Solution manual fundamentals of electric circuits 3rd edition chapter13

Calculate the maximum possible equivalent inductance if: a the two coils are connected in series b the coils are connected in parallel Chapter 13, Solution 5... After transforming the c

L T = 4 – 1 + 7 = 10H

or L T = L 1 + L 2 + L 3 – 2M 12 – 2M 23 + 2M 12

L T = 6 + 8 + 10 = 10H Chapter 13, Problem 2

Determine the inductance of the three series-connected inductors of Fig 13.73

Two coils connected in series-aiding fashion have a total inductance of 250 mH When connected in a series-opposing configuration, the coils have a total inductance of 150

mH If the inductance of one coil (L1) is three times the other, find L1, L2, and M What is

the coupling coefficient?

L 1 = 3L 2 = 150 mH

From (2), 150 + 50 – 2M = 150 leads to M = 25 mH

k = M/ L1L2 =25/ 50x150 = 0.2887

M L

its dotted terminal Therefore, the mutually induced voltages have the same sign as the self-induced voltages Thus,

ωω

1 s

s

I

ILjMj

MjLjV

Chapter 13, Problem 5

Two coils are mutually coupled, with L1 = 25 mH, L2 = 60 mH, and k = 0.5 Calculate the

maximum possible equivalent inductance if:

(a) the two coils are connected in series

(b) the coils are connected in parallel

Chapter 13, Solution 5

(a) If the coils are connected in series,

=+

+

=++

−

=

−+

−

36.19x6025

36.1960x25M2LL

MLL

2 1

2 2

The coils in Fig 13.75 have L1 = 40 mH, L2 = 5 mH, and coupling coefficient k = 0.6 Find i1 (t) and v2(t), given that v1(t) = 10 cos ω t and i2(t) = 2 sin ω t, ω = 2000 rad/s

Figure 13.75

For Prob 13.6

j4j8

I1

I2

Find Vx in the network shown in Fig 13.78

or I 1 = (3 – j2)i 2 – 2 (2) Substituting (2) into (1), 8∠30° + (2 + j4)2 = (14 + j7)I 2

I2 = (10.928 + j12)/(14 + j7) = 1.037∠21.12°

Vx = 2I2 = 2.074∠21.12°

2 Ω

+ –

Use mesh analysis to find i x in Fig 13.80, where

i s = 4 cos(600t) A and v s = 110 cos(600t + 30º)

jC

j

1F

ω

→

After transforming the current source to a voltage source, we get the circuit shown below

150

•

I2

j360 j720

Determine the equivalent L eq in the circuit of Fig 13.81

1 1

I

We can also use the equivalent T-section for the transform to find the equivalent

inductance

4)

52(j4

Zin

+++

=+

−++++

Obtain the Thevenin equivalent circuit for the circuit in Fig 13.83 at terminals a-b

Figure 13.83

For Prob 13.14

Chapter 13, Solution 14

To obtain VTh, convert the current source to a voltage source as shown below

Note that the two coils are connected series aiding

ωL = ωL1 + ωL2 – 2ωM jωL = j6 + j8 – j4 = j10 Thus, –j10 + (5 + j10 – j3 + 2)I + 8 = 0

I = (– 8 + j10)/ (7 + j7) But, –j10 + (5 + j6)I – j2I + VTh = 0

VTh = j10 – (5 + j4)I = j10 – (5 + j4)(–8 + j10)/(7 + j7)

VTh = 5.349∠34.11°

+ –

To obtain Z Th , we set all the sources to zero and insert a 1-A current source at the terminals a–b as shown below

Clearly, we now have only a super mesh to analyze

(5 + j6)I1 – j2I2 + (2 + j8 – j3)I2 – j2I1 = 0 (5 + j4)I 1 + (2 + j3)I 2 = 0 (1) But, I 2 – I 1 = 1 or I 2 = I 1 – 1 (2) Substituting (2) into (1), (5 + j4)I 1 +(2 + j3)(1 + I 1 ) = 0

