False, because it can be operated both as Current Source Inverter (CSI) or a VSI (D) False, because MOSFETs can be operated as excellent constant current sources in the saturation region. YEAR 2007

MCQ 3.38

TWO MARKS

A single-phase voltages source inverter is controlled in a single

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Page 454

pulse-width modulated mode with a pulse width of 150c in each half cycle. Total harmonic distortion is defined as 2 - V 12 V rms # 100 V1 where V1 is the rms value of the fundamental component of the output voltage. The THD of output ac voltage waveform is (A) 65.65% (B) 48.42% (C) 31.83% (D) 30.49%

THD =

MCQ 3.39

A three-phase, 440 V, 50 Hz ac mains fed thyristor bridge is feeding a 440 V dc, 15 kW, 1500 rpm separately excited dc motor with a ripple free continuos current in the dc link under all operating conditions, Neglecting the losses, the power factor of the ac mains at half the rated speed is (A) 0.354 (B) 0.372 (C) 0.90 (D) 0.955

MCQ 3.40

A single-phase, 230 V, 50 Hz ac mains fed step down transformer (4:1) is supplying power to a half-wave uncontrolled ac-dc converter used for charging a battery (12 V dc) with the series current limiting resistor being 19.04 W. The charging current is (A) 2.43 A (B) 1.65 A (C) 1.22 A (D) 1.0 A

MCQ 3.41

In the circuit of adjacent figure the diode connects the ac source to a pure inn ductance L. o.i

A I D

O N

.c

ia od

n

. ww

w

The diode conducts for (A) 90c (C) 270c MCQ 3.42

(B) 180c (D) 360c

The circuit in the figure is a current commutated dc-dc chopper where, Th M is the main SCR and Th AUX is the auxiliary SCR. The load current is constant at 10 A. Th M is ON. Th AUX is trigged at t = 0 . Th M is turned OFF between.

(A) 0 ms < t # 25 ms (C) 50 ms < t # 75 ms

(B) 25 ms < t # 50 ms (D) 75 ms < t # 100 ms

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Page 455

Common Data for Question 37 and 38. A 1:1 Pulse Transformer (PT) is used to trigger the SCR in the adjacent figure. The SCR is rated at 1.5 kV, 250 A with IL = 250 mA, IH = 150 mA, and IG max = 150 mA, IG min = 100 mA. The SCR is connected to an inductive load, where L = 150 mH in series with a small resistance and the supply voltage is 200 V dc. The forward drops of all transistors/ diodes and gate-cathode junction during ON state are 1.0 V

MCQ 3.43

MCQ 3.44

The resistance R should be (A) 4.7 kW (C) 47 W

A I D (B) 470 kW (D) 4.7 W

The minimum approximate volt-second rating of pulse transformer suitable for triggering the SCR should be : (volt-second rating is the maximum of product n of the voltage and the width of the pulse that o.i may applied) c . (A) 2000 mV-s dia(B) 200 mV-s o n (C) 20 mV-s (D) 2 mV-s w.

O N ww

YEAR 2006

ONE MARK

MCQ 3.45

The speed of a 3-phase, 440 V, 50 Hz induction motor is to be controlled over a wide range from zero speed to 1.5 time the rated speed using a 3-phase voltage source inverter. It is desired to keep the flux in the machine constant in the constant torque region by controlling the terminal voltage as the frequency changes. The inverter output voltage vs frequency characteristic should be

MCQ 3.46

A single-phase half wave uncontrolled converter circuit is shown in figure. A 2-winding transformer is used at the input for isolation. Assuming the load cur-

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Page 456

rent to be constant and V = Vm sin wt , the current waveform through diode D2 will be

A I D

YEAR 2006 MCQ 3.47

TWO MARKS

A single-phase inverter is operated in PWM mode generating a single-pulse of width 2d in the centre of each in halfth cycle as shown in figure. . o It is found that the output voltage is free a.c from 5 harmonic for pulse width rddi 144c. What will be percentage of n3o harmonic present in the output voltage . (Vo3 /Vo1 max) ? ww

O N w

(A) 0.0% (C) 31.7%

(B) 19.6% (D) 53.9%

MCQ 3.48

A 3-phase fully controlled bridge converter with free wheeling diode is fed from 400 V, 50 Hz AC source and is operating at a firing angle of 60c. The load current is assumed constant at 10 A due to high load inductance. The input displacement factor (IDF) and the input power factor (IPF) of the converter will be (B) IDF = 0.867; IPF = 0.552 (A) IDF = 0.867; IPF = 0.828 (C) IDF = 0.5; IPF = 0.478 (D) IDF = 0.5; IPF = 0.318

MCQ 3.49

A voltage commutation circuit is shown in figure. If the turn-off time of the SCR is 50 msec and a safety margin of 2 is considered, then what will be the approximate minimum value of capacitor required for proper commutation ?

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(A) 2.88 mF (C) 0.91 mF

Page 457

(B) 1.44 mF (D) 0.72 mF

MCQ 3.50

A solar cell of 350 V is feeding power to an ac supply of 440 V, 50 Hz through a 3-phase fully controlled bridge converter. A large inductance is connected in the dc circuit to maintain the dc current at 20 A. If the solar cell resistance is 0.5 W ,then each thyristor will be reverse biased for a period of (A) 125c (B) 120c (C) 60c (D) 55c

MCQ 3.51

A single-phase bridge converter is used to charge a battery of 200 V having an internal resistance of 0.2 W as shown in figure. The SCRs are triggered by a constant dc signal. If SCR 2 gets open circuited, what will be the average charging current ?

A I D

O N

no w.

