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+ -

+

-

++

o

12 mA

|| 12 mA

12mA

-

-

12mA ^

O

''''~Io 11'210>|

!>~n ~

|

---1------

11> - 4.5m A

-

R1

R 2 = 12

R eq -

>

o + + o -

2 kW

-o + + -o

-

+

^

+

-

20V

1

+

+

(-

4 - -

5

5


1

+ -

l

10kW

+ -

s

+

-

6kW

-

+

R

R

3

Vs

s

|

3

+ -

+ +-

3

2.88 Given I"

=

2 mA in the network in Fig. P2.88,

find V.I'

VA 3

v.. ..•.

1 kn

Iq

+ VJ1 kfl

-'t 1 kfl

~

-

+

2kf!

.,. _I

KC,.L""

::.

~,2)C.

1,).. t I~.= I~

VA ::. -2,h - 2-D

|

V2

3

+

-

+ -

+ -

c

s

=

+

= =

=

-

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: Is

>

R1 12V

Ix

R2

R3

R4

R5

R6 The correct answer is d. Req = { [[( R5 + R6 )

R4 ] + R3 ]

Req = { [[(8 + 2) 10] + 1]

R2 } + R1

3} + 1

Req = (6 3) + 1 = 3Ω Is =

12 12 = = 4A Req 3

R ' = [( R5 + R6 )

R4 ] + R3 = 6Ω

24 8 ⎛ 6 ⎞ = A Ix = ⎜ ⎟(4) = 9 3 ⎝3+ 6⎠

Chapter 2: Resistive Circuits

Pr oblem 2.FE-10

3k W

| -

1 O

2 kW

Vo

6 kW

4 kW

l

2

3

o

3

\/

> -+

+

Vo -

>

> '

'

+ \/

o

+ -

\/

+ -

'

-

+

o \/

+

-

-+

-o +

o

\/ x

1

\/

'

'''

>

\/

6Vx

'

"

'''

\/

'

^ \/

\/ 2

o\/

2

-O

-O

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is c. ⎛ − R2 ⎞ ⎛ R ⎞ Vo = ⎜ ⎟(4) − ⎜ 2 ⎟(−2) ⎝ 4k ⎠ ⎝ 12k ⎠ Vo = −3V 1 − 3 = −1m( R2 ) + m( R2 ) 6 5 − m ( R 2 ) = −3 6 R2 = 3.6kΩ

Chapter 4: Operational Amplifiers

Problem 4.FE-1

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: 36kΩ 18kΩ

-

V1

+ 6kΩ

+ 6kΩ

12 kΩ

12kΩ

2V

3V 1V

The correct answer is b.

⎛ 18k ⎞ ⎛ 18k ⎞ V1 = −2⎜ ⎟(4) + 1⎜ ⎟ = −4.5V ⎝ 6k ⎠ ⎝ 12k ⎠ ⎛ 36k ⎞ ⎛ 36k ⎞ V0 = −V1 ⎜ ⎟ ⎟ − 3⎜ ⎝ 6k ⎠ ⎝ 12k ⎠ ⎛ 36k ⎞ ⎛ 36k ⎞ V0 = 4.5⎜ ⎟ ⎟ − 3⎜ ⎝ 6k ⎠ ⎝ 12k ⎠ V0 = 18V

Chapter 4: Operational Amplifiers

Problem 4.FE-2

Vo

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: if 2Ω

i=0 -

Vo

+

3Ω 4Ω

6V

2V

The correct answer is a. V1 V2 6 2 + = + = 2 .5 A R1 R2 3 4 V0 = −i f R f = −(2.5)(2) if =

V0 = −5V

Chapter 4: Operational Amplifiers

Problem 4.FE-3

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is d. The op-amp is a noninverting op-amp. Rf A = 1+ R1 R f = ( A − 1) R1