I 1 = –(2 + j3)/(7 + j7) Now, ((5 + j6)I 1 – j2I 1 + V o = 0

Find the Norton equivalent for the circuit in Fig 13.84 at terminals a-b

Figure 13.84

For Prob 13.15

a j20 Ω

j5

a j20 Ω

j10 Ω

20 Ω

Obtain the Norton equivalent at terminals a-b of the circuit in Fig 13.85

Figure 13.85

For Prob 13.16

80∠0o V I1

-

b

80)

28(0

)42

+

=

=

To find ZN, insert a 1-A current source at terminals a-b Transforming the current source

to voltage source gives the circuit below

1

j

jI I

jI I

0)

M32

j

10

+++

=++

ω++

We replace the transformer by its equivalent T-section

5,

25520,

1055)

=+

++

=+

+

74

)4(627)6//(

)4

(

j

j j

j j j j

Th

j j

H,5525

Determine currents I1, I2, and I3 in the circuit of Fig 13.89 Find the energy

stored in the coupled coils at t = 2 ms Take ω = 1,000 rad/s

o

For mesh 1, j12 = (4 + j10 – j5)I 1 + j5I 2 + j5I 2 = (4 + j5)I 1 + j10I 2 (1)

For mesh 2, 0 = 20 + (8 + j10 – j5)I 2 + j5I 1 + j5I 1

++

10j5j420

12j

w = 0.5L 1 i 1 + 0.5L 2 i 2 – Mi 1 i 2 Since ωL1 = 10 and ω = 1000, L1 = L2 = 10 mH, M = 0.5L1 = 5mH

w = 0.5(10)(–2.445) 2 + 0.5(10)(–0.8391) 2 – 5(–2.445)(–0.8391)

w = 43.67 mJ

Find I1 and I2 in the circuit of Fig 13.90 Calculate the power absorbed by the 4-Ω resistor

j26j70

3036

coupling as sources in terms of currents that enter or leave the dot side of the coil Figure 13.85 then becomes,

Note the following,

-50 + j20(I3 – I2) j 40(I1 – I3) + j10(I2 – I1) – j30(I3 – I2) + j80(I1 – I2) – j10(I1 – I3) = 0

j100I1 – j60I2 – j40I3 = 50 Multiplying everything by (1/j10) yields 10I1 – 6I2 – 4I3 = - j5 (1)

+

−

− +

− ++ −

+ −

j10I b j40

-j50I3 +j20(I1 –I3) +j60(I3 –I2) +j30(I2 –I1) –j10(I2 –I1) +j40(I3 –I1) –j20(I3 –I2) = 0

-j40I1 – j20I2 + j10I3 = 0 Multiplying by (1/j10) yields, -4I1 – 2I2 + I3= 0 (3) Multiplying (2) by (1/j20) yields -3I1 + (4 – j5)I2 – I3 = 0 (4) Multiplying (3) by (1/4) yields -I1 – 0.5I2 – 0.25I3= 0 (5) Multiplying (4) by (-1/3) yields I1 – ((4/3) – j(5/3))I2 + (1/3)I3 = -j0.5 (7) Multiplying [(6)+(5)] by 12 yields (-22 + j20)I2 + 7I3 = 0 (8) Multiplying [(5)+(7)] by 20 yields -22I2 – 3I3 = -j10 (9) (8) leads to I2 = -7I3/(-22 + j20) = 0.2355∠42.3o = (0.17418+j0.15849)I3 (10) (9) leads to I3 = (j10 – 22I2)/3, substituting (1) into this equation produces,

I3 = j3.333 + (-1.2273 – j1.1623)I3

or I3 = Io = 1.3040∠63 o amp

If M = 0.2 H and v s = 12 cos 10t V in the circuit of Fig 13.92, find i1 and i2

Calculate the energy stored in the coupled coils at t = 15 ms

Figure 13.92

For Prob 13.23

Chapter 13, Solution 23

ω = 10 0.5 H converts to jωL 1 = j5 ohms

1 H converts to jωL 2 = j10 ohms 0.2 H converts to jωM = j2 ohms

25 mF converts to 1/(jωC) = 1/(10x25x10 -3 ) = –j4 ohms The frequency-domain equivalent circuit is shown below

