Solution manual fundamentals of electric circuits 3rd edition chapter09

If the input voltage is vt = 10 cos 2t find the currrent flowing through the circuit... A series RL circuit is connected to a 110-V ac source.. Find current Io in the circuit shown in Fi

Given the sinusoidal voltage v(t) = 50 cos (30t + 10o) V, find: (a) the amplitude Vm,(b)

the period T, (c) the frequency f, and (d) v(t) at t = 10 ms

(a) What is the amplitude of the current?

(b) What is the angular frequency?

(c) Find the frequency of the current

Is(2 ms) = 8 cos(( 500 π )( 2 × 10- 3) − 25 ° )

= 8 cos(π − 25°) = 8 cos(155°)

= -7.25 A

(a) Express v = 8 cos(7t = 15o) in sine form

(b) Convert i = -10 sin(3t - 85o) to cosine form

Chapter 9, Solution 4

(a) v = 8 cos(7t + 15°) = 8 sin(7t + 15° + 90°) = 8 sin(7t + 105°)

(b) i = -10 sin(3t – 85°) = 10 cos(3t – 85° + 90°) = 10 cos(3t + 5°)

For the following pairs of sinusoids, determine which one leads and by how much

(a) v(t) = 10 cos(4t - 60o

) and i(t) = 4 sin (4t + 50o

) (b) v1(t) = 4 cos(377t + 10o) and v2(t) = -20 cos 377t

(c) x(t) = 13 cos 2t + 5 sin 2t and y(t) = 15 cos(2t -11.8o)

Chapter 9, Solution 6

(a) v(t) = 10 cos(4t – 60°)

i(t) = 4 sin(4t + 50°) = 4 cos(4t + 50° – 90°) = 4 cos(4t – 40°)

Thus, i(t) leads v(t) by 20°

Thus, y(t) leads x(t) by 9.24°

d

df

φ

= φ + φ

= φ + φ

Calculate these complex numbers and express your results in rectangular form:

10

j

+

− (c) 10 + (8 ∠ 50o) (5 – j12)

Chapter 9, Solution 8

(a)

4 j 3

45 15

−

°

∠ + j2 =

°

∠

°

∠ 53.13 - 5

45 15

+ j2 = 3∠98.13° + j2 = -0.4245 + j2.97 + j2

= -0.4243 + j4.97

(b) (2 + j)(3 – j4) = 6 – j8 + j3 + 4 = 10 – j5 = 11.18∠-26.57°

j4) - j)(3 (2

20 - 8 +

°

∠

+ j12 5 -

10

°

∠ 26.57 - 11.18

20 - 8

+

144 25

) 10 )(

12 j 5 - +

−

= 0.7156∠6.57° − 0.2958 − j0.71 = 0.7109 + j0.08188 − 0.2958 − j0.71

= 0.4151 − j0.6281

(c) 10 + (8∠50°)(13∠-68.38°) = 10+104∠-17.38°

= 109.25 – j31.07

Evaluate the following complex numbers and leave your results in polar form:

−

j j

o

2

60 3 8 6

(b)

) 5 ( ) 6

2

(

) 50 35 ( 60

10

(

j j

o o

+

− +

30

5

(

) 261 7 j 13 7 )(

30 5 ( ) 7392 0 j 1197 1 8 j 6 )(

3 2

Find the phasors corresponding to the following signals:

Let X = 8 ∠ 40o and and Y = 10 ∠ − 30o Evaluate the following quantities and express your results in polar form:

(a) (X + Y)X* (b) (X -Y)* (c) (X + Y)/X

Chapter 9, Solution 12

Let X = 8∠40° and Y = 10∠-30° Evaluate the following quantities and express

your results in polar form

45 39 31 118 ) 40 8 )(

55 0 789 14 (

) 40 8 )(

142 0 j 788 14 (

91 77 318 14 17

−

−

7 2134 06

246

9 6944 24186

) 5983 10 96 16 )(

84 67

(

) 80 60 )(

80 56 138 82 231 116

.

62

(

j j

j j

j j

j

−

−

= +

−

= +

+

−

− +

+

(c) ( )

221 4

256 24

139 46

.

