Solution manual fundamentals of electric circuits 3rd edition chapter11

11.40, determine the average power absorbed by the 40- Ω resistor... 11.44, determine the load impedance Z for maximum power transfer to Z.. Calculate the maximum power absorbed by the l

Chapter 11, Problem 1

If v(t) = 160 cos 50t V and i(t) = –20 sin(50t – 30°) A, calculate the instantaneous power

and the average power

Chapter 11, Solution 1

) t 50 cos(

160 ) t (

) 90 180 30 t 50 cos(

2 ) 30 t 50 sin(

20 - ) t (

) 60 t 50 cos(

20 ) t (

) 60 t 50 cos(

) t 50 cos(

) 20 )(

160 ( ) t ( i ) t ( v ) t (

[ cos( 100 t 60 ) cos( 60 ) ] W 1600

) t (

= ) t (

p 800 + 1600 cos( 100 t + 60 ° ) W

) 60 cos(

) 20 )(

160 ( 2

1 ) cos(

I V 2

1

=

P 800 W

Given the circuit in Fig 11.35, find the average power supplied or absorbed by each element.

Chapter 11, Problem 3

A load consists of a 60- Ω resistor in parallel with a 90 µ F capacitor If the load is

connected to a voltage source v (t) = 40 cos 2000t, find the average power delivered to s

the load

Chapter 11, Solution 3

I

C 40∠0˚ R

I = 40/60 = 0.6667A or Irms = 0.6667/1.4142 = 0.4714A

The average power delivered to the load is the same as the average power absorbed by

the resistor which is

Pavg = |Irms|260 = 13.333 W

+ –

Find the average power dissipated by the resistances in the circuit of Fig 11.36

Additionally, verify the conservation of power

Figure 11.36

For Prob 11.4

20 30

2.438 3.0661 A 5

o

o o

V I

j

− The average power dissipated by the resistor is

Assuming that v = 8 cos(2t – 40º) V in the circuit of Fig 11.37, find the average power s

delivered to each of the passive elements

6828 1 P

38 25 6828 1 2 j 2 6 j

) 2 j 2 ( 6 j 1

40 8 I

21

− +

P3H = P0.25F = 0

W 097 5 2 2

3 3

20 mH ⎯⎯ → j L ω = j 10 20 10 x x − = j 20

25 j 10

x 40 x 10 j

1 C

j

1 F

40

6

= ω

→

We apply nodal analysis to the circuit below

Vo 20Ix + –

0 V 20 j 10

I 20 V

−

− + +

− +

−

But

25 j 50

1 )

57 26 9 55 )(

43 63 36 22 (

20 43

63 36

.

22

1

6 V 25 j 50

1 )

25 j 50 (

1 ) 20 j 10 (

20 20

j 10

1

o

oo

= +

+ +

− +

(0.0232 – j0.0224)Vo = 6 or Vo = 6/(0.03225∠–43.99˚ = 186.05∠43.99˚

For power, all we need is the magnitude of the rms value of Ix

|Ix| = 186.05/55.9 = 3.328 and |Ix|rms = 3.328/1.4142 = 2.353

We can now calculate the average power absorbed by the 50-Ω resistor

Pavg = (2.353)2x50 = 276.8 W

Applying KCL to the right side of the circuit,

0 5 j 10 5 j

− +

j 10

10

V V

j

5 j 10

8 ∠ ° = Vo +

j 1

20 80

1 R 2

1

In the circuit of Fig 11.40, determine the average power absorbed by the 40- Ω resistor

Figure 11.40

For Prob 11.8

Chapter 11, Solution 8

We apply nodal analysis to the following circuit

At node 1,

20 j - 10 j

.

o o

V I

I + = But,

j20 -

2 1 o

V V

=

Hence,

40 j20

-) (

5

j − V2 − V2 + V2 =

j6) -1 ( 37

360 j 6

360 j

9 40

1 R 2

1 P

2 2

2

40 Ω j10 Ω 0.5 Io

6 ∠0° A

I2 -j20 Ω

For the op amp circuit in Fig 11.41, Vs = 10 ∠ 30 ° V rms Find the average power

absorbed by the 20-k Ω resistor

3 1

Chapter 11, Problem 11

For the network in Fig 11.43, assume that the port impedance is

RC C

10 )(

377 (

) 754 0 ( 1

k 10 Z

2ab

mA ) 68 t 377 cos(

2 ) 22 t 377 sin(

2 ) t (

°

∠

= 2 - 68 I

3

23-ab

2rms ( 7 985 - 37.02 ) 10

2

10 2 Z

97 15

=

= S cos( 37 02 )