For mesh 1, 12 = (j5 – j4)I 1 + j2I 2 – (–j4)I 2

For mesh 2, 0 = (5 + j10)I 2 + j2I 1 –(–j4)I 1

0 = (5 + j10)I 2 + j6I 1 (2) From (1), I 1 = –j12 – 6I 2

Substituting this into (2) produces,

–j4

+

j10

In the circuit of Fig 13.93,

(a) find the coupling coefficient,

For the network in Fig 13.94, find Zab and Io

Figure 13.94

For Prob 13.25

Chapter 13, Solution 25

m = k L1L2 = 0.5 H

We transform the circuit to frequency domain as shown below

12sin2t converts to 12∠0°, ω = 2 0.5 F converts to 1/(jωC) = –j

Find Io in the circuit of Fig 13.95 Switch the dot on the winding on the right

and calculate Io again

Figure 13.95

For Prob 13.26

Chapter 13, Solution 26

M = k L1L2

ωM = k ωL1ωL2 = 0.6 20x40 = 17 The frequency-domain equivalent circuit is shown below

For mesh 1, 200∠60° = (50 – j30 + j20)I1 + j17I2 = (50 – j10)I1 + j17I2 (1)

17j10j500

60200

∆ = 900 + j100, ∆1 = 2000∠60°(1 + j4) = 8246.2∠136°, ∆2 = 3400∠–30°

I2 = ∆2/∆ = 3.755∠–36.34°

Io = I2 = 3.755∠–36.34° A

Switching the dot on the winding on the right only reverses the direction of Io This can

be seen by looking at the resulting value of ∆2 which now becomes 3400∠150° Thus,

Find the average power delivered to the 50-Ω resistor in the circuit of

To make the problem easier to solve, let us have I3 flow around the outside loop as

00

40I6050

0

5040j5010

j

010j20

In the circuit of Fig 13.97, find the value of X that will give maximum power

transfer to the 20-Ω load

Figure 13.97

For Prob 13.28

I I

j I

Substituting (2) into (1) leads to

X j j

X j

I

5.1812

1.08.02.1

++

−

=

X j

X j j I

Z Th

1.08.02.1

5.18121

−+

=

−

=

62472

75.108

.0)1.02.1(

)5.18(1220

|

2 2

2 2

−+

=

⎯→

⎯+

−

−+

127.38j30j100

(a) Find the input impedance of the circuit in Fig 13.99 using the concept of reflected impedance

(b) Obtain the input impedance by replacing the linear transformer by its T equivalent

Chapter 13, Problem 31

For the circuit in Fig 13.100, find:

(a) the T-equivalent circuit,

(b) the Π -equivalent circuit

* Two linear transformers are cascaded as shown in Fig 13.101 Show that

)(

)(

(

(

2 2 2

2 2

2 2

3

2 2

2

in

b a b

b b a

a b b a b a b a a b a a

L L R j M L L L

M L M L L L L L j

M L L L

R

+

−

−+

−

−+

=

ωω

Chapter 13, Solution 32

We first find Zin for the second stage using the concept of reflected impedance

Zin’ = jωLb + ω2Mb2/(R + jωLb) = (jωLbR - ω2Lb2 + ω2Mb2)/(R + jωLb) (1) For the first stage, we have the circuit below

Zin = jωLa + ω2Ma2/(jωLa + Zin) = (–ω2La2 + ω2Ma2 + jωLaZin)/( jωLa + Zin) (2) Substituting (1) into (2) gives,

