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EE6201-Circuit Theory

MAY– 2016

Answer key

PART A 1.The resistance of two wires is 25Ω when connected in series and 6Ω when connected in parallel. Calculate the resistance of each wire.

Rs = R1 + R2 = 25 Ω Rp =

R1 R2 150 = 6 Ω  R1 R2  6  25  150  R1  R1 + R2 R2

Hence we have, R2 2  25R2  150  0  R2  15 or 10

R1 = 10Ω or 15Ω 2. Distinguish between mesh and loop of a circuit.

A loop is any closed path of a network. A mesh is the most elementary form of a loop. In the circuit shown in Fig ABEFA and BCDEB are meshes but ABCDEFA is not a mesh because it encloses the above two loops.

3. State Reciprocity theorem. The reciprocity theorem states that in a linear, bilateral single source network the ratio of excitation (input) to response (output) is constant when the positions of excitation and response are interchanged. 4. What is the condition for maximum power transfer in DC and Ac circuits? In DC circuits maximum power is transferred to the load when RL = Rg In AC circuits maximum power is transferred to the load when RL = z g or Z L = Zg*

5. Define co efficient of coupling. The fraction of the magnetic flux produced by one oil linking with the other coil is known as coefficient of coupling between two coils. 6. A series RLC circuit has L  100 mH , and C  0.1 f . Find the resonant frequency. The resonant frequency is 0 

1 1   10 K rad / sec LC 0.1 0.1 106

Prepared by Prof.S.Nagammai, HOD/EIE, KLNCE

1

EE6201-Circuit Theory

MAY– 2016

Answer key

7. In a series RLC circuit has L  2H , and C  5  f . Determine the value of R to give critical damping. The condition for the critically damped response is, 2

1  R     2 L  LC

R2

L  1265 farad C

8.Define time constant of RL circuit. The time constant of RL transient circuit is, 



L sec R

.It is defined as the time taken by

the current response to reach 63.2 % of its final value. 9. A three phase, 400 Volts is given to balanced star connected load of 8  j 6  . Find the line current. The line voltage is,

(EE1151 MAY 2012) VL  440Volts

In a star connected system, line current is equal to phase current. Hence the line currents are, IR 

VRN V  00 400  L   23.1  370 Amps 0 Z 3 (8  j 6) 3 (1037 )

10. What are the advantages of three phase system?

(EE1151 NOV 13,15)

 The output of a three phase machine generating electricity is more than the output of a single phase machine of the same size.  Three phase power transmission is more economical than single power transmission.  The power factor of three phase system is better than single phase system.  The power output of a symmetrical three phase system is steady while that of single phase system is fluctuating. PART B 11.a.i) Determine the current

I L in

the circuit shown in Fig below.

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(8)

1

EE6201-Circuit Theory

MAY– 2016

Answer key

Solution: Let i1 , i2 and i3 be the mesh currents as indicated in Fig

By inspection the loop basis matrix is, 9 3 3 9   3 5

3 5  9 

i1   4      i2    6  i3   8

Using Grammar’s rule the above matrix is solved for current i2 and i3 .

  252  9 4 3 i2   3 6 5   168  3 8 9 

i2 

i2 168    0.65 Amps  258

 9 3 4 i3   3 9 6  420  3 5 8 

i3 

i3 420   1.63 Amps  258

The current I L  i2  i3  0.65  1.63  0.978 Amps Prepared by Prof.S.Nagammai, HOD/EIE, KLNCE

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EE6201-Circuit Theory

MAY– 2016

Answer key

11.a.ii). What is the voltage across terminals A and B in the circuit shown in Fig below. [8]

Solution: The given circuit is redrawn and shown in Fig….. Referring to Fig….., i1 