(A) 23.8 A (C) 11.9 A MCQ 3.52

ww

.in

co ia.

d

(B) 15 A (D) 3.54 A

An SCR having a turn ON times of 5 msec, latching current of 50 A and holding current of 40 mA is triggered by a short duration pulse and is used in the circuit shown in figure. The minimum pulse width required to turn the SCR ON will be

(A) 251 msec (C) 100 msec

(B) 150 msec (D) 5 msec

Data for Q. 53 and Q. 54 are given below. Solve the problems and choose the correct answers. A voltage commutated chopper operating at 1 kHz is used to control the speed of dc as shown in figure. The load current is assumed to be constant at 10 A

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MCQ 3.53

The minimum time in msec for which the SCR M should be ON is. (A) 280 (B) 140 (C) 70 (D) 0

MCQ 3.54

The average output voltage of the chopper will be (A) 70 V (B) 47.5 V (C) 35 V (D) 0 V YEAR 2005

MCQ 3.55

MCQ 3.56

The conduction loss versus device current characteristic of a power MOSFET is best approximated by (A) a parabola (B) a straight line (C) a rectangular hyperbola (D) an exponentially decaying function in

A I D

O N

co. . a A three-phase diode bridge rectifier disi fed from a 400 V RMS, o 50 Hz, three-phase AC source.wIf.nthe load is purely resistive, then peak instantaneous output voltage is equal wwto (A) 400 V

MCQ 3.57

ONE MARK

(B) 400 2 V (C) 400 2 V (D) 400 V 3 3 The output voltage waveform of a three-phase square-wave inverter contains (A) only even harmonics (B) both odd and even harmonic (C) only odd harmonics (D) only triple harmonics YEAR 2005

MCQ 3.58

TWO MARKS

The figure shows the voltage across a power semiconductor device and the current through the device during a switching transitions. If the transition a turn ON transition or a turn OFF transition ? What is the energy lost during the transition?

(A) Turn ON, VI (t1 + t2) 2

(B) Turn OFF, VI (t1 + t2)

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Page 459

(D) Turn OFF, VI (t1 + t2) 2 An electronics switch S is required to block voltage of either polarity during its OFF state as shown in the figure (a). This switch is required to conduct in only one direction its ON state as shown in the figure (b) (C) Turn ON, VI (t1 + t2)

MCQ 3.59

Which of the following are valid realizations of the switch S?

(A) Only P (C) P and R MCQ 3.60

A I D

O N w

no w.

d

n o.iP and Q c . (B) ia (D) R and S

w The given figure shows a step-down chopper switched at 1 kHz with a duty ratio D = 0.5 . The peak-peak ripple in the load current is close to

(A) 10 A (C) 0.125 A

(B) 0.5 A (D) 0.25 A

MCQ 3.61

An electric motor, developing a starting torque of 15 Nm, starts with a load torque of 7 Nm on its shaft. If the acceleration at start is 2 rad/sec2 , the moment of inertia of the system must be (neglecting viscous and coulomb friction) (A) 0.25 kg-m2 (B) 0.25 Nm2 (C) 4 kg-m2 (D) 4 Nm2

MCQ 3.62

Consider a phase-controlled converter shown in the figure. The thyristor is fired at an angle a in every positive half cycle of the input voltage. If the peak value of the instantaneous output voltage equals 230 V, the firing angle a is close to

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(A) 45c (C) 90c

Page 460

(B) 135c (D) 83.6c

YEAR 2004

ONE MARK

MCQ 3.63

A bipolar junction transistor (BJT) is used as a power control switch by biasing it in the cut-off region (OFF state) or in the saturation region (ON state). In the ON state, for the BJT (A) both the base-emitter and base-collector junctions are reverse biased (B) the base-emitter junction is reverse biased, and the base-collector junction is forward biased (C) the base-emitter junction is forward biased, and the base-collector junction is reverse biased (D) both the base-emitter and base-collector junctions are forward biased

MCQ 3.64

The circuit in figure shows a full-wave rectifier. The input voltage is 230 V (rms) single-phase ac. The peak reverse voltage across the diodes D 1 and D 2 is

A I D

O N

no w.

.in

co ia.

d

ww

MCQ 3.65

(A) 100 2 V

(B) 100 V

(C) 50 2 V

(D) 50 V

The triggering circuit of a thyristor is shown in figure. The thyristor requires a gate current of 10 mA, for guaranteed turn-on. The value of R required for the thyristor to turn on reliably under all conditions of Vb variation is

(A) 10000 W (C) 1200 W MCQ 3.66

(B) 1600 W (D) 800 W

The circuit in figure shows a 3-phase half-wave rectifier. The source is a symmetrical, 3-phase four-wire system. The line-to-line voltage of the source is 100

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Page 461

V. The supply frequency is 400 Hz. The ripple frequency at the output is

(A) 400 Hz (C) 1200 Hz

(B) 800 Hz (D) 2400 Hz

YEAR 2004 MCQ 3.67

A MOSFET rated for 15 A, carries a periodic current as shown in figure. The ON state resistance of the MOSFET is 0.15 W. The average ON state loss in the MOSFET is

(A) 33.8 W (C) 7.5 W MCQ 3.68

(B) 5290 W (D) 10580 W

Figure shows a chopper operating from a 100 V dc input. The duty ratio of the main switch S is 0.8. The load is sufficiently inductive so that the load current is ripple free. The average current through the diode D under steady state is

(A) 1.6 A (B) 8.0 A MCQ 3.70

O N

A I D (B) 15.0 W n W (D).i3.8

.co a i d The triac circuit shown in figure controls no the ac output power to the resistive . win the load is load. The peak power dissipation ww

(A) 3968 W (C) 7935 W MCQ 3.69

TWO MARKS

(B) 6.4 A (D) 10.0 A

Figure shows a chopper. The device S 1 is the main switching device. S 2 is the auxiliary commutation device. S 1 is rated for 400 V, 60 A. S 2 is rated for 400 V,

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30 A. The load current is 20 A. The main device operates with a duty ratio of 0.5. The peak current through S 1 is

(A) 10 A (C) 30 A

(B) 20 A (D) 40 A

MCQ 3.71

A single-phase half-controlled rectifier is driving a separately excited dc motor. The dc motor has a back emf constant of 0.5 V/rpm. The armature current is 5 A without any ripple. The armature resistance is 2 W. The converter is working from a 230 V, single-phase ac source with a firing angle of 30c. Under this operating condition, the speed of the motor will be (A) 339 rpm (B) 359 rpm (C) 366 rpm (D) 386 rpm

MCQ 3.72

A variable speed drive rated for 1500 rpm, 40 Nm is reversing under no load. Figure shows the reversing torque and the speed during the transient. The moment of inertia of the drive is

A I D

O N

no w.