=

b

2 b 2 2 b

2 b a

b

2 b 2 2 b

2 b a

2 a 2 2 a 2

LjR

ML

RLjLj

LjR

)ML

RLj(LjML

ω+

ω+ω

−ω+ω

ω+

ω+ω

−ωω+ω+ω

Determine the input impedance of the air-core transformer circuit of Fig 13.102

j12Ω j10 Ω j4 Ω

1<0o V I1 I2

- -j2Ω

For loop 1,

2

)101(

For loop 2,

2 1

1 1

)2104

=

j I

Z 1 1.6154 9.077 9.219 79.91

1

Alternatively, an easier way to obtain Z is to replace the transformer with its equivalent

T circuit and use series/parallel impedance combinations This leads to exactly the same

result

* Find currents I1, I2, and I3 in the circuit of Fig 13.104

For mesh 2, 0= j2I1+(30+ j26)I2 − j12I3 (2) For mesh 3, 0=−j12I2 +(5+ j11)I3 (3)

We may use MATLAB to solve (1) to (3) and obtain

A 41.214754.15385.03736.1

.00549.00547.0

.00721.00268

A 480/2,400-V rms step-up ideal transformer delivers 50 kW to a resistive load

Calculate:

(a) the turns ratio

(b) the primary current

(c) the secondary current

,

2 2 2 1

Chapter 13, Problem 39

A 1,200/240-V rms transformer has impedance 60∠−30°Ω on the high-voltage side If the transformer is connected to a 0.8∠10°- Ω load on the low-voltage side, determine the primary and secondary currents when the transformer is connected to 1200 V rms

= (1200/240)( 15.7 ∠20.31°) = 78.5∠20.31° A

The primary of an ideal transformer with a turns ratio of 5 is connected to a voltage

source with Thevenin parameters vTh = 10 cos 2000t V and RTh = 100Ω Determine the average power delivered to a 200-Ω load connected across the secondary winding

Chapter 13, Solution 40

Consider the circuit as shown below

We reflect the 200-Ω load to the primary side

I1 = 14∠0°/28 = 0.5 A and I2 = I1/n = 0.5/(–1/3) = –1.5 A

For the circuit in Fig 13.107, determine the power absorbed by the 2-Ω

resistor Assume the 80 V is an rms value

Figure 13.107

For Prob 13.42

I j

V V

Figure 13.108

For Prob 13.43

Chapter 13, Solution 43

Transform the two current sources to voltage sources, as shown below

Using mesh analysis, –20 + 10I1 + v1 = 0

10 Ω

+ –

1 : 4

+

v 2 +

v 1

I2” = –I1”/n = –vm/(Rn2) Hence, i1(t) = (v m /Rn)cos ωt A, and i2(t) = (–v m /(n 2 R))cos ωt A

R

+ –

1 : n

+

v 2 +

v 1

absorbed by the 8-Ω resistor

Figure 13.110

For Prob 13.45

Chapter 13, Solution 45

4j8C

j8

90436

j7248

904I

36j72Z9n

Z

2 L

We now have some choices, we can go ahead and calculate the current in the second loop and calculate the power delivered to the 8-ohm resistor directly or we can merely say that the power delivered to the equivalent resistor in the primary side must be the same as the power delivered to the 8-ohm resistor Therefore,