338

) 38 12 923 16 )(

86 126 20

( ) 120 260

( 4

j

j j

−

Evaluate these determinants:

(a)

j

j j

+

−

−

− +

1 5

3 2 6

16

10 4 30

20

(c)

j j

j j

1

0 1

Chapter 9, Solution 15

(a)

j 1 - 5 -

3 j 2 6 j 10

+

− +

16

10 - 4 - 30 20

0 j j 1

j 1 j 1

j 1 j

0 j j 1

Transform the following sinusoids to phasors:

The phasor form is 5∠-100°

(c) 4 cos(2t) + 3 sin(2t) = 4 cos(2t) + 3 cos(2t – 90°)

The phasor form is 4∠0° + 3∠-90° = 4 – j3 = 5∠-36.87°

Obtain the sinusoids corresponding to each of the following phasors:

v2 = 10 cos(40t + 53.13°)

(c) i1( t ) = 2.8 cos(377t – π/3)

(d) I2 = -0.5 – j1.2 = 1.3∠247.4°

) t (

i2 = 1.3 cos(103t + 247.4 °)

Using phasors, find:

(a) 3cos(20t + 10º) – 5 cos(20t- 30º)

Therefore, 3 cos(20t + 10°) – 5 cos(20t – 30°) = 3.32 cos(20t + 114.49°)

(b) 40∠-90° + 30∠-45° = -j40 + 21.21 – j21.21

= 21.21 – j61.21

= 64.78∠-70.89°

Therefore, 40 sin(50t) + 30 cos(50t – 45°) = 64.78 cos(50t – 70.89°)

(c) Using sinα = cos(α − 90°),

20∠-90° + 10∠60° − 5∠-110° = -j20 + 5 + j8.66 + 1.7101 + j4.699

= 6.7101 – j6.641 = 9.44∠-44.7°

Therefore, 20 sin(400t) + 10 cos(400t + 60°) – 5 sin(400t – 20°)

= 9.44 cos(400t – 44.7°)

Chapter 9, Problem 20

A linear network has a current input 4cos(ω t + 20º)A and a voltage output 10

cos(ω t +110º) V Determine the associated impedance

Simplify the following:

(a) f(t) = 5 cos(2t + 15(º) – 4sin(2t -30º)

324 8 )

t t

565 5 )

t t

v ( ) 4 2 ( )

10

o

V j

V V j

V

F = 10 + 4 − 2 , ω = 5 , = 20 ∠ − 30

ω ω

o

89 33 4 454 ) 10 j 32 17 )(

4 20 j 10 ( V 4 0 j V 20

(a) v = 50 cos( ω t + 30o) + 30 cos(ω t + 90o)V

v t

0 10

j = ∠ ° ω = ω

+ V

V

10 ) j 1 ( − =

V

°

∠

= +

=

−

= 5 j 5 7 071 45 j

90 10 ( 20 j

4 5

ω + +

4 5 4 j

V

°

∠

= +

°

∠

= 3 43 - 110 96

3 j 5

80 - 20

V

Therefore, v(t) = 3.43 cos(4t – 110.96°)

Using phasors, determine i(t) in the following equations:

(a) 2 3 ( ) 4 cos( 2 45o)

t t

i dt

di dt

i

Chapter 9, Solution 25

(a)

2 ,

45 - 4 3 2j ω I + I = ∠ ° ω =

°

∠

= + j 4 ) 4 - 45 3

°

∠

13 53 5

45 - 4 j4 3

45 - 4 I Therefore, i(t) = 0.8 cos(2t – 98.13°)

(b)

5 ,

22 5 6 j j

°

∠

56 26 708 6

22 5 3

j 6

22 5

Assuming that the value of the integral at t = - ∞ is zero, find i(t) using the phasor

method

Chapter 9, Solution 26

2 ,

0 1 j 2

ω + +

1 2 j

1 2 2

I

°

∠

= +

= 0 4 - 36 87

5 1 j 2

1

I

Therefore, i(t) = 0.4 cos(2t – 36.87°)

A parallel RLC circuit has the node equation

10 - 110 j

100 50

ω +

380 (

110 R

) t ( v ) t (

What is the instantaneous voltage across a 2- µ F capacitor when the current through it is

i =4 sin(106t +25o) A?