.in

co ia.

d

ww

(A) 0.048 kg-m2 (C) 0.096 kg-m2

(B) 0.064 km-m2 (D) 0.128 kg-m2

YEAR 2003 MCQ 3.73

ONE MARK

Figure shows a thyristor with the standard terminations of anode (A), cathode (K), gate (G) and the different junctions named J1, J2 and J3. When the thyristor is turned on and conducting

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(A) J1 and J2 are forward biased and J3 is reverse biased (B) J1 and J3 are forward biased and J2 is reverse biased (C) J1 is forward biased and J2 and J3 are reverse biased (D) J1, J2 and J3 are all forward biased MCQ 3.74

Figure shows a MOSFET with an integral body diode. It is employed as a power switching device in the ON and OFF states through appropriate control. The ON and OFF states of the switch are given on the VDS - IS plane by

A I D

O N

.in

no w.

co ia.

d

ww

MCQ 3.75

The speed/torque regimes in a dc motor and the control methods suitable for the same are given respectively in List-II and List-I List-I P.

Field Control

1.

Below base speed

Q.

Armature Control

2.

Above base speed

3.

Above base torque

4.

Below base torque

Codes: (A) P-1, Q-3 (C) P-2, Q-3 MCQ 3.76

List-II

(B) P-2, Q-1 (D) P-1, Q-4

A fully controlled natural commutated 3-phase bridge rectifier is operating with a firing angle a = 30c, The peak to peak voltage ripple expressed as a ratio of the peak output dc voltage at the output of the converter bridge is

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(A) 0.5

(B)

3 /2

(C) c1 - 3 m 2

(D)

3 -1

YEAR 2003 MCQ 3.77

Page 464

TWO MARKS

A phase-controlled half-controlled single-phase converter is shown in figure. The control angle a = 30c

The output dc voltage wave shape will be as shown in

A I D

O N

no w.

.in

co ia.

d

ww

MCQ 3.78

A chopper is employed to charge a battery as shown in figure. The charging current is 5 A. The duty ratio is 0.2. The chopper output voltage is also shown in the figure. The peak to peak ripple current in the charging current is

(A) 0.48 A (C) 2.4 A MCQ 3.79

(B) 1.2 A (D) 1 A

An inverter has a periodic output voltage with the output wave form as shown in figure

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Page 465

When the conduction angle a = 120c, the rms fundamental component of the output voltage is (A) 0.78 V (B) 1.10 V (C) 0.90 V (D) 1.27 V MCQ 3.80

With reference to the output wave form given in above figure , the output of the converter will be free from 5 th harmonic when (A) a = 72c (B) a = 36c (C) a = 150c (D) a = 120c

MCQ 3.81

An ac induction motor is used for a speed control application. It is driven from an inverter with a constant V/f control. The motor name-plate details are as follows (no. of poles = 2) V: 415 V VPh: 3 V f: 50 Hz N: 2850 rpm The motor runs with the inverter output frequency set at 40 Hz, and with half the rated slip. The running speed of the motor is (A) 2400 rpm (B) 2280 rpm (C) 2340 rpm (D) 2790 rpm YEAR 2002

MCQ 3.82

MCQ 3.83

in co.

ONE MARK

A six pulse thyristor rectifier bridge is connected to a balanced 50 Hz three ia. d o phase ac source. Assuming that the .ndc output current of the rectifier is conw stant, the lowest frequency harmonic component in the ac source line current is ww (A) 100 Hz (B) 150 Hz (C) 250 Hz (D) 300 Hz A step-down chopper is operated in the continuous conduction mode is steady state with a constant duty ratio D . If V0 is the magnitude of the dc output voltage and if Vs is the magnitude of the dc input voltage, the ratio V0 /Vs is given by (A) D (B) 1 - D (C) 1 (D) D 1-D 1-D YEAR 2002

MCQ 3.84

O N

A I D

TWO MARKS

In the chopper circuit shown in figure, the input dc voltage has a constant value Vs . The output voltage V0 is assumed ripple-free. The switch S is operated with a switching time period T and a duty ratio D . What is the value of D at the boundary of continuous and discontinuous conduction of the inductor current iL ?

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(A) D = 1 - Vs V0 (C) D = 1 - 2L RT MCQ 3.85

(B) D = 2L RT (D) D = RT L

Figure(a) shows an inverter circuit with a dc source voltage Vs . The semiconductor switches of the inverter are operated in such a manner that the pole voltage V10 and V20 are as shown in figure(b). What is the rms value of the poleto-pole voltage V12 ?

(A)

Vs f p 2

(C) Vs MCQ 3.86

Page 466

(B) Vs

f p

A I D

f 2p

(D) Vs p

In the single phase diode bridge rectifier shown in figure, the load resistor is R = 50 W . The source voltage is V = 200 sin (wt), where w = 2p # 50 radians per .in R is second. The power dissipated in the loadcoresistor

O N

.

ia od

n

. ww

w

MCQ 3.87

(A) 3200 W p

(B) 400 W p

(C) 400 W

(D) 800 W

*The semiconductor switch S in the circuit of figure is operated at a frequency of 20 kHz and a duty ratio D = 0.5 . The circuit operates in the steady state. Calculate the power transferred from the dc voltage source V2 .

YEAR 2001 MCQ 3.88

ONE MARK

The main reason for connecting a pulse transformer at the output stage of thy-

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Page 467

ristor triggering circuit is to (A) amplify the power of the triggering pulse (B) provide electrical isolation (C) reduce the turn on time of thyristor (D) avoid spurious triggering of the thyristor due to noise MCQ 3.89

AC-to-DC circulating current dual converters are operated with the following relationship between their triggering angles( a1 and a2 ) (A) a1 + a2 = 180c (B) a1 + a2 = 360c (C) a1 - a2 = 180c (D) a1 + a2 = 90c YEAR 2001

MCQ 3.90

A half-wave thyristor converter supplies a purely inductive load as shown in figure. If the triggering angle of the thyristor is 120c, the extinction angle will be

(A) 240c (C) 200c MCQ 3.91

O N

A I D (B) 180c (D) 120c

in

A single-phase full bridge voltage source inverter co. feeds a purely inductive load . a as shown in figure, where T1 , T2 , T3o ,d Ti4 are power transistors and D 1 , D 2 , D 3 n , D 4 are feedback diodes. The inverter w. is operated in square-wave mode with a w w load current is zero, what is the time durafrequency of 50 Hz. If the average tion of conduction of each feedback diode in a cycle?