6  0.6 Amps 10

i2 

12  0.86 Amps 14

Applying KVL to loop ABCDA, VAB  4 i1  4 i2  12  0

VAB  4  0.86  4 0.6  12  10.96Volts

11.b.i) Three loads A, B and C are connected in parallel to a 240 V source. Load A takes 9.6 KW, Load B takes and Load C has a resistance of 4.8. Calculate (1) RA and RB (2) The total current (3) The total power and (4) equivalent resistance.(8) Solution: i)

V2 2402  RA   6 RA 9600

P

RB 

V 240   4 IB 60

ii) I A 

IC 

240  40 Amps 6

240  50 Amps 4.8

The total current is, IT  I A  I B  IC  150 Amps iii) The total power is, PT  VIT  240 150  36000 watts

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EE6201-Circuit Theory iv) The equivalent resistance is, Req 

MAY– 2016

Answer key

1 1 1    1.846  6 4 4.8

11.b.ii) For the circuit shown in Fig below find the total current IT ,phase angle and power factor. (8) (EE1151 NOV 2012/R 2008)

Solution: X L  L  2 100  0.1=62.83 

XC 

1 1  = 15.92  C 2  100  100  10-6

ZT  10  j15.92   30 j 62.83  34.697.05o  IT 

50 50   1.44  7.05o Amps ZT 34.697.05o

The phase angle is, 7.05o The power factor is, cos  cos7.05  0.9925

12.a. Find the voltage across 5Ω resistor in the circuit shown in Fig., using source transformation technique and verify the results using mesh analysis. [16] (EE1151 MAY 2012)

Solution: Let i1 , i2 and i3 be the mesh currents as indicated in Fig……. By inspection the loop basis matrix is,

Prepared by Prof.S.Nagammai, HOD/EIE, KLNCE

1

EE6201-Circuit Theory

MAY– 2016

Answer key

7 4 0  i1   67   4 15 6  i   152    2    0 6 13 i3   74 

Using Grammar’s rule the above matrix is solved for current i2 .   905

7 67 0  i2   4 152 6   7240  0 74 13

i2 

i2  8 Amps 

The voltage across 5Ω resistor is, V4  5i2  40Volts Source transformation technique: All the voltage sources are converted into current sources & the resulting circuit is shown in Fig….,

The reduced circuit diagram is shown in Fig below

Further simplifying the resulting circuit is,

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EE6201-Circuit Theory

The voltage across 5Ω resistor is, V4  5 

MAY– 2016

 35.86  57  13.25 3.23  5  1.71

Answer key

 40Volts

12.b.Obtain the Norton’s equivalent circuit and find the maximum power that can be transferred to the 100 load resistance in the network shown in Fig below

(16)

Solution: The 100Ω is removed and the terminals are made open. To find Rth , the batteries are short circuited.

Rth   220 || 470   200 || 330  274.4 

To find Isc: The output terminals are short circuited and the current is Isc.

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EE6201-Circuit Theory

MAY– 2016

Answer key

The loop equations are written in matrix form results in,

 690 470 0  i1  10   470 800 330  i    0     2    0 330 530  i3   5  Using Grammar’s rule the matrix is solved for current i2 .  100342000

 690 10 0  i 2   470 0 330   1352500  0 5 530  I sc = i2 = 13.5 mAmps

The Norton’s equivalent circuit is,

For maximum power transfer RL  Rth  274.4  The load current is, I L  13.5 

274.4  6.75 mAmps 274.4  274.4

The maximum power transferred to the load resistance is, PL  6.752  274.4  12.5 mwatts

13.a. Determine the resonant frequency, band width and quality factor of the coil for the series resonant circuit with R  10  , L  0.1H and

C  10 f

. Derive the formula used for

(16)

bandwidth.