R f = (50 − 1)5k = 245kΩ

Chapter 4: Operational Amplifiers

Problem 4.FE-4

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: R1 = 6kΩ + -

+

R2 = 2kΩ

5V R3 = 1kΩ

8kΩ

Vo

2 kΩ

−

The correct answer is c. The 8kΩ and 2kΩ resistors make up a noninverting op-amp. ⎛ 8k ⎞ V1 = ⎜1 + ⎟5 = 25V ⎝ 2k ⎠ Use nodal analysis at node A: Vo Vo − Vi Vo − 25 + + =0 R3 R1 2k

Chapter 4: Operational Amplifiers

Problem 4.FE-5

2

Irwin, Basic Engineering Circuit Analysis, 9/E

Vo Vo − 5 Vo − 25 + + =0 1k 6k 2k 6Vo + Vo − 5 + 3Vo − 75 = 0

10Vo = 80 Vo = 8V

Problem 4.FE-5

Chapter 4: Operational Amplifiers

+ –

_ |

|

| .

+

.

+

-

6k

K

-T'0

'

"

-

-

'

A^

eq/

eq/

-

'

0

'''

+-

+ -

+ _

+ –

+ -

-

"

T -

"

+ –

+ -

+ –

+ -

-+

+ -

–+

>

=

-+ +

-+ +

–+

+

-

+ –

– +

2

x

+ -

+ x

-

x

+–

–+

–+ – +

–+ + –

2

+–

+ -

+ -

=

+ + –

– +

–

0

+

-

+ + -

-

+ -

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:

Chapter 5: Additional Analysis Techniques

Problem 5.113

2

Problem 5.113

Irwin, Basic Engineering Circuit Analysis, 9/E

Chapter 5: Additional Analysis Techniques

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:

Chapter 5: Additional Analysis Techniques

Problem 5.114

2

Problem 5.114

Irwin, Basic Engineering Circuit Analysis, 9/E

Chapter 5: Additional Analysis Techniques

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:

Chapter 5: Additional Analysis Techniques

Problem 5.115

2

Problem 5.115

Irwin, Basic Engineering Circuit Analysis, 9/E

Chapter 5: Additional Analysis Techniques

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:

Chapter 5: Additional Analysis Techniques

Problem 5.116

2

Problem 5.116

Irwin, Basic Engineering Circuit Analysis, 9/E

Chapter 5: Additional Analysis Techniques

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:

I' R1

R2

R4



Voc'

4mA

R3



The correct answer is d. Use superposition.   R1 (4m) I '    R1  R2  R3  1k   I'  (4m)  1mA 1 k  1 k  2 k   ' Voc  I ' R3  (1m)(2k )

Voc'  2V

Chapter 5: Additional Analysis Techniques

Problem 5.FE-1

2

Irwin, Basic Engineering Circuit Analysis, 9/E

R2

R1



R4

12V

Voc"

R3



  R3 (12) Voc"    R1  R2  R3  2k   Voc"   (12)  1k  1k  2k  Voc"  6V Voc  Voc'  Voc"  2  6  8V

R1

R2

R4

R3

RTH

RTH  R1  R2  R3   R4 RTH 

2k (2k )  1k  2k  2k  2k

Problem 5.FE-1

Chapter 5: Additional Analysis Techniques

Irwin, Basic Engineering Circuit Analysis, 9/E

3

RTH  2k

Voc  8V

RL

PLmax  I L2 RL

R L  RTH for maximum power.  V PLmax   oc  2 RTH PLmax

2

  RTH  V2 82  oc   8mW 4 RTH 4(2k )

Chapter 5: Additional Analysis Techniques

Problem 5.FE-1

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:  Vx 

2k

1k

I 

12V

Voc 

+ -

2V x

The correct answer is c. 12  2kI  1kI  2V x V x  I ( 2k ) 12  2kI  1kI  2(2kI ) 12 I  mA 7 12  2kI  Voc  12  12  2k  m   Voc 7  Voc  8.57V

Chapter 5: Additional Analysis Techniques

Problem 5.FE-2

2

Irwin, Basic Engineering Circuit Analysis, 9/E

 Vx 

2k

I1

12V

1k

I sc

I2

+ -

2V x

 12  I 1     6mA  2k 

1kI 2  2V x  0

1kI 2  2(2kI1 )  0 I 2  24mA I 1  I 2  I sc I sc  6m  (24m)  30mA Voc 8.57   285.7 I sc 30m RL  RTH for maximum power.