(A) 5 msec (C) 20 msec MCQ 3.92

TWO MARKS

(B) 10 msec (D) 2.5 msec

*A voltage commutated thyristor chopper circuit is shown in figure. The chopper is operated at 500 Hz with 50% duty ratio. The load takes a constant current of 20 A. (a) Evaluate the circuit turn off time for the main thyristor Th 1 . (b) Calculate the value of inductor L, if the peak current through the main thyristor Th 1 is limited to 180% of the load current. (c) Calculate the maximum instantaneous output voltage of chopper.

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MCQ 3.93

Page 468

*A separately excited dc motor is controlled by varying its armature voltage using a single-phase full-converter bridge as shown in figure. The field current is kept constant at the rated value. The motor has an armature resistance of 0.2 W, and the motor voltage constant is 2.5 V/(rad/sec). The motor is driving a mechanical load having a constant torque of 140 Nm. The triggering angle of converter is 60c. The armature current can be assumed to be continuous and ripple free.

A I D

O N

in

(a) Calculate the motor armature constant. co. . a di (b) Evaluate the motor speed in n rad/sec. o . wthe (c) Calculate the rms value of fundamental component of the input current w w to the bridge. ***********

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SOLUTION

SOL 3.1

Page 469

Option (C) is correct. Given, L = 100 mH p C = 100 mF p When the circuit is triggered by 10 ms pulse, the thyristor is short circuited and so, we consider IC = Im sin wt Therefore, voltage stored across capacitor is VC = 1 IC dt C = Vm ^1 - cos wt h where w is angular frequency obtained as 1 w = 1 = = p # 10 4 100 -6 LC b p l # 10 So, T = 1 = 2p = 200 ms w f

#

A I D

O N

.in o c . As IC = Im sin wt oscillates ia between - ve and - ve half cycle so, d o circuit is conducting for only half .of n cycle and thyristor is open after half cycle. w T = 100 ms i.e., the conduction period = ww 2 SOL 3.2

Option (D) is correct. Given, the rated armature current Ia^rated h = 20 A as rated armature voltage Va^rated h = 150 volt Also, for the armature, we have La = 0.1 mH , Ra = 1 W and T = 50% of Trated ^T " Torqueh So, we get I = 6Ia^rotatedh@^0.5h = 10 A N = Nrated , I f = I f rated " rated field current At the rated conditions, E = V - Ia^ratedh Ra = 150 - 20 ^1 h = 130 volt For given torque, V = E + Ia Ra = 130 + ^10h^1 h = 140 V Therefore, chopper output = 140 V or, D ^200h = 140 or, (D " duty cycle) D = 140 = 0.7 200

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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia SOL 3.3

Page 470

Option (C) is correct. Here, as the current from source of 12 V is the same as that pass through inductor. So, the peak to peak current ripple will be equal to peak to peak inductor current. Now, the peak to peak inductor current can be obtained as IL (Peak to Peak) = Vs D Ts L where,

Vs " source voltage = 12 volt , L " inductance = 100mH = 10-4 H , D " Duty ration = 0.4 ,

TS " switching time period of MOSFET = 1 fS and fs " switching frequency = 250 kHz Therefore, we get 1 IL^Peak to Peakh = 12-4 # 0.4 # 250 # 103 10 = 0.192 A This is the peak to peak source current ripple. SOL 3.4

Option (B) is correct. Here, the average current through the capacitor will be zero. (since, it is a boost converter). We consider the two cases : Case I : When MOSFET is ON (i 0 is output current) ic =- i 0 (since, diode will be in cut off mode) n Case II : When MOSFET is OFF o.i c . Diode will be forward biased and so ia d (Is is source current) ic = Is - i 0 .no w Therefore, average current through capacitor ww ic + Ic Ic, avg = 2 DTs ^- io h + ^1 - D h Ts ^Is - io h (D is duty ratio) & 0= 2 Solving the equation, we get i0 ....(1) Is = ^1 - D h Since, the output load current can be given as V/ 12/0.6 i 0 = V0 = s 1 - D = = 1A 20 R R Hence, from Eq. (1) Is = i 0 = 1 = 5 A 0.6 3 1-D

A I D

1

O N 1

1

SOL 3.5

2

Option (D) is correct. We consider the following two cases : Case I : When Q1, Q2 ON In this case the + ve terminal of V0 will be at higher voltage. i.e. V0 > 0 and so i 0 > 0 (i.e., it will be + ve ). Now, when the Q1 , Q2 goes to OFF condition we consider the second case. Case II : When Q 3 , Q 4 ON and Q , Q2 OFF : In this condition, - ve terminal of applied voltage V0 will be at higher potential i.e., V0 < 0 and since, inductor opposes the change in current so, although the

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Page 471

polarity of voltage V0 is inversed, current remains same in inductor i.e. I 0 > 0 . This is the condition when conduction have been asked. In this condition ^V0 > 0, I 0 > 0h since, IGBT’s can’t conduct reverse currents therefore current will flow through D 3, D 4 until ID becomes negative. Thus, D 3 and D 4 conducts. SOL 3.6

Option (D) is correct. When Q 3, Q 4 is switched ON, initially due to the reverse current it remain in OFF state and current passes through diode. In this condition the voltage across Q 3 and Q 4 are zero as diodes conduct. Hence, it shows zero voltage switching during turn-ON

SOL 3.7

Option (D) is correct. The circuit of a single-phase half controlled bridge rectifier with RL load and free wheel diode is shown as below.

A I D

The voltage current wave forms are shown in figure below.

O N

no w.

.in

co ia.

d

ww

We note that, for continuous load current, the flywheel diode conducts from p

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Page 472

to p + a in a cycle. Thus, fraction of cycle that freewheel diode conducts is a/p. Thus fraction of cycle that freewheel diode conducts is a/p. SOL 3.8

Option (B) is correct. The latching current is higher than the holding current. Usually, latching current is taken two to three times the holding currents.

SOL 3.9

Option (C) is correct.

SOL 3.10

IS = I 0 + Tic = 5 + 0.8 = 5.8 A 2 Option (B) is correct. For a three-phase bridge inverter, rms value of output line voltage is VL =

2V = 2 300 3 dc 3 #

Vdc = 300 V

= 141.4 V SOL 3.11

SOL 3.12

Option (D) is correct. 2 (141.4) 2 P = 3 # VL = 3 # - 3 kW 20 R Option (C) is correct. Only option C allow bi direction power flow from source to the drive

A I D

O N

no w.