For the circuit shown in Fig 11.44, determine the load impedance Z for maximum power

transfer (to Z) Calculate the maximum power absorbed by the load

=

39 46 1936 1

22 15 86 4

) 61 61 4502 1 ( 4 2758

1 j 69 4

) 2758 1 j 6896 0 ( 4 2 j 5

2 xj 5 3 j 4

2 j 5

2 xj 5 3 j 4

VTh 5 Ω

–

40 ) 17 67 4123 0 ( V ) 55 46 5235 0 (

40 ) 5 0 j 12 0 j 16 0 ( V ) 2 0 5 0 j 12 0 j 16 0 (

0 5

0 V 2

j

40 V 3 j 4

40 V

22

22

= +

− +

=

− +

− +

−

−

Thus, V2 = 31.5∠–20.62˚V = 29.48 – j11.093V

I = (40 – V2)/(4 – j3) = (40 – 29.48 + j11.093)/(4 – j3) = 15.288∠46.52˚/5∠–36.87˚ = 3.058∠83.39˚ = 0.352 + j3.038

VThev = 40 – 4I = 40 – 1.408 – j12.152 = 38.59 – j12.152V = 40.46∠–17.479˚V

+ –

solve for V1

At node 1,

10 V ) 3333 0 j 2 0 ( V ) 3333 0 j 25 0 ( 0 5

0 V 3 j

V V

4

40

V

21

221

−

− +

−

At node 2,

20 j V ) 1667 0 j ( V 3333 0 j 0 2

j

40 V 3 j

V V

21

21

To calculate the maximum power to the load,

|IL|rms = (40.46/(2x0.8233))/1.4141 = 17.376A

Pavg = (|IL|rms)20.8233 = 248.58 W

Chapter 11, Problem 13

The Thevenin impedance of a source is ZTh = 120 + j 0 Ω , while the peak Thevenin voltage is VTh = 110 + j 0 V Determine the maximum available average power from the source

Chapter 11, Solution 13

For maximum power transfer to the load, ZL = 120 – j60Ω

ILrms = 110/(240x1.4142) = 0.3241A

Pavg = |ILrms|2120 = 12.605 W

It is desired to transfer maximum power to the load Z in the circuit of Fig 11.45 Find Z

and the maximum power Let is = 5 cos 0 t A

To find VTh, consider the circuit below

I1 + 5∠0o j0.3 12 Ω VTh

In the circuit of Fig 11.46, find the value of ZL that will absorb the maximum power and the value of the maximum power

2 o o

j j

j

V V V V

o 2

j - 2

Substituting (1) into (2),

2 2

j

j 1

1

V

5 0 j 5 0 2

j 1 1

We now obtain VTh from Fig (b)

j 1

12

o

V V

j 1

12 -

V

– Vo − - j × 2 Vo) + VTh = 0

j 1

) 2 j 1 )(

12 ( j2) -

8 ( 2

5 12 R

8 P

2

L

2 Th max

V

W 90

For the circuit of Fig 11.47, find the maximum power delivered to the load ZL

Figure 11.47

For Prob 11.16

Chapter 11, Solution 16

5 20 / 1 4

1 1

F 20 / 1 , 4 H

1

,

x j C j j

2111

4

V V V 5 0 5

=

⎯→

⎯

− + +

0 0 4

j

V V V

V

+

− +

=

⎯→

⎯

= +

−

(2)

Solving (1) and (2) leads to

2 1 1 1

4 25

V

V

− +

0 1 4

25 0

4

1 2

1

j V

V j

V V V

V

+

− +

4072 9 8

|

x R

Calculate the value of ZL in the circuit of Fig 11.48 in order for ZL to receive maximum average power What is the maximum average power received by ZL?