RTH 

RTH  285 .7

Voc  8.57V

PLmax 

Problem 5.FE-2

RL

Voc2 (8.57) 2   64.3mW 4 RTH 4(285.7)

Chapter 5: Additional Analysis Techniques

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:

Ix 

R1

12V

I R2

2I x

text

Voc



The correct answer is a. I  I x  2I x I  3I x 12  3I x  12 I 12  3I x  12(3I x ) 4 Ix  A 13

4 Voc  12 I  12(3I x )  12(3)   11.08V  13 

Chapter 5: Additional Analysis Techniques

Problem 5.FE-3

2

Irwin, Basic Engineering Circuit Analysis, 9/E

Ix R1

I2 12V

R2

2I x

I sc

text

I sc  I x  I 2  2 I x  3I x  I 2 I2  0A I sc  3I x 12  3I x

I x  4A I sc  3(4)  12 A V 11.08  0.92 RTH  oc  12 I sc

RTH  0.92

12

Voc  11.08V

RL

RL  0.92  12  12.92 R L  12.92 for maximum power transfer.

Problem 5.FE-3

Chapter 5: Additional Analysis Techniques

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:

2

I' 3

10 A

2

4

The correct answer is c. Use superposition.  2  I'   (10)  4 A  2  3

I" 3

Chapter 5: Additional Analysis Techniques

2

4

20V

2

Problem 5.FE-4

2

Irwin, Basic Engineering Circuit Analysis, 9/E

20  4A 5 I  I'  I" I  4  4 I  0A I" 

Problem 5.FE-4

Chapter 5: Additional Analysis Techniques

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: Use superposition.

4

a I

2

3

'

12 A



Voc'



b The correct answer is b.

 4  I'  (12)  8 A 2 4 Voc'  I ' (2)  8(2)  16V

Chapter 5: Additional Analysis Techniques

Problem 5.FE-5

2

Irwin, Basic Engineering Circuit Analysis, 9/E

a 3

4 12V

2



Voc"



b  2  Voc"   (12)  4V 2 4 Voc  16  4  12V

Problem 5.FE-5

Chapter 5: Additional Analysis Techniques

-

S S

s s

S S

t

2.24

/

c

3

t

t

t

3

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is a. Yes. The capacitors should be connected as shown.

6 F 2 F

4 F

C eq 

6  (6  )  3 F 6  6

Chapter 6: Capacitance and Inductance

Ceq

Problem 6.FE-1

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is c. q (t )   i (t ) dt

0 C , t  0  q (t )  6t C , 0  t  1 s 6  C , t  1 s  q (t ) C 0 V , t  0  v(t )  6 x10 6 t V , 0  t  1 s 6 V , t  1 s  v(t ) 

1 C v 2 (t ) 2 0 J , t  0  w(t )  18 x10 6 t 2 J , 0  t  1 s 18 J , t  1 s 

w(t ) 

Chapter 6: Capacitance and Inductance

Problem 6.FE-2

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is b. The voltage across the unknown capacitor Cx is (using KVL): 24  8  V x V x  16V q  Cv The capacitors are connected in series and the charge is the same. q  60  (8)  480  C q 480   30  F Cx   v 16

Chapter 6: Capacitance and Inductance

Problem 6.FE-3

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: 2 mH

Leq

3mH

3mH

6mH

12mH

9mH

The correct answer is d. Leq  [ [(3m  9m) 12m] 6m 3m]  2m Leq  [ [(12m) 12m] 6m 3m]  2m