.in

co ia.

d

ww

SOL 3.13

Option (C) is correct. Once the SCR start conducting by an forward current, the gate has no control on it and the device can be brought back to the blocking state only by reducing the forward current to a level below that of holding current. This process of turn-off is called commutation. This time is known as the circuit turn-off time of an SCR.

SOL 3.14

Option (A) is correct. Maximum current through main thyristor C = 10 + 200 L Maximum current through auxiliary thyristor IM (max) = I 0 + Vs

0.1 # 10-6 = 12 A 1 # 103

IA (max) = I 0 = 10 A SOL 3.15

Option (A) is correct. Output voltage of 3-phase bridge converter V0 = 3 3 Vph cos a p Maximum output (V0) max = 3 3 Vph cos a = 1 p = 3 3 # 400 # 2 p 3

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Page 473

= 540.6 V Resistance of filter choke is 10 W, So (V0) max = E + IR chock 540.6 = 400 + I (10) I - 14 A SOL 3.16

Option (D) is correct. kVA rating =

3 # 400 # 6 # 14 p

3 VL IL =

= 7.5 kVA SOL 3.17

Option (A) is correct. The figure shows a step down chopper circuit. a Vout = DVin where, D = Duty cycle and D < 1

SOL 3.18

Option (C) is correct. Given figure as

A I D

The I -V characteristic are as

O N

no w.

.in

co ia.

d

w

w so I can never be negative. Since diode connected in series When current flows voltage across switch is zero and when current is zero than there may be any voltage across switch. SOL 3.19

Option (A) is correct. Given fully-controlled thyristor converter, when firing angle a = 0 , dc output voltage Vdc = 300 V If a = 60c, then Vdc = ? For fully-controlled converter 0

Vdc = 0

2 2 Vdc cos a p 1

a a = 0 , Vdc = 300 V 0

2 2 Vdc cos 0c p = 300p 2 2

300 = Vdc

1

1

At a = 60c, Vdc = ? 2

Vdc = 2 2 # 300p cos 60c = 300 # 1 = 150 V p 2 2 2 Option (C) is correct. SCR has the property that it can be turned ON but not OFF with a gate pulse, So SCR is being considered to be a semi-controlled device. 2

SOL 3.20

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Page 474

Option (D) is correct.

Current wave form for iL vL = LdiL dt iL = 1 # vL dt 2 vL = vin = 10 sin wt = diL dt iL = 1 # vL dt =- cos 100pt + C 2

for 0 < wt +p,

iL = 0 , C = 0 iL =- 100 cos pt iL (peak) = 1 Amp

at 100pt = p/2 ,

A I D

O N

no w.

for p < wt vL = vin = 0

.in

co ia.

d

ww

SOL 3.22

Option (C) is correct. In CSI let T3 and T4 already conducting at t = 0 At triggering T1 and T2 , T3 and T4 are force cumulated. Again at t = T , T1 and T2 are force cumulated. This completes a cycle. 2

Time constant t = RC = 4 # 0.5 = 2 m sec 1 Frequency f = 1 = = 500 kHz t 2 # 10- 6 SOL 3.23

Option (A) is correct. duty ratio TM = 0.8 Maximum dv on TM = 50 V/msec dt Minimum value of C1

=?

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Given that current ripple through L 0 is negligible. Current through TM = Im = duty ratio # current a

SOL 3.24

= 0.8 # 12.5 = 10 A Im = C1 dv dt 10 = C1 # 50- 6 10 C1 = 50 # 10- 6 = 0.2 mF 10

Option (C) is correct. Characteristics are as

A I D

O N

no w.

.in

co ia.

d

ww

SOL 3.25

Option (A) is correct.

`

R + jXL = 50 + 50j tan f = wL = 50 = 1 50 R

f = 45c so, firing angle ‘a’ must be higher the 45c, Thus for 0 < a < 45c, V0 is uncontrollable. SOL 3.26

Option (D) is correct. A 3-f voltage source inverter is operated in 180c mode in that case third harmonics

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Page 476

are absent in pole voltage and line voltage due to the factor cos (np/6). so both are free from 3rd harmonic components. SOL 3.27

Option (B) is correct. In this case and,

SOL 3.28

1 TON1 + TON 2 TON 2 D = TON1 + TON 2 f =

Option (B) is correct. Given a = 30c, in a 1-f fully bridge converter we know that, Power factor = Distortion factor # cos a D.f. (Distortion factor) = Is(fundamental) /Is = 0.9 power factor = 0.9 # cos 30c = 0.78

SOL 3.29

Option (A) is correct. Output of this

A I D

O N

.in

co ia.

od n . Here the inductor makes T1 and w T3 in ON because current passing through w w and is more than the holding current. T3

SOL 3.30

SOL 3.31

T1

Option (C) is correct. Input is given as

Here load current does not have any dc component Peak current occur at (p/w) ` ` Vs = L di dt 200 = 0.1 # di dt Here di = a p kb 1 l = 1 2p 50 100 So di(max) = 200 # 1 # 1 = 20 A 100 0.1 Option (C) is correct. Here for continuous conduction mode, by Kirchoff’s voltage law, average load

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Page 477

current

V - 2Ia + 150 = 0 Ia = V + 150 2 ` I1 = 10 A, So V =- 130 V 2Vm cos a =- 130 p 2#

2 # 230 cos a =- 130c p a = 129c

SOL 3.32

A I D

Option (B) is correct.

2 10 = 8.16 A 3#

Total rms current Ia =

Fundamental current Ia1 = 0.78 # 10 = 7.8 A

O N

1 - 1 o.in c DF2 dia.

THD =

where

DF = w

` SOL 3.33

no

. wIa1w

THD =

Ia

= 0.78 # 10 = 0.955 0.816 # 10

1 2 b 0.955 l - 1 = 31%

Option (C) is correct.