Chapter 9, Solution 29

5 0 j - ) 10 2 )(

10 ( j

1 C

j

1

6 -

×

= ω

A series RLC circuit has R = 80 Ω , L = 240 mH, and C = 5 mF If the input voltage is v(t) = 10 cos 2t find the currrent flowing through the circuit

A series RL circuit is connected to a 110-V ac source If the voltage across the resistor is

85 V, find the voltage across the inductor

Chapter 9, Solution 33

2 L

1

ω

= ω

5 (

1

3

Find current i in the circuit of Fig 9.42, when vs(t) = 50 cos200t V

in

V I

In the circuit of Fig 9.43, determine i Let vs = 60 cos(200t - 10o)V

100 200 mH

j x

x j L

⎯→

500 200

10 10

1 1

F

x x j C

⎯→

ω µ

1

t

Determine the admittance Y for the circuit in Fig 9.44

Find i(t) and v(t) in each of the circuits of Fig 9.45

3 ( j

1 C

j

1 F

6

1

=

= ω

j 4

2 j -

4 ( j

1 C

j

1 F

12

1

=

= ω

⎯→

⎯

12 j ) 3 )(

4 ( j L j H

0 50

Z

V I

Hence, i(t) = 10 cos(4t + 36.87°) A

= ( 50 0 ) 41 6 33 69 j12

8

12 j

V

Hence, v(t) = 41.6 cos(4t + 33.69°) V

For the circuit shown in Fig 9.46, find Zeg and use that to find current I Let ω = 10 rad/s

In the circuit of Fig 9.47, find io when:

(a) ω = 1 rad/s (b) ω = 5 rad/s

(c) ω = 10 rad/s

Figure 9.47

For Prob 9.40.