Leq  [ 6m 6m 3m]  2m Leq  [ 3m 3m]  2m Leq  1.5m  2m Leq  3.5mH

Chapter 6: Capacitance and Inductance

Problem 6.FE-4

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is a. di(t ) v(t )  L dt di(t )  20te  2t (2)  20e  2t  20e  2t  40te  2t dt





v(t )  10m 20e  2t  40te  2t 0 V , t  0 v(t )   2t  2t 0.2e  0.4te V , t  0

Chapter 6: Capacitance and Inductance

Problem 6.FE-5

2

3

c

\/

i(t)

i(t)

\/

t

____

____

.

(

-

)

-

n

^

>

0

|

:

-

C

C

^ i(t)

3

-

-

2

o o

-

-

c

-

.

.

.

.

.

.

.

Note: Please go to next page to see the graph on Output Voltage vs Time

^ i L (0+)

-

=

2

R2

-

-

-

-

- -

-

> i(t)

-

Note: Please go to next page to see the graph on Output Voltage vs Time

-

12V

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:

8k

6 k

 vc (0)

6k

12V



6 k

The correct answer is b.

R' 

12k (6k )  4k 12k  6k

8k  12V

R '  4 k

Chapter 7: First- and Second- Order Transient Circuits

v c ( 0 ) 

Problem 7.FE-1

2

Irwin, Basic Engineering Circuit Analysis, 9/E

 4k  v c ( 0 )    (12)  4V  4 k  8k 

The t  0 circuit:

6k  vc (t )

6k



vc (t) 12k x i c (t)  0 dv (t ) ic (t )  C c dt dv (t ) vc (t )  12k (100 ) c 0 dt dvc (t ) 1  vc (t )  0 dt 1.2 1 r 0 1.2 1 r 1.2 vc (t )  Ae

t 1.2

vc (0)  4V A4 vc (t )  4e

t 1. 2

V, t  0

vc (2)  0.756V

Problem 7.FE-1

Chapter 7: First- and Second- Order Transient Circuits

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: Find the initial condition:

12k

4k 

12 k

12V

v0 (0) 

The correct answer is d. vo (0)  0V

The t  0 circuit:

Chapter 7: First- and Second- Order Transient Circuits

Problem 7.FE-2

2

Irwin, Basic Engineering Circuit Analysis, 9/E

R3  4k

R1  12k



R 2  12 k

v0 (t )

12V



R 3  4 k

R1  12k

 R 2  12 k

voc 

12V

voc 

R2 12k (12)  (12)  6V R1  R2 12k  12k

R1  12k R 2  12 k

R3  4k

RTH

RTH  ( R1 R2 )  R3 RTH 

Problem 7.FE-2

12k (12k )  4k  10k 12k  12k

Chapter 7: First- and Second- Order Transient Circuits

Irwin, Basic Engineering Circuit Analysis, 9/E

3

RTH  10k

100 F

voc  6V

 v0 (t ) 

10k x i(t) o v (t)  6 dv (t ) ic (t )  C o dt dv (t ) RTH C c  v o (t )  voc dt v dv o (t ) 1  v o (t )  oc RTH C dt RTH C 1 0 r RTH C 1 r RTH C The natural solution is: t

von (t )  Ae RTH C t

von (t )  Ae 10 k (100  )  Ae t The forced solution is: vo f (t )  k

dv o f (t ) dt

0

1

0

k

RTH C k  voc vo f (t )  6V

voc RTH C

vo (t )  6  Ae  t vo (0)  0V 0  6 A A  6

Chapter 7: First- and Second- Order Transient Circuits

Problem 7.FE-2

4

Irwin, Basic Engineering Circuit Analysis, 9/E

vo (t )  6  6e  t V , t  0 vo (1)  6  6e 1

vo (1)  3.79V

Problem 7.FE-2

Chapter 7: First- and Second- Order Transient Circuits

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: Find the initial condition: 12 k

 v c (0  )

12V

6k



The correct answer is a.