Figure 11.48

For Prob 11.17

Chapter 11, Solution 17

We find RTh at terminals a-b following Fig (a)

10 j 70

) 20 j 40 )(

10 j 30 ( 40 20 j

||

30 10 j

+

−

= + +

−

=

We obtain VTh from Fig (b)

Using current division,

3 2 j 1 1 - ) 5 j ( 10 j 70

20 j 30

10 j 40

8 (

5000 R

8

P

L

2Thmax

Find the value of ZL in the circuit of Fig 11.49 for maximum power transfer

(80)(-j10) 20

j20 -j10) (

||

80 40

||

40 20 j

Z

154 10 j 23 21

Z

=

= * Th

Chapter 11, Problem 19

The variable resistor R in the circuit of Fig 11.50 is adjusted until it absorbs the

maximum average power Find R and the maximum average power absorbed

j) (6)(3 -j2

) j 3 (

||

6 2 j -

+ +

= + +

=

Z

561 1 j 049 2

382

+

The load resistance RL in Fig 11.51 is adjusted until it absorbs the maximum average

power Calculate the value of RL and the maximum average power

Figure 11.51

For Prob 11.20

Chapter 11, Solution 20

Combine j20 Ω and -j10 Ω to get j 20 || -j10 = -j20

To find ZTh, insert a 1-A current source at the terminals of RL, as shown in Fig (a)

At the supernode,

10 j - 20 j - 40

1 V1 V1 V2

+ +

=

2

) 2 j 1 (

1 1

Substituting (2) into (1),

2 2

4 j 1 1 ) 2 j 1 (

=

4 6 j 1

2 Th

V Z

12

2 1

°

∠

= +

−

=

818 5 j 9091 0

82 21 j 09 109

893

Assuming that the load impedance is to be purely resistive, what load should be

connected to terminals a-b of the circuits in Fig 11.52 so that the maximum power is

transferred to the load?

||

100 10 j -

) 30 j 40 )(

100 ( ) 30 j 40 (

634 4 j 707 81

) 634 4 j 707 31 )(

50 ( ) 634 4 j 707 31 (

=

Z

73 1 j 5 19

16 4

2 sin 2

16 sin

16

1

0 0

T

Vrms = 5.774 V

Determine the rms value of the waveform in Fig 11.55

1 t 0 , 5 ) t ( v

2

25 dt -5) ( dt 5 2

1

1 2 1

0 2 2

32

f

3

32 ] 16

1 dt ) t ( f

+

=

+ +

−

=

2 t 0 5 ) t ( v

4

1 dt ) 10 ( dt 5 4

1

2 2 2

0 2 2

125 3

t 5

1 dt t 5

1

0

3 5

0 2 2

=

I 2 887 A

Find the rms value of the voltage waveform of Fig 11.59 as well as the average power absorbed by a 2- Ω resistor when the voltage is applied across the resistor

0 2 2

5

1 V

533 8 ) 8 ( 15

16 3

t 16 5

1

0

3 2

V P

2

15 t 5 t 2 20 )

t ( i

15

2 15

5

2 2

20

1 I

5

2 2

5

1 I

3 15 5

3 2 2

3

t 3

t t 10 t 100 5

1 I

332 33 ] 33 83 33 83 [ 5

Compute the rms value of the waveform depicted in Fig 11.61

2 t 0 t )

1 dt -1) ( dt t 4

1

2 2 2

0 2 2

4 2

1 )

4 ( )

2 ( 2

1 ) ( 2

12 0

=

= ∫ v t dt ∫ t dt ∫ dt

V rms

V 944 2

=

rms

V

2 2 2

2

1 I

10 5

t 50 dt t 50

0

5 1

0 4 2

Find the effective value of f(t) defined in Fig 11.65

20 36 3

t 9 3 1

dt 6 dt ) t 3 ( 3

1 dt ) t ( f T

1 f

rms

203

One cycle of a periodic voltage waveform is depicted in Fig 11.66 Find the effective

value of the voltage Note that the cycle starts at t = 0 and ends at t = 6 s

4 2 4

2 2 2

1 2 1

0 2 2

6

1 V

67 466 ] 100 400 1800 400 100 [ 6

Chapter 11, Problem 36

Calculate the rms value for each of the following functions:

(a) i(t) = 10 A (b) v(t) = 4 + 3 cos 5t V

(c) i(t) = 8 – 6 sin 2t A (d) v(t) = 5 sint + 4 cos t V

3 4

2 2

16 2

25

= +

6 ) 10 sin(

4 8

3 2

1

o o

t t

i i

i

A 487 9 90 2

36 2

16 64

3 2

I

For the power system in Fig 11.67, find: (a) the average power, (b) the reactive power, (c) the power factor Note that 220 V is an rms value