Solution: The resonant frequency is 0 

The quality factor is,

Q

1 1   1000 rad / s LC 0.1  10  106

0 L 1000  0.1   10 R 10

Prepared by Prof.S.Nagammai, HOD/EIE, KLNCE

1

EE6201-Circuit Theory

MAY– 2016

Answer key

At the half-power frequencies the net reactance is equal to the resistance. At the lower half-power frequency the capacitive reactance exceeds the inductive reactance. Then,

1  1 L  R 1C 1  12 LC  1CR

12  1

R 1  0 L LC 2

From which, 1  

R 1  R     2L  2 L  LC

2

1  R  If    then, LC  2L 

1  

R 1 R   o  2L 2L LC

At the upper half power frequency the inductive reactance exceeds the capacitive reactance. Then

2 L 

1 R 2C

22 LC  1  2CR 2 2  2

R 1  0 L LC 2

From which, 2 

R 1  R      2L LC  2L 

2

1  R  If    then, 2 L LC  

2 

R 1 R    0 2L LC 2L

The distance between 1 and 2 measured in rad/sec, is called the band width (BW). Band width is, BW  2  1 

R  rad / sec  L

The band width is, BW  2  1 

R 100  rad / sec  L

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EE6201-Circuit Theory

MAY– 2016

Answer key

13.b. i) Derive the expression for equivalent inductance of the parallel resonant circuit as shown in Fig below.

[8]

Consider two coils with self inductances inductance of

M

L1 and L2 mutually

coupled with the mutual

are connected in parallel as shown in Fig....

The loop equations are V  jX1i1  j X mi2 V  jX 2i2  jX mi1

In matrix form,  jX 1   jX m 

 jX m   i1  V   jX 2  i2  V 

   X1 X 2  X m 2 V i1   V

 jX m   jV  X 2  X m  jX 2 

 jX 1 V  i2     jV  X 1  X m    jX m V 

i1 

i1 jV  X 2  X m    X m2  X1 X 2

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EE6201-Circuit Theory i2 

MAY– 2016

Answer key

i2 jV ( X 1  X m )   X m2  X1 X 2

I  i1  i2  Z eq 

jV  X 1  X 2  2 X m  X m2  X1 X 2

V X m2  X1 X 2  I j ( X1  X 2  2 X m )

j Leq 

 j  2 M 2   2 L1L2 

 L1   L2  2M 



j  L1L2  M 2  L1  L2  2M

L1L2  M 2 Leq  L1  L2  2M For the coils in which current enters both the coils,the equivalent inductance is Leq 

L1L2  M 2 L1  L2  2M (Parallel aiding connection)

13.b.ii) Write the mesh equations and obtain the conductively coupled equivalent circuit for the magnetically coupled circuit shown in Fig below.

[8]

Solution: The loop equations are, j5i1   3  j 4 (i1  i2 )  j 6  i2  50

5  j10 i2  3  j 4 (i2  i1)  j6  i1  0 In matrix form

3  j 2  i1  50  3 j  3  j 2 8  j 6  i    0    2  

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EE6201-Circuit Theory

MAY– 2016

Answer key

Z a   3  j   ; Zb   3  j 2   ; ZC   5  j 4   14.a. A series RL circuit with current transient.

R  10  , L  0.1 H

excited by V  10 sin100 t . Determine the resulting

[16]

Solution: The differential equation is, 10 i(t )  0.1

di(t )  10 Sin100 t dt

Taking Laplace transform on both sides,

10  0.1S  I (S )  I (S ) 

10  100 S 2  (100)2

10000 [S  10000][ S  100] 2

Appling partial fraction expansion, 10000 A BS  C   2 [ S  10000][S  100] S  100 S  10000 2

10000  A S 2  10000   BS  C  S  100

Comparing the coefficients of like powers of ‘S’

A B  0

S 2 terms,

100 B  C  0

S terms,

Constant terms,

10000  10000 A  100 C

Solving for the constants

A  0.5 ; I (S ) 

B   0.5 ;

C  50

0.5 0.5S 50  2  2 2 S  100 S  (100) S  (100)2

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EE6201-Circuit Theory I (S ) 

MAY– 2016

Answer key

0.5 0.5  S 0.5 100  2  2 2 S  100 S  (100) S  (100)2

i(t )  0.5 e100t  0.5 Co s 100 t  0.5 Sin 100t  0.5 e100t  0707 Sin (100t   )

Where   tan 1 

10  0   45  10 

i(t )  0.5 e100t

 0.707

Sin (100t  450 ) Amps

14.b. Consider a series RC circuit with

R  15  , C  200 f excited by 50Volt DC source. The

capacitor has an initial charge q0  100 C with an upper plate negative. Find the transient current.