 6k  v c ( 0)   (12)  4V  6k  12k  The t  0 circuit:

 vc (t )

6k



Chapter 7: First- and Second- Order Transient Circuits

Problem 7.FE-3

2

Irwin, Basic Engineering Circuit Analysis, 9/E

vc (t )  6kic (t )  0 dv (t ) ic (t )  C c dt dv (t ) vc (t )  6k (100  ) c 0 dt dvc (t ) 1  vc (t )  0 dt 0.6 1 r 0 0.6 1 r 0.6 t

vc (t )  Ae 0.6

vc (0)  4V A4 t

vc (t )  4e 0.6 V , t  0 t

0.5(4)  4e 0.6 t

1  e 0.6 2 1 t ln  2 0.6 t  0.6 ln t  0.416 s

Problem 7.FE-3

1 2

Chapter 7: First- and Second- Order Transient Circuits

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: Find the initial condition:

i L (0)

2

2

10V

The correct answer is c.

i L ( 0 ) 

10  5A 2

Chapter 7: First- and Second- Order Transient Circuits

Problem 7.FE-4

2

Irwin, Basic Engineering Circuit Analysis, 9/E

1

5A

 voc

2

1A



1 2

2  3

2  3

6A

 voc 

2 voc  6   4V 3

2

2

Problem 7.FE-4

RTH

2

Chapter 7: First- and Second- Order Transient Circuits

Irwin, Basic Engineering Circuit Analysis, 9/E

RTH  2 2  2  1 2 

2  3

RTH 

2  3

voc  4V

3

iL (t )

4H

di L (t ) 2  i L (t )  4 dt 3 di L (t ) 1  i L (t )  1 dt 6 1 r 0 6 1 r 6 4

t

i Ln (t )  Ae 6 i L f (t )  k

di L f (t )

0 dt 1 0 k 1 6 k 6 i L f (t )  6

i L ( 0 )  5 A t

i L (t )  6  Ae 6 56 A A  1 t

i L (t )  6  e 6 A, t  0

Chapter 7: First- and Second- Order Transient Circuits

Problem 7.FE-4

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: Find the initial condition: i L (0)

3 1

12V

The correct answer is d. i L ( 0 ) 

12  3A 4

Chapter 7: First- and Second- Order Transient Circuits

Problem 7.FE-5

2

Irwin, Basic Engineering Circuit Analysis, 9/E

L

iL (t )

R1 R2 Vs

di L (t )  R1  R2 i L (t )  Vs dt V di L (t ) R1  R2 i L (t )  s  dt L L R1  R2 0 r L R  R2 r 1 L L

i Ln (t )  Ae i Ln (t )  Ae

 R R   1 2  t  L  5  t 3

i L f (t )  k di L f (t )

0 dt 5 0 k  4 3 12 k 5

i L (0  )  3 A 5

 t 12 i L (t )   Ae 3 5 3  2. 4  A A  0. 6

i L (t )  2.4  0.6e

Problem 7.FE-5

5  t 3

A, t  0

Chapter 7: First- and Second- Order Transient Circuits

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l

ll

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t

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AC ANALYSIS

***************************************************************************************************************************

FREQ

IM(V_PRINT1)

4.000E+02

6.058E+00

IP(V_PRINT1) -1.361E+01

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:

2

j1

120V

+ 40 V

 j1

1

V0 -

The correct answer is d.

j1

+ 80 A

2

 j1

1

V0 -

Chapter 8: AC Steady-State Analysis

Problem 8.FE-1

2

Irwin, Basic Engineering Circuit Analysis, 9/E

Z 

2( j1)  0.4  j 0.8 2  j1 I0 j1

80  A

Z

1

  0.4  j 0.8 80  1.6  j 4.8 A I 0    0.4  j 0.8  1  j1  V0  1.6  j 4.81  5.06  71.6V

Problem 8.FE-1

Chapter 8: AC Steady-State Analysis

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:

text

2I x

1

V1 I1

 j1

2

V2

V0

I2

120V

20 A

Ix The correct answer is c.