(a) For ω = 1 ,

j ) 1 )(

1 ( j L j H

20 j - ) 05 0 )(

1 ( j

1 C

j

1 F

05

40 j - j ) 20 j -

||

2

− +

= +

0 4 j0.802

1.98

0 4

V I

Hence, io( t ) = 1.872 cos(t – 22.05°) A

(b) For ω = 5 ,

5 j ) 1 )(

5 ( j L j H

4 j - ) 05 0 )(

5 ( j

1 C

j

1 F

05

4 j - 5 j ) 4 j -

||

2 5

− +

= +

0 4 j4

1.6

0 4

V I

Hence, io( t ) = 0.89 cos(5t – 69.14°) A

(c) For ω = 10 ,

10 j ) 1 )(

10 ( j L j H

2 j - ) 05 0 )(

10 ( j

1 C

j

1 F

05

4 j - 10 j ) 2 j -

||

2 10

− +

= +

0 4 9

j 1

0 4

V I

Hence, io( t ) = 0.4417 cos(10t – 83.66°) A

Find v(t) in the RLC circuit of Fig 9.48

1 ( j L j H

j - ) 1 )(

1 ( j

1 C j

1 F

1 j - 1 ) j -

||

) j 1 (

=

Z

j 2

j 2

) 10 )(

j 1 ( ) j 1 ( ) j 1 )(

j

V

Thus, v(t) = 6.325 cos(t – 18.43 °) V

Calculate v o(t) in the circuit of Fig 9.49

100 j - ) 10 50 )(

200 ( j

1 C

j

1 F

×

= ω

⎯→

⎯ µ

20 j ) 1 0 )(

200 ( j L j H

1

20 j 40 j2 - 1

j100 - j100 50

) (50)(-j100 -j100

70

20 j ) 0 60 ( 20 j 40 30 20 j

20 j

o

V

Thus, vo( t ) = 17.14 sin(200t + 90 °) V

or vo( t ) = 17.14 cos(200t) V

Find current Io in the circuit shown in Fig 9.50

2 j ) 10 10 )(

200 ( j L j mH

j - ) 10 5 )(

200 ( j

1 C

j

1 mF

×

= ω

⎯→

⎯

4 0 j 55 0 10

j 3 5 0 j 25 0 j 3

1 2 j

1 4

1

−

=

+ +

−

=

− + +

=

Y

865 0 j 1892 1 4 0 j 55 0

1 1

°

∠

= +

°

∠

= +

°

∠

865 0 j 1892 6

0 6 5

0 6

Z I

Thus, i(t) = 0.96 cos(200t – 7.956 °) A

Find current Io in the network of Fig 9.52

1 =

j2 - 2

j4 - j4 2

||

-j2) ( 4 j

Z

j 1

j10 - ) 0 5 ( 3 j 1 2 j -

2 j -2

1

1

Z I

= +

10 - j 1

j10 - j - 1

j - j2

2

-j2 -

j - ) 1 0 )(

10 ( j

1 C

j

1 F

1

10 ( j L j H

2

2 j 4

8 j 2 j

j1.6 0.8s 2 1

Z Z

Z I

) 40 5 )(

43 63 789 1 (

o

I

Thus, io( t ) = 2.325 cos(10t + 94.46 °) A

In the circuit of Fig 9.54, determine the value of is(t)

= + +

−

+

− +

63 52 854 10

5 626

8 j 588 4 2

5 4

j 20 10 j

) 4 j 20 ( 10 j 2

Given that vs(t) = 20 sin(100t - 40o

) in Fig 9.55, determine ix(t)

Figure 9.55

For Prob 9.48.

Chapter 9, Solution 48

Converting the circuit to the frequency domain, we get:

We can solve this using nodal analysis

A ) 4 9 t 100 sin(

j 30

29 24 643

.

15

I

29 24 643 15 03462 0 j 12307

.

0

40 2

V

40 2 ) 01538 0 j 02307 0 05

0 V 20 j

0 V 10

1

° +

− +

Find vs(t) in the circuit of Fig 9.56 if the current ix through the 1- Ω resistor is 0.5 sin

) j 1 )(

2 j ( 2 ) j 1 (

||

2 j 2

+

− +

=

− +

=

Z

I I

I

j 1

2 j j 1 2 j

2 j

1 0 5 0

I

4 j

j 1 2

j

j 1

j 1

Determine vx in the circuit of Fig 9.57 Let is(t) = 5 cos(100t + 40o)A

cos(

50

v

50 50

I

20

V

50 5 2 40 5 2 j 40 5 10 j 20

If the voltage vo across the 2- Ω resistor in the circuit of Fig 9.58 is 10 cos2t V, obtain

2 ( j

1 C

j

1 F

1

2 ( j L j H

5

The current I through the 2- Ω resistor is

4 j 3 2 j 5 j 1

+ +

I

Therefore,

= ) t (

is 25 cos(2t – 53.13 °) A

If Vo = 8 ∠ 30oV in the circuit of Fig 9.59, find I .

5 j 5 j 5

25 j 5 j

||

+

= +

s 2 1

1

4 5

2 j 5 12

10

I I

I Z Z

Z I

−

=

−

= +

=

) 5 2 j 5 2

5 2 ( j 5

4 30

−

+

= +

) j 5 )(

30 8

Find Io in the circuit of Fig 9.60

4 x 6 j Z , 1 j 1 45 6569 5

90 8 4 j 4

4 x

−

=

5 1 j 5 1 4 j 4

) 5 1 j 5 9 ( ) 10 Z //(

2

A 67 21 873 8 33 8 7623 6

30 60 Z

30 60

In the circuit of Fig 9.61, find Vs if Io= 2 ∠ 0o

A

Figure 9.61

For Prob 9.54.

Chapter 9, Solution 54

Since the left portion of the circuit is twice as large as the right portion, the

equivalent circuit is shown below

) j 1 ( 2 ) j 1 (

o

V

) j 1 ( 4

V

) j 1 ( 6

21

Vs

* Find Z in the network of Fig 9.62, given that Vo = 4 ∠ 0oV

j

4 8 j

o

I

j 8 j4

-) 8 j ( (-j0.5) j4

-) 8 j (

1

I

5 0 j 8

j 8

) 8 j ( 2

j - 2

j 8 12 j20

3 j26

25 261 31 26 2

1 j 2

3

j26 - 4

+

Vo

−

At ω = 377 rad/s, find the input impedance of the circuit shown in Fig 9.63

2 ⎯ ⎯→ j ω L = j

j C

2 1 j 6 2 j 2 2 j

) j 2 ( 2 j 1 ) j 2

− +

=

− +

=

S 1463 0 j 3171

) 57 4 016 5 )(

90 5 ( ) 4 0 j 5 ( 5 j

Zin

= 2.707+j2.509 Ω

Obtain Zin for the circuit in Fig 9.67

Figure 9.67

For Prob 9.60.