Chapter 13, Problem 46

(a) Find I1 and I2 in the circuit of Fig 13.111 below

(b) Switch the dot on one of the windings Find I1 and I2 again

Hence, I1 = 1.072 ∠5.88° A, and I2 = –0.5I1 = 0.536 ∠185.88° A

(b) Switching a dot will not effect Zin but will effect I1 and I2

Find v(t) for the circuit in Fig 13.112

Figure 13.112

For Prob 13.47

00041

140

00

11j202

10050

0120

3

2 1 3 2 1

Chapter 13, Problem 48

Find Ix in the ideal transformer circuit of Fig 13.113

Figure 13.113

For Prob 13.48

8Ω 2:1 10 Ω

+ +

• • + V1 V2

)48(

2 1

)210(

1

n I

2 2

)410(

Solving (1)a and (2)a leads to I2 = -3.5503 +j1.4793

A 4.157923.15

0 2

2 1

o

Chapter 13, Problem 49

Find current i x in the ideal transformer circuit shown in Fig 13.114

Figure 13.114

For Prob 13.49

At node 1,

2 1

1 1

2 1

102

12

V j j

V I I

j

V V V

−+

+

=

⎯→

⎯+

2 2

2 1

6

V j

V V

Substituting these in (1) and (2),

)4.23(6

0),8.01(6

12=− I2 +V1 + j = I2 +V1 + j

Adding these gives V1=1.829 –j1.463 and

o x

j

V j

V V

10

410

1 2

937

For the circuit in Fig 13.117, determine the turns ratio n that will cause maximum

average power transfer to the load Calculate that maximum average power

Chapter 13, Problem 53

Refer to the network in Fig 13.118

(a) Find n for maximum power supplied to the 200- Ω load

(b) Determine the power in the 200-Ω load if n = 10

Figure 13.118

For Prob 13.53

Chapter 13, Solution 53

(a) The Thevenin equivalent to the left of the transformer is shown below

The reflected load impedance is ZL’ = ZL/n2 = 200/n2

For maximum power transfer, 8 = 200/n2 produces n = 5

A transformer is used to match an amplifier with an 8- Ω load as shown in Fig

13.119 The Thevenin equivalent of the amplifier is: VTh = 10 V, ZTh = 128 Ω

(a) Find the required turns ratio for maximum energy power transfer

(b) Determine the primary and secondary currents

(c) Calculate the primary and secondary voltages

1.667

L L

Z

Z = = = 1.6669 Ω Ω

Find the power absorbed by the 10-Ω resistor in the ideal transformer circuit of Fig 13.121

Figure 13.121

For Prob 13.56

Chapter 13, Solution 56

We apply mesh analysis to the circuit as shown below

At the terminals of the transformer,

For the ideal transformer circuit of Fig 13.122 below, find:

For mesh1, 80 = 20I1 – 20I3 + v1 (1)

For mesh 3, 0 = 40I3 – 20I1 which leads to I1 = 2I3 (3)

At the transformer terminals, v2 = –nv1 = –5v1 (4)

p20(the one between 1 and 3) = 0.5(20)(I1 – I3)2 = 10(20/7)2 = 81.63 watts

p20(at the top of the circuit) = 0.5(20)I32 = 81.63 watts

p100 = 0.5(100)I22 = 65.31 watts

20 Ω

+ –

Chapter 13, Problem 59

In the circuit of Fig 13.124, let v s = 40 cos 1000t Find the average power

delivered to each resistor

+ _

40 ∠ 0°

1400

103212

011222

Chapter 13, Problem 60

Refer to the circuit in Fig 13.125 on the following page

(a) Find currents I1, I2, and I3

(b) Find the power dissipated in the 40-Ω resistor

We transfer this to the primary side

Chapter 13, Problem 62

For the network in Fig 13.127, find

(a) the complex power supplied by the source,

(b) the average power delivered to the 18-Ω resistor

I1 = 40/Zin = 40/7.767∠11.89° = 5.15∠–11.89°

S = 0.5vsI1* = (20∠0°)(5.15∠11.89°) = 103∠11.89° VA (b) I2 = –I1/n, n = 2.5

I3 = –I2/n’, n = 3

I3 = I1/(nn’) = 5.15∠–11.89°/(2.5x3) = 0.6867∠–11.89°

p = 0.5|I2|2(18) = 9(0.6867)2 = 4.244 watts

Find the mesh currents in the circuit of Fig 13.128

Zin = 1 + Zin’/n2 = 1 + 2 – j = 3 – j, where n = 2

I1 = 12∠0°/(3 – j) = 12/3.162∠–18.43° = 3.795∠18.43A

I2 = I1/n = 1.8975 ∠18.43° A

I3 = –I2/n2 = 632.5 ∠161.57° mA

The Thevenin equivalent to the left of the transformer is shown below

The reflected load impedance is

'

30

L L

24 ∠ 0° V