[16]

Solution The differential equation of the circuit is, 15 i(t ) 

106 i(t )  50 200 

15I (s) 

106 I (s) 100  106 50   200 S S (200  106 ) S

 106  50 0.5 15    I ( s)   200S  S S 

50.5 3.37  333.3   1   I ( s)  S 15S S  

I (s)  S  333.3  3.37 I ( s) 

3.37 S  333.3

i(t )  3.37 e333.3t Amps

for t  0 sec

15.a. Obtain the readings of two wattmeters connected to a three phase, 3 wire, 120 V system feeding a balanced delta connected load with a load impedance 12300  each. Assume RYB phase sequence. Determine the phase power and compare the total power to the sum of wattmeter readings. [16] Solution: The line voltage is VL  VP  120Volt The phase voltages are,

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EE6201-Circuit Theory

MAY– 2016

Answer key

VRY  120 00 ; VYB  120   1200 ; VBR  120 1200 The phase currents are, I RY

VRY 120 00    10   300 Amps 0 Z 12 30

VYB 120   1200 IYB    10   1500 Amps 0 Z 12 30 I BY 

VBY 120 1200   10 900 Amps Z 12 300

For a delta load, the line current always lags the corresponding phase current by 30o & has a magnitude

3 times that of a phase current.

Hence the line currents are, I R  I RY  I BY  10   300 10   1500  17.32   600 Amps 0 0 0 Or I R  3I RY   30  3  10    30  30 

I R  17.32   600 Amps

IY  IYB  I RY  10   1500  10   300  17.32   1800 Amps 0 0 0 Or IY  I R  120  17.32   120  60 

IY  17.32   1800 Amps I B  I BY  IYB  10 900  10   1500 I B  17.32 600 Amps 0 0 0 Or I B  I R120  17.32  120  60 

I B  17.32 600 Amps

The power factor, cos  cos300  0.866 The total active power is,

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EE6201-Circuit Theory

MAY– 2016

Answer key

PT  3 VL I L cos  3 120 17.32  0.866  3.12 KW

The watt meter readings are, (  is the angle of load impedance) w1  VL I L cos 30 0 

0

w2  VL I L cos 30 0 

0

  120 17.32cos 30  30   1039 watts 0

0

  120 17.32cos 30  30   2078 watts 0

0

15.b.i). If W1 and W2 are the readings of the two wattmeters which measures power in the three phase balanced system and if of the circuit is given by cos 

W1  a , show that the power factor of the circuit W2

a 1

[8]

2 a2  a  1

For a three phase balanced system, W  W2 tan   3  1 W1  W2 As ,

W1  a , we get tan   W2

sec2   1  tan 2   cos 

cos 

1 1  tan  2



3

a 1 a 1

1  1  tan 2  cos2  1

a  1  1  3  a  1  

2

1



 a  1 2  a  1 2

1 3

 a  1 2 a2  a  1

15.b.ii) A symmetrical, three phase, three wire 440 V ABC system feeds a balanced Y 0 connected load with impedance 1030  in each phase. Obtain the line currents. [8] The line voltage is, VL  440Volts In a star connected system, line current is equal to phase current. Hence the line currents are, IR 

VRN VL  00 440    25.4  300 Amps Z 3 (10300 ) 3 (10300 )

IY 

VYN V   1200 440  1200  L   25.4  1500 Amps Z 3 (10300 ) 3 (10300 )

IB 

VBN 4401200   25.4900 Amps 0 Z 3 (1030 )

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