Chapter 8: AC Steady-State Analysis

Problem 8.FE-2

2

Irwin, Basic Engineering Circuit Analysis, 9/E

V2  120V V Ix  1  j1 2I x  I x  I1  0  V  V V  120 2 1   1  1 0 1   j1   j1 2V1  V1   j1V1  j1(120)  0  j1(120) V1   1.2  j 3.6V 3  j1 KCL at node 0: 2 I x  I 2  20  0  V  V  V0 2 1   2  20  0 2   j1   1.2  j 3.6  120  V0   2  20  0 2   j1  V0  30.88.97V

Problem 8.FE-2

Chapter 8: AC Steady-State Analysis

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:

I

2

j 2

 j1

I1 120V

2V0

I2 +

 4

V0 

-

The correct answer is a.

I1  I  I 2 I  I1  I 2 KVL around the left loop: 120  2 I 1  j 2 I  2V0 V0  4I 2 (2  j 2) I 1  (8  j 2) I 2  120 KVL around the right loop: 2V0   j 2 I  j1I 2  4 I 2 j 2 I 1  (4  j1) I 2  0

Chapter 8: AC Steady-State Analysis

Problem 8.FE-3

2

Irwin, Basic Engineering Circuit Analysis, 9/E

Two equations and two unknowns: (2  j 2) I 1  (8  j 2) I 2  120 j 2 I 1  (4  j1) I 2  0 It follows that: I 1  4.2445 A

I 2  2.06  30.96 A V0  4( 2.06  30.96) V0  8.24  30.96V

Problem 8.FE-3

Chapter 8: AC Steady-State Analysis

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is b. Z c is small at midband. Vx 5k 5   Vs 5k  1k 6

V0  6k (12k )   40 x10 3    160 Vx  6k  12k  V0  V x  V0    Vs  Vs  V x

 5    160  133.33  6

Chapter 8: AC Steady-State Analysis

Problem 8.FE-4

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:

Is 60V Zeq

The correct answer is c. (1  j1)(1  j3)  3  1.58  18.43  3  4.53  6.34  1  j1  1  j 3 60 Is   1.32456.34 A 4.53  6.34

Z eq 

Chapter 8: AC Steady-State Analysis

Problem 8.FE-5

2

Irwin, Basic Engineering Circuit Analysis, 9/E

3

 j 1

Io j 3

1

60 V

1

KVL around the outer loop: 60  3(1.32456.34)  I 0 (1  j1) I 0  1.4832.92 A

Problem 8.FE-5

Chapter 8: AC Steady-State Analysis

-

c

>

I2

>I

4

-

>

-I

I

1

>

3

-

supplied

W

>I

-

-

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39

2 x

1 2

\/

|

2

2

2

2

Unknown element

.

.

.

.

-

.

.

.

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is c. From the power triangle: Qold  Pold tan  old

Qold  120k tan 45  120k var S old  (120k  j120k ) VA  17045 kVA S new  Pold  jQnew

 new  cos 1 (0.95)  18.19 Qnew  Pold tan  new Qnew  120k tan 18.19  39.43k var S new  120k  j 39.43k VA S new  126.3118.19 kVA S cap  S new  S old   j80.57 kVA Qcap   CV 2  80.57 k  2 60(C )(480 2 ) C  928 F