Chapter 9, Solution 60

Ω +

=

− +

+

= +

− + +

1 5 j

1 2 j 1

1 j 1

1 1

Z

4 0 j 8 0 ) 3 0 j 1 0 ( ) 2 0 j - ) 4 0 j 2 0 ( ) 5 0 j 5 0 ( 1

eq

−

=

− + +

− + +

1

eq

For the circuit in Fig 9.69, find the input impedance Zin at 10 krad/s

10 10 ( j

1 C

j

1 F

×

×

= ω

⎯→

⎯ µ

50 ) 50 )(

0 1

=

V

) 50 )(

2 ( ) 100 j 20 j 50 )(

0 1 (

V

80 j 150 100 80 j 50

in in

For the circuit in Fig 9.70, find the value of ZT⋅.

450 j 200 z

, 333 13 j 30 15

j

450 j 200

z

45 j 20 10

300 j 150 j 200

z

32

− +

j

8

Z

821 3 j 70 21 )

29 j 40

) 16 j 10 )(

333 13 j 30 ( )

30

I

5 j 19 2

j 6

) 8 j 6 ( 10

||

) 6 j 4 ( 2

Z

2 j 7

) 4 j 3 )(

6 j 4 ( 2

+

− +

10 120

T

Z

V

For the circuit in Fig 9.73, calculate ZT and Vab⋅

Figure 9.73

For Prob 9.66.

) j 12 ( 145

170 5

j 60

) 10 j 40 )(

5 j 20 ( ) 10 j 40 (

||

) 5 j 20 (

90 60

I

j 12

2 j 8 5 j 60

10 j 40

+

= +

+

=

I I

I

j 12

j 4 5 j 60

5 j 20

−

= +

−

=

2 1

I I

V

j 12

40 j 10 j

12

) 40 j (160 -

+ + +

+

=

I I

V

145

j)(150) -12

( j 12

150 -

ab

+

= +

=

) 76 97 25 4 )(

24 175 457 12 (

At ω = 103 rad/s find the input admittance of each of the circuits in Fig 9.74

10 ( j

1 C

j

1 F

5

×

= ω

⎯→

⎯ µ

) 80 j 60 (

||

20 j 60

Z

60 j 60

) 80 j 60 )(

20 j ( 60

− +

= 63 33 j 23 33 67 494 20 22in

Z

=

=

in in

10 ( j

1 C

j

1 F

×

= ω

⎯→

⎯ µ 20 60

||

) 10 j 40 (

||

20 50 j -

Z

10 j 60

) 10 j 40 )(

20 ( 50 j -

+ +

Z

=

=

in in

1

Z

Y 19.7 ∠74.56° mS = 5.24 + j18.99 mS

Determine Yeq for the circuit in Fig 9.75

Figure 9.75

For Prob 9.68.

Chapter 9, Solution 68

4 j -

1 j 3

1 2 j 5

Find the equivalent admittance Yeq of the circuit in Fig 9.76

Figure 9.76

For Prob 9.69.

Chapter 9, Solution 69

) 2 j 1 ( 4

1 2 j -

1 4

1 1

o

+

= +

=

Y

6 1 j 8 0 5

) 2 j 1 )(

4 ( 2 j 1

Y

) 6 0 j 8 0 ( ) 333 0 j ( ) 1 ( 6 0 j 8 0

1 3

j -

1 1

1 1

o

+ + +

=

− + +

=

′ 1 . 8 j 0 . 933 2 . 028 27 . 41 1

o

Y

2271 0 j 4378 0 41 27 - 4932 0

Y

773 4 j 4378 0 5 j

Y

97 22

773 4 j 4378 0 5 0 773 4 j 4378 0

1 2

1 1eq

− +

= +

+

=

Y

2078 0 j 5191 0 1eq

2078 0 j 5191 0

eq

Find the equivalent impedance of the circuit in Fig 9.77

Figure 9.77

For Prob 9.70.