Chapter 11, Problem 39

An ac motor with impedance ZL = 4.2 + j3.6 Ω is supplied by a 220-V, 60-Hz source (a)

Find pf, P, and Q (b) Determine the capacitor required to be connected in parallel with

the motor so that the power factor is corrected to unity

{It is important to note that this capacitor will see a peak voltage of 220 2 =

311.08V, this means that the specifications on the capacitor must be at least this or greater!}

Chapter 11, Problem 40

A load consisting of induction motors is drawing 80 kW from a 220-V, 60-Hz power line

at a pf of 0.72 lagging Find the capacitance of a capacitor required to raise the pf to 0.92

Chapter 11, Solution 40

0 1

Obtain the power factor for each of the circuits in Fig 11.68 Specify each power factor

||

-j2 ) 2 j 5 j (

||

2 j

) j 4 )(

2 j ( ) j 4 (

°

∠

= +

+

=

− +

44 0 j 64 1

44 0 j 64 0 ) j 52 1 j 64 0 (

Chapter 11, Problem 42

A 110-V rms, 60-Hz source is applied to a load impedance Z The apparent power

entering the load is 120 VA at a power factor of 0.707 lagging

(a) Calculate the complex power

(b) Find the rms current supplied to the load

) 10 cos(

(a) Calculate the rms value of the voltage

(b) Determine the average power dissipated in the resistor

Chapter 11, Solution 43

2

1 2

9 25

3 2

P rms

Find the complex power delivered by vs to the network in Fig 11.69

V I j

= Solving for Vo leads to

(a) the rms values of the voltage and of the current

(b) the average power dissipated in the load

Chapter 11, Solution 45

2

60 20

2 2

=

(b) p(t) = v(t)i(t) = 20 + 60cos100t – 10sin100t – 30(sin100t)(cos100t); clearly

the average power = 20W

For the following voltage and current phasors, calculate the complex power, apparent power, real power, and reactive power Specify whether the pf is leading or lagging (a) rms V = 220 ∠ 30 ° V rms, I = 0.5 ∠ 60 ° A

(b)

rms

A 5 2

.

6

, rms V 10 250

Reactive power = 55 VAR

pf is leading because current leads voltage

Reactive power = 401 2 VAR

pf is lagging because current lags voltage

Reactive power = 74 54 VAR

pf is lagging because current lags voltage

(d) S = V I* = ( 160 ∠ 45 ° )( 8 5 ∠ - 90 ° ) = 1360 ∠ - 45 °

=

S 961.7 – j961.7 VA

Apparent power = 1360 VA Real power = 961.7 W

Reactive power = - 961 7 VAR

pf is leading because current leads voltage

For each of the following cases, find the complex power, the average power, and the reactive power:

Average power = 226.3 W Reactive power = –226.3 VAR

) 80 ( 2

Chapter 11, Problem 48

Determine the complex power for the following cases:

(a) P = 269 W, Q = 150 VAR (capacitive)

84 25 sin(

2000 sin

Q S sin

S

°

= θ

=

⎯→

⎯ θ

=

48 4129 cos

Q sin sin

S

59 48

=

86 396 ) 6614 0 )(

600 ( cos S

2 2

=

=

=

Z V

8264 0 1210

1000 S

P cos cos

S

=

S 1000 + j 681 2 VA

Find the complex power for the following cases:

4 j

P

= θ

14400 60

j 40

) 120

1000 j 1000 jQ

=

S

750 j

Z

V

S =

23 23 j 98 30 750 j 1000

) 220

2 rms

2 (

) 120 ( 2

2 2

2 rms

*

S

V S

V Z

For the entire circuit in Fig 11.70, calculate:

(a) the power factor

(b) the average power delivered by the source

(c) the reactive power

(d) the apparent power

(e) the complex power

20 j 110 2 j

18

) 6 j 8 )(

5 j 10 ( 2

+ +

= +

+

− +

2 (

) 16 ( 2

Chapter 11, Problem 52

In the circuit of Fig 11.71, device A receives 2 kW at 0.8 pf lagging, device B receives 3 kVA at 0.4 pf leading, while device C is inductive and consumes 1 kW and receives 500