Chapter 9: Steady – State Power Analysis

Problem 9.FE-1

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is d. T

Vrms 

1 2 v (t ) dt T 0

Vrms 

2 3 1 4  1 2 2 2 2   1 dt   2 dt   1 dt  0 dt  4 0 1 2 3 

Vrms  Vrms 





1 1 2 3 t 0  4t 1  t 2  0 4 1 1  (8  4)  (3  2)  1.22V 4

Chapter 9: Steady – State Power Analysis

Problem 9.FE-2

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: 2

 j1 j 2

ZTH

The correct answer is b. Z TH  2  j1  j 2  0.4  j1.2 * Z L  Z TH  0.4  j1.2

Chapter 9: Steady – State Power Analysis

Problem 9.FE-3

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is a. 36k Vs 12k 1.414 Vs   10 V rms 2 36k 10  30 V rms V0   12k V0  

The meter will read 3V.

Chapter 9: Steady – State Power Analysis

Problem 9.FE-4

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is b. T

I rms 

1 2 i (t ) dt T 0

I rms 

1 2  1 2 2   t dt   (2) dt  2 0 1 

I rms 

1 1   (8  4)  1.47 A  2 3 

2 P  I rms R  (1.47) 2 ( 4)  8.64 W

Chapter 9: Steady – State Power Analysis

Problem 9.FE-5

.

.

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.

.

-

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-

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-

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Ideal 1:1

1:2

Ideal

j

.

.

1:4

Ideal

2:1

.

.

.

.

.

.

o

o

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:

M 4

24 0V

I1

 j 5 j 8

j 2

I2

5

k=0.5

The correct answer is a.

M  k L1 L2  1H 240  I 1 (4  j 2)  j 2 I 2  j 2 I 1  (5  j 3) I 2  0 I 1  4.92  19.75 A

I 2  1.6939.29 A 240 Z1   4.8819.75 4.92  19.75

Chapter 10: Magnetically Coupled Networks

Problem 10.FE-1

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:

48 n 2 

j 32n 2 

 j 2

R L  3

V2

The correct answer is b.

Z TH  3 for maximum power transfer. Reflecting the primary into the secondary: Z TH  48n 2  j 32n 2  j 2  3 48n 2  j32n 2  j 2  3 1 if n  : 4 Z TH 

48  32   j   2   3 16  16 

Chapter 10: Magnetically Coupled Networks

Problem 10.FE-2

2

Irwin, Basic Engineering Circuit Analysis, 9/E

3

I RL  3

300V

300  50 A 6 1 1 PL  I M2 RL  (5) 2 (3)  37.5W 2 2 I 

Problem 10.FE-2

Chapter 10: Magnetically Coupled Networks

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:

1k

+ +

5k

Vs

4V x 100

Vx

10 k

Voc

-

The correct answer is d.   5  4 Voc   Vs   x10 4    6  100

1k + Vs

5k

Vx

4V x 100

10 k

I sc

-

 5  4  I sc   Vs     6  100  Find Rout of the amplifier: 5 V x  Vs 6

Chapter 10: Magnetically Coupled Networks

Problem 10.FE-3

2

Irwin, Basic Engineering Circuit Analysis, 9/E

 5  4  x10 4   Vs   V 6  100   10k Rout  oc   I sc  5  4   Vs     6  100  16a 2  10k a  25

Problem 10.FE-3

Chapter 10: Magnetically Coupled Networks

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is c. Z1 

ZL  2 2 (1  j1)  4  j 4 n2

1200  11.7711.31 A 6  j2  4  j4 I I2  1 n I 2  23.5411.31 A I1 

Chapter 10: Magnetically Coupled Networks

Problem 10.FE-4

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is a. I1 

V1 Z1

ZL n2 N n 2 2 N1 10  j10 Z1   2.5  j 2.5 22 1200 I1   24  j 24 A 2.5  j 2.5 I 24  j 24 I2  1  n 2 I 2  12  j12 A  16.97  45 A Z1 

Chapter 10: Magnetically Coupled Networks

Problem 10.FE-5

\

-

I aA

>

.

-

> 3

-

AN

AN

-

an

PF

36

o

A

/

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is d.