Make a delta-to-wye transformation as shown in the figure below

9 j 7 5

j 15

) 10 j 15 )(

10 ( 15 j 10 10 j 5

) 15 j 10 )(

10 j -

+

−

= + +

) 15 j 10 )(

5 (

) 10 j - )(

5 (

||

) 2

||

) 5 3 j 5 6 ( 9 j 7

Z

5 4 j 5 13

) 8 j 7 )(

5 3 j 5 6 ( 9 j 7

− +

Obtain the equivalent impedance of the circuit in Fig 9.78

Figure 9.78

For Prob 9.71.

We apply a wye-to-delta transformation

j 1 2 j

2 j 2 2

j

4 j 2 j 2

Z

j 1 2

2 j 2

Z

j2 -2 j -

2 j 2

Z

8 0 j 6 1 3 j 1

) j 1 )(

4 j ( ) j 1 (

||

4 j

) j 1 )(

1 ( ) j 1 (

=

Z

6 0 j 2 2

1 2

j 2 -

1 2 j -

1 1

Calculate the value of Zab in the network of Fig 9.79

Figure 9.79

For Prob 9.72.

Transform the delta connections to wye connections as shown below

6 j - 18 j -

10 20 20

) 20 )(

20 (

50

) 10 )(

20 (

20 (

R3

4 4) j6 (j2

||

) 8 2 j ( j2

Z

j4) (4

||

) 2 j 8 ( j2 4

Z

j2 - 12

) 4 j j2)(4 (8

j2 4

ab

− +

+ +

=

Z

4054 1 j 567 3 j2 4

Determine the equivalent impedance of the circuit in Fig 9.80

Figure 9.80

For Prob 9.73.

Chapter 9, Solution 73

Transform the delta connection to a wye connection as in Fig (a) and then

transform the wye connection to a delta connection as in Fig (b)

8 4 j - 10 j 6 j 8 j 8 j

− +

=

Z

-j4.81

Z

4 6 j j10

64 - 10 j

) 8 j )(

8 j (

Z

= +

+ +

+ +

2

6 9 j 4 46 ) 4 6 j )(

8 4 j 2 ( ) 4 6 j )(

8 4 j 4 ( ) 8 4 j 4 )(

8 4 j 2

25 7 j 5 1 4

6 j

6 9 j 4 46

Z

688 6 j 574 3 8 4 j 4

6 9 j 4 46

6 9 j 4 46

12 j 574 3

) 88 61 583 7 )(

90 6 (

j7.25) -j4)(1.5

20 j 727 1

) 07 79 11 9 )(

90 12 (

||

4 j -

||

)

||

6 j

) 5673 2 j 7494 0 (

||

) 3716 3 j 7407 0 (

Design an RL circuit to provide a 90o

leading phase shift

Chapter 9, Solution 74

One such RL circuit is shown below

We now want to show that this circuit will produce a 90° phase shift

) 3 j 1 ( 4 2 j 1

20 j 20 - 40

j 20

) 20 j 20 )(

20 j ( ) 20 j 20 (

+

= +

=

Z

) j 1 ( 3

1 3 j 6

3 j 1 ) 0 1 ( 12 j 24

12 j 4

+

= +

Z

Z V

= +

3

j ) j 1 ( 3

1 j 1

j 20

j 20

20 j

Design a circuit that will transform a sinusoidal voltage input to a cosinusoidal voltage output

Chapter 9, Solution 75

Since ) cos( ω t ) = sin( ω t + 90 ° , we need a phase shift circuit that will cause the output to lead the input by 90° This is achieved by the RL circuit shown

below, as explained in the previous problem

This can also be obtained by an RC circuit

Chapter 9, Problem 76

For the following pairs of signals, determine if v1 leads or lags v2 and by how much

(a) v1 = 10 cos(5t - 20o

), v2 = 8 sin5t (b) v1 = 19 cos(2t - 90o

), v2 = 6 sin2t (c) v1 = - 4 cos10t , v2 = 15 sin10t