In the given diagram when switch S is open I 0 = IL = 4 A, Vs = 20 V when switch S is closed ID = 0, V0 = 0 V Duty cycle = 0.5 so average voltage is Vs 1-d Average current = 0 + 4 = 2 amp 2 Average voltage = 20 = 40 V 1 - 0.5 SOL 3.34

Option (A) is correct. Firing angle a = 25c Overlap angle m = 10c

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so,

I 0 = Vm [cos a - cos (a + m)] wLs

`

20 =

`

Page 478

230 2 [cos 25c - cos (25c + 10c)] 2p # 50Ls

Ls = 0.0045 H V0 = 2Vm cos a - wLsI 0 p p -3 = 2 # 230 2 cos 25c - 2 # 3.14 # 50 # 4.5 # 10 # 20 3.14 3.14

= 187.73 - 9 = 178.74c Displacement factor = V0 I 0 = 178.25 # 20 = 0.78 230 # 20 Vs Is SOL 3.35

SOL 3.36

Option (C) is correct. Given that P = 50 # 1000 W Vd = 420 So P = Vd # Id Id = 50 # 1000 = 119.05 420 RMS value of thyristor current = 119.05 = 68.73 3 Option (B) is correct. Single phase full wave half controlled bridge in converter feeds an Inductive load. . o The two SCRs in the converter are connected c to a common dc bus. The converter ia. the converter does not provide for free d has to have free wheeling diode because .no angles. w wheeling for high values of triggering w

A I D

O N w

SOL 3.37

Option (D) is correct. If we connect the MOSFET with the VSI, but the six MOSFETs are connected in bridge configuration, in that case they also operated as constant current sources in the saturation region so this statement is false.

SOL 3.38

Option (C) is correct. Given that, total harmonic distortion THD =

Vrms - V 12 # 100 V1 2

Pulse width is 150c Here

SOL 3.39

150 V = 0.91V s 180 l s V1 = Vrms(fundamental) = 0.4Vs sin 75c = 0.8696Vs p# 2 2 (0.91Vs) - (0.87Vs) 2 THD = = 31.9% (0.87Vs) 2 Vrms = b

Option (A) is correct. When losses are neglected, 2 # 440 cos a = K 750 # 2p m# p 60 Here back emf e with f is constant 3#

e = V0 = Km wm

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Page 479

440 = Km # 1500 # 2p 60 Km = 2.8 cos a = 0.37 at this firing angle Vt = 3 2 # 440 # (0.37) = 219.85 V p Ia = 1500 = 34.090 440 Isr = Ia 2/3 = 27.83 p.f. = SOL 3.40

Vt Is = 0.354 3 Vs Isr

Option (D) is correct. Vs = 230 = 57.5 4 Here charging current = I Vm sin q = 12 q1 = 8.486 = 0.148 radian Vm = 81.317 V

A I D

e = 12 V There is no power consumption in battery due to ac current, so average value of charging current. n o.i c 1 . Iav(charging) = [2V acos q1 - e (p - 2q1)] 2p # 19.04 o dmi .n w 1 = w [2 V cos q1 - 12 (p - 2q1)] 2p #w19.04 # m #

O N

= 1.059 W/A

SOL 3.41

Option (C) is correct. Conduction angle for diode is 270c as shown in fig.

SOL 3.42

Option ( ) is correct.

SOL 3.43

Option (C) is correct. Here, Vm = maximum pulse voltage that can be applied so = 10 - 1 - 1 - 1 = 7 V Here 1 V drop is in primary transistor side, so that we get 9V pulse on the secondary side. Again there are 1 V drop in diode and in gate cathode junction each. I g max = 150 mA

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So SOL 3.44

Page 480

7 R = Vm = = 46.67 W Ig max 150 mA

Option (A) is correct. We know that the pulse width required is equal to the time taken by ia to rise upto iL so, Vs = L di + Ri (VT . 0) dt ia = 200 [1 - e- t/0.15] 1 Here also

t = T, 0.25 = 200 [1 - e- T/0.5]

ia = iL = 0.25

T = 1.876 # 10- 4 = 187.6 ms Width of pulse = 187.6 ms Magnitude of voltage = 10 V Vsec rating of P.T. = 10 # 187.6 ms = 1867 mV-s is approx to 2000 mV-s SOL 3.45

Option (D) is correct. If we varying the frequency for speed control, V/f should be kept as constant so that, minimum flux density (Bm ) also remains constant So, V = 4.44NBm Af

A I D

O N

no w.

.in

co ia.

d

ww

SOL 3.46

Option (D) is correct. In first half cycle D 1 will conduct and D 2 will not and at q = 0 there is zero voltage. So current wave form is as following

SOL 3.47

Option (B) is correct. In the PWM inverter V0 = output voltage of inverter 3 V0 = / 4Vs sin nd sin nwt sin np/2 n = 1 np So the pulse width = 2d = 144c V01 = 4Vs sin 72c sin wt p V03 = 4Vs sin ^3 # 72ch sin 3wt 3p 4Vs sin (3 # 72c) V 3p # 03 so, = 19.61% bV01 max l = 4Vs sin 72c p

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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia SOL 3.48

Option (C) is correct. Given that 400 V, 50 Hz AC source, a = 60c, IL = 10 A so, Input displacement factor = cos a = 0.5 and, input power factor = D.F. # cos a

so, SOL 3.49

SOL 3.50

Page 481

4 # 10 sin 60c Is(fundamental) distortion factor = = p# 2 Is 10 # 2/3 = 0.955 input power factor = 0.955 # 0.5 = 0.478

Option (A) is correct. We know that T = RC ln 2 T 100 So C = = 2.88 mF = 50 # 0.693 R # 0.693 Option (A) is correct. Let we have R solar = 0.5 W , I 0 = 20 A so Vs = 350 - 20 # 0.5 = 340 V `

A I D

340 = 3 # 440 # p

2 cos a

O N

in

cos a = 55c co. . a di 180c - 55c = 125c. So each thyristor will reverse biased ofor SOL 3.51

.n

Option (C) is correct. ww w In this circuitry if SCR gets open circuited, than circuit behaves like a half wave rectifier.

So I avg = Average value of current p-q = 1 # (Vm sin wt - E) dq 2pR q I 0(avg) = 1 62Vm cos q - E (p - 2q1)@ 2p R 1

1

a

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Page 482

1 [2 (230 # 2 ) cos q - 200 (p - 2q1)] 2p # 2 # 200 q1 = sin- 1 b E l = sin- 1 c = 38c = 0.66 Rad Vm 230 # 2 m 1 [2 2 I 0 (avg) = # 230 cos 38c - 200 (p - 2 # 0.66)] 2p # 2 =

`

= 11.9 A SOL 3.52

Option (B) is correct.