VAR

(a) Determine the power factor of the entire system

(b) Find I given that Vs = 120 ∠ 45 ° V rms

Figure 11.71

For Prob 11.52

Chapter 11, Solution 52

749 j 4200 S

S S S

500 j 1000 S

2749 j 1200 9165

0 x 3000 j 4 0 x 3000 S

1500 j 2000 6

0 8 0

2000 j 2000 S

CBACBA

−

= + +

=

749 4200

4200 pf

749 j 4200 I

I V

Irms = 35.55∠55.11˚ A

In the circuit of Fig 11.72, load A receives 4 kVA at 0.8 pf leading Load B receives 2.4 kVA at 0.6 pf lagging Box C is an inductive load that consumes 1 kW and receives 500

(a)

A 8 29 97 93 88 29 46 66 x 2 I

8 29 46 66 2

30 120

2 0 5640 V

S V

S S V

S I

rmsrms

CArms

Brms

=

∗

(b) pf = cos(0.2˚) ≈ 1.0 lagging

20 - 8

20 - 8

2

I

) 4643 0 j 531 1 ( ) 504 1 j -0.5472 (

2

= I I I

1 )(

20 - 8 ( 2

1 2

Find the complex power absorbed by each of the five elements in the circuit of Fig 11.74

Figure 11.74

For Prob 11.55

I j 1 (

2

I -2 j5

1 - j 1 j5 - 2 j

3 j 4

5 j 1 -

=

−

− +

=

−

= I I ( 3 5 j 0 5 ) ( 2 j 3 ) 1 5 j 3 5 3 808 66 8

I3 1 2For the 40-V source,

- * 1

I V

Obtain the complex power delivered by the source in the circuit of Fig 11.75

Figure 11.75

For Prob 11.56

Chapter 11, Solution 56

8 1 j 6 0 2 j 6

) 2 j - )(

6 ( 6

||

2 j

−

=

j2.2 3.6 6

||

-j2) ( 4 j

+

= ( 2 30 ) 0 95 47.08

2 2 j 6 8

2 2 j 6 3

4 ( 2

1 2

s

= V I S

Vo

−

3.6 + j2.2 Ω

Consider the circuit as shown below

At node o,

j - 1

At node 1,

2 j

2 j -

1 o 1

V V V

= +

j4 11

24 -

j4 11

j4) - (-24)(2

V

The voltage across the dependent source is

o 1 o 1

4 j 11

) 4 j 6 )(

24 - ) 4 4 j 2 ( j4 11

24 -

−

= +

−

⋅ +

=

V

) 2 ( 2

1 2

o 2

576 j4

11

-24 - 4 j 11

) 4 j 6 )(

24 -

2 0

I

mA 8

20 Io = From the right portion of the circuit,

mA j 7

16 ) mA 8 ( 3 j j 10 4

4

I

) 10 10 ( 50

) 10 16 (

2 -3 2

= I S

Calculate the reactive power in the inductor and capacitor in the circuit of Fig 11.78

Figure 11.78

For Prob 11.59

Chapter 11, Solution 59

Let Vo represent the voltage across the current source and then apply nodal

analysis to the circuit and we get:

30 j 40 j20 - 50

240

+ +

°

∠

= +

38 0 j 36 0

-o 1

V I

°

∠

= +

= 3 363 - 83.42

30 j 40

o 2

V I

Reactive power in the inductor is

0

20 j 20

749 7 j 16 )) 9 0 ( sin(cos 9

0

16 j 16

= +

Given the circuit in Fig 11.80, find Io and the overall complex power supplied

Figure 11.80

For Prob 11.61

Chapter 11, Solution 61

Consider the network shown below

kVA 8 0 j 2 1

S

kVA 937 1 j 4 )) 9 0 ( sin(cos 9

0

4 j

1

I V

S =

104 j 74 22 90

100

10 ) 137 1 j 2 5 )(

2 (

=

=

V

S I

104 j 74 22

I

Similarly, sin(cos ( 0 707 )) 2 ( 1 j ) kVA

707 0

2 j

1

I V

S =

284 28 j 8.284 2 - 100

j

10 ) 8284 2 j 8284 2 ( V

5.54 -

21

I

* o o

90 100 ( 2

For the circuit in Fig 11.81, find Vs

Figure 11.81

For Prob 11.62