I  6 A rms VR  84.85V rms VL  84.85V rms

R

VR

L 

I VL I



84.85  14.14 6



84.85  14.14 6

Z load  14.14  j14.14  2

S 3  3 I Z load S 3  36  14.14  j14.14  2

S 3  2.1645 kVA

Chapter 11: Polyphase Circuits

Problem 11.FE-1

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is a.

Z   12  j12  Finding the equivalent per-phase wye: Z Y  4  j 4  5.6645 V AB  230 V rms

V AN 

I aA 

230 3

V AN ZY

 132.79V rms



132.79  23.5 A rms 5.66

P3  3 V AN I aA cos Z  3(132.79)(23.5) cos 45  6.62kW

Chapter 11: Polyphase Circuits

Problem 11.FE-2

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is b. 24  j18  8  j 6 3  6  j 4

ZY1  ZY 2

V AB  208V rms

V AN 

208

 120V rms 3 1200 I aN 1   12  36.87 A 8  j6 1200 I aN 2   16.64  33.69 A 6  j4 I aA  I aN 1  I aN 2  28.63  35.02 A rms

Chapter 11: Polyphase Circuits

Problem 11.FE-3

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is c. S 3  2430 kVA  20784.61  j12000 VA P3  20.78kW P1  6.93kW

Chapter 11: Polyphase Circuits

Problem 11.FE-4

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is d. By use of the power triangle: S 2  P2  Q2

Q  S 2  P2

Q  (100k ) 2  (90k ) 2 Q  43.59k var

Chapter 11: Polyphase Circuits

Problem 11.FE-5

|

-

4w

)

2

-

l

-

/

)

_

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

/ -

-/

-

4

4

4

. .

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION:

The correct answer is a.

0 

1 LC



1 1m(10  )

 10,000

rad s

    1 1  Z ( j 0 )    60   60  90V V0 ( j 0 )  120 c  60  2    10,000(10  )90    0 C90 

Chapter 12: Variable – Frequency Network Performance

Problem 12. FE-1

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is c.

0 

BW 

1 LC



1 20m(50  )

 1000

rad s

R rad  200 L s

R  200(20m)  4

Chapter 12: Variable – Frequency Network Performance

Problem 12. FE-2

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is d. 1 1 V0 ZC jC RC     Gv ( j ) 1 1 VI Z C  R  R j  jC RC at DC, Gv=1=0dB at 3dB down,  



1 RC

rad 1  200 s (5k )(1 )

  2f f 

  31.83  32 Hz 2

Chapter 12: Variable – Frequency Network Performance

Problem 12. FE-3

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is b.

0  L

1 LC

 1000

rad s

1 1   100mH 2  0 C (1000) 2 (10  )

BW 

R rad  100 L s

R  100m(100)  10

Chapter 12: Variable – Frequency Network Performance

Problem 12. FE-4

Irwin, Basic Engineering Circuit Analysis, 9/E

1

SOLUTION: The correct answer is a.

f  8 Hz ,  1  16

rad s

1   j 2 k jC 1 1 j C RC Gv ( j 1 )   1 1  R j  jC RC ZC 

For f=8Hz and 1=16 V0 ZC  j 2000  G v ( j 1 )    0.707  45 VS Z C  R 2000  j 2000 Gv ( j16 )  0.707  45

fC 

1 1   7.96  8 Hz 2RC 2 (2k )(10  )

Chapter 12: Variable – Frequency Network Performance

Problem 12. FE-5

s

s

u

C C C

S

S

C

C

S

C

S

C t

l

C

S

sC

C

S

C

C

S

C

S

C

C

S

__________________________

l

+

+

+

+

+

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-

+

+

+

-

+

+

+

+

+

-

+

+

+

s

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

-

+

+

-

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+

+

+

+

+

+

+

+

+

+

+

+

l

l

+

+

+

+

2

+

+

l

+

+

+

+

u

A

V