In this given circuit minimum gate pulse width time= Time required by ia rise up to iL i2 = 100 3 = 20 mA 5 # 10 i1 = 100 [1 - e- 40t] 20 `

O N T = 150 ms

SOL 3.53

o

co

. dia

.n Option (B) is correct. w w Given IL = 10 A . So in thew+ ve half cycle, it will charge the capacitor, minimum time will be half the time for one cycle. so min time required for charging = p = p LC w0 = 3.14 #

SOL 3.54

A I D

anode current I = I1 + I2 = 0.02 + 5 [1 - e- 40t] 0.05 = 0.05 + 5 [1 - e- 40t] 1 - e- 40t = 0.03 5 .in

2 # 10- 3 # 10- 6 = 140 m sec

Option (C) is correct. Given Ton = 140 m sec Average output = Ton # V Ttotal Ttotal = 1/f = 1 = 1 msec 103 -6

so average output = 140 # 10-3 # 250 = 35 V 1 # 10 SOL 3.55

Option (A) is correct. The conduction loss v/s MOSFET current characteristics of a power MOSFET is best approximated by a parabola.

SOL 3.56

Option (B) is correct. In a 3-f bridge rectifier Vrms = 400 V , f = 50 Hz This is purely resistive then

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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia

instantaneous voltage

V0 =

Page 483

2 Vrms = 400 2 V

SOL 3.57

Option (C) is correct. A 3-f square wave (symmetrical) inverter contains only odd harmonics.

SOL 3.58

Option (A) is correct. In Ideal condition we take voltage across the device is zero. average power loss during switching = VI (t1 + t2) (turn ON) 2 Option (C) is correct. So in P thyristor blocks voltage in both polarities until gate is triggered and also in R transistor along with diode can do same process.

SOL 3.59

SOL 3.60

Option (C) is correct. Duty ratio a = 0.5 1 -3 sec - 3 = 10 1 # 10 Ta = L = 200 mH = 40 msec 5 R T =

here

(1 - e- aT/Ts) (1 - e- (1 - a) T/Ta) Ripple = Vs = G R 1 - e- T/Ts 100 (TI) max = Vs = 4fL 4 # 103 # 200 # 10- 3 = 0.125 A SOL 3.61

Tst TL a T T

= 15 Nm = 7 Nm

.no w = 2 rad/sec ww = Ia 2

.in

co ia.

d

= Tst - TL = 8 Nm I = 8 = 4 kgm2 2 Option (B) is correct. We know that Vrms = 230 V so

SOL 3.62

A I D

O N

Option (C) is correct.

so, If whether Then

Vm = 230 # 2 V a 1 90c Vpeak = Vm sin a = 230

230 2 sin a = 230 sin a = 1 2 angle a = 135c SOL 3.63

Option (D) is correct. When we use BJT as a power control switch by biasing it in cut-off region or in the saturation region. In the on state both the base emitter and base-collector junction are forward biased.

SOL 3.64

Option (A) is correct. Peak Inverse Voltage (PIV) across full wave rectifier is 2Vm Vm = 50 2 V

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so, SOL 3.65

Page 484

PIV = 100 2 V

Option (D) is correct. Vb = 12 ! 4 V Vb max = 16 V Vb min = 8 V V (min) 8 Required value of R = b = = 800 W Ig 10 # 10- 3

SOL 3.66

Option (C) is correct.

Ripple frequency = 3f = 3 # 400 = 1200 Hz So from V0 ripple frequency = 1200 Hz SOL 3.67

A I D

Option (C) is correct. Given that

R = 0.15 W I = 15 A 1 # p/woI.i2n So average power losses = c Rdt 0. (2p/w)ia od 2 n w . = w # 10 # 0.15 # p/w w 2p

O N w

= 7.5 W

SOL 3.68

Option (D) is correct. Output dc voltage across load is given as following 1

Vdc = =

SOL 3.69

SOL 3.70

2 V ; 1 &(2p - a) + sin 2a 0E2 ap 2 1

1

sin p/2 2 p 'a2p - 4 k + b 2 l1H p

2 # 230 2 >p # 4 = 317.8 V 2 (317.8) 2 losses = V dc = = 10100 W 100 R Option (C) is correct. Vs = 100 V , duty ratio = 0.8 , R = 10 W

So average current through diode = aVs R = 0.8 # 100 = 8 A 10 Option (D) is correct. Peak current through S 1 I = I 0 + VS C/L = 20 + 200

2 # 10- 6 = 40 A 200 # 10- 6

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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia SOL 3.71

Option ( ) is correct.

SOL 3.72

Option (C) is correct.

Page 485

so a =; and

500 - (- 1500) 2p 2 E # 60 = 418.67 rad/sec 0.5

T = 40 Nm T = Ia I = T # 40 = 0.096 kgm2 a 418.67

SOL 3.73

Option (D) is correct. When thyristor turned on at that time J2 junction will break. So J1, J2, J3 all are in forward bias.

SOL 3.74

Option (D) is correct. The ON-OFF state of switch is given on VDS - IS plane as following

A I D

O N

no w.

.in

co ia.

d

ww

When VDS =+ ve , diode conducts and IS = 0 VDS =- ve , diode opens, but IS = 0 , D "- ve potential. SOL 3.75

Option (B) is correct. P. Field control-Above base speed Q. Armature control-below base torque

SOL 3.76

Option (A) is correct. As we know in fully controlled rectifier. or or

VPP = Vm - Vm cos (p/6 + a) VPP = Vm [1 - cos (p/6 + 30c)] VPP = 0.5 Vm

a a = 30c

SOL 3.77

Option ( ) is correct.

SOL 3.78

Option (A) is correct. In the chopper during turn on of chopper V -t area across L is, T T di dt = # imax Ldi #0 onVL dt = #0 on L b dt l i min = L (i max - i min) = L ^DI h

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Page 486

V -t are applied to ‘L’ is = (60 - 12) Ton = 48Ton So now volt area -3 DI = 48Ton = 48 # 0.2 #-10 = 0.48 A L 20 # 10 3 SOL 3.79

Option (A) is correct.

A I D

O N

no w.

.in

co ia.

d

ww

4VS b np l^sin nd h^sin nwt h^sin np/2h n = 1, 3, 5 ` RMS value of fundamental component Vrms(fundamental) = 4VS sin d # 1 2p a = 120c, 2d = 120c & d = 60c Output voltage V0 =

3

/

Vrms(fundamental) = 4VS # sin 60c 2p = 0.78VS = 0.78 V SOL 3.80

Option (A) is correct. After removing 5 th harmonic 5d = 0, p, 2p `

Pulse width = 2d = a = 0, 2p , 4p 5 5 = 0c, 72c, 144c

SOL 3.81

Option (C) is correct. NSa = 3000 rpm Na = 2850 rpm SFL = 3000 - 2850 = 0.05 3000

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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 487

where by (V/f) control

SOL 3.82

SOL 3.83

Nsb = 3000 b 40 l = 2400 rpm 50 ` N2 = new running speed of motor = 2400 b1 - 0.05 l = 2340 rpm 2 Option (C) is correct. For six pulse thyristor rectifier bridge the lowest frequency component in AC source line current is of 250 Hz. Option (A) is correct. Given a step down chopper is operated in continuous conduction mode in steady state with a constant duty Ratio D . V0 " dc output voltage . Vs " dc input voltage V0 = D = duty ratio Vs

SOL 3.84

Option ( ) is correct.

SOL 3.85

Option (B) is correct. From figure

A I D

(V12) rms = : 1 # V s2 dwD p0 f

O N

1/2

= Vs # f = Vs p

SOL 3.86

f p .in

.co

Option (C) is correct. dia o n Given that, V = 200 sin wtw. ww f = 50 Hz Power dispatched in the load resistor R = ? First we have to calculate output of rectifier. 1/2 p (V0) rms = : 1 # (200 sin wt) 2 dwtD p0 1/2 p = 200 ; # b 1 - cos 2wt l dwtE 2 p 0 p 1/2 = 200 ;1 b wt - sin 2wt l E 2 p 2 0 1/2 200 1 200 = : # pD = 2 p 2 Power dissipiated to resistor

2 ^V0h2rms e 200 o = 400 W = = 50 R 2

PR SOL 3.87

* Given

f = 20 kHz D = 0.5 Power transferred from source V1 to V2 = ? 1 Time period t = 1 = = 50 m sec f 20 # 10- 3 D = 0.5

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Page 488

tON = 25 m sec , t off = 25 m sec at tON , energy will stored in inductor circuit v = L di dt 100 = 100 # 10- 6 di dt di = 106 dt i = 106 t + i (0) i = 106 t

a i (0) = 0 ...(1)

E = 1 Li2 2 E = 1 # 100 # 10- 6 # 1012 # 25 # 25 # 10- 12 2

SOL 3.88

SOL 3.89

E = 3.1250 # 10- 2 J Now power transferred during t off -2 Pt = 3.1250 # 10 = 12.5 # 102 W -6 25 # 10 Option (B) is correct. For providing electrical isolation it is necessary to connect a pulse transformer at the output stage of a thyristor triggering circuit.

O N

a1 + a2 = 180c SOL 3.90

A I D

Option (A) is correct. In ac to dc circulating current dual converters if triggering angles are a1 and a 2 n , than it is necessary that o.i

.c

ia od

.n

Option (D) is correct. ww w Given a half wave Thyristor converter supplies a purely inductive load Triggering angle a = 120c than extinction angle b = ?

First we have to draw its output characteristics as shown below

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output is given by i 0 = Vm sin (wt - f) - Vm sin (a - f) exp b - R - a l Z wL Z We know at extinction angle i.e. wt = b , i 0 = 0 from equation (1), at (wt = b) 0 = Vm sin (b - f) - Vm sin (a - f) ec Z Z or sin (b - f) = sin (a - f) or b-f = a-f or b = a = 120c SOL 3.91

.in

no w.

co ia.

d

f = 50 Hz ww So total time = 1 = 1 = 20 msec 50 f Conduction time for each feedback diode in a cycle is being given by t conduction = 20 = 2.5 msec 8

a

SOL 3.92

...(1)

A I D

O N

Option (D) is correct.

Page 489

* Given a voltage commulated thyristor chopper circuit in figure which is operated at 500 Hz, with 50% duty ratio. IL = 20 A (constant) We have to evaluate (a) Toff for thyristor Th 1 (b) L = ? if peak current through Th 1 is 180% limited (c) Maximum instantaneous output voltage -6 Turn off time Toff = CVs = 6 # 10 # 100 = 30 m sec 20 IL Peak current through Th 1 i Th = I 0 + Vdc C L 1

a

i Th = 1.8IL = 1.8 # 20 = 36 A 1

36 = 20 + 100

6 # 10- 6 L

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or

Page 490

6 # 10- 6 L

0.16 =

-6 L = 6 # 10 2 = 2.34 # 10- 4 H (0.16) Maximum instantaneous output voltage

Vm = 2Vdc = 200 V SOL 3.93

* Given in figure separately excited dc motor is controlled by varying its armature voltage using 1-f full converter bridge. Motor voltage constant Kv = 2.5 V/rad/sec Motor Torque T = 140 Nm , a = 60c armature current continuous and ripple free. (a) Ia = ? (b) Nm = ? (c) rms of fundamental component of input current. T = Eb Ia

(a) a Motor Torque Than

A I D

and Eb = Kv w

Kv wIa = Tw

Ia = T = 140 = 50 Amp 25 Kv

O N

(b) In dc motor we know

in

o. Ia = V0 - Ebia.c Ra d no

. ww

Ew b = V0 - Ia Ra

V0 = 2Vm cos a p

Eb = 500 2 # 2 - 20 (0.2) p

= 2 # 250 2 cos 60c p

Eb = 215.2 V w = Ea Ia = 215.2 # 20 = 30.74 rad/sec T 140 (c) Rms value of fundamental component of input current Ior Isr = 1/2 2 ; 1 b(p - a) + 1 sin 2a lE p 2 Ior = 56 Amp , a = 60c 56 Isr = 1/2 1 p 2 : ap - k + 1 sin 120cD p 3 2 Isr =

39.6 = 61.34 Amp 2 - 1 1/2 b3 4l ***********

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