Solution manual fundamentals of electric circuits 3rd edition chapter03

Apply nodal analysis to solve for Vx in the circuit in Fig... Using nodal analysis, determine Vo in the circuit in Fig... Apply nodal analysis to find io and the power dissipated in each

Determine Ix in the circuit shown in Fig 3.50 using nodal analysis

+ _

For the circuit in Fig 3.51, obtain v1 and v2

v 10

+ +

= 36 = - 2v1 + 3v2 (2)

Solving (1) and (2),

v1 = 0 V, v2 = 12 V

Find the currents i1 through i4 and the voltage vo in the circuit in Fig 3.52

v 20

v 10

+ + +

i1 = = 10

Given the circuit in Fig 3.53, calculate the currents i1 through i4

Obtain v0 in the circuit of Fig 3.54

v 20 k

v

=

− +

v 4

12

=

− + +

−

or v0 = 8.727 V

Apply nodal analysis to solve for Vx in the circuit in Fig 3.56

0 V 10

0 V

Using nodal analysis, find v0 in the circuit in Fig 3.57

3 v 5

=

− +

− +

Determine Ib in the circuit in Fig 3.58 using nodal analysis

0 I 300 V 5 V 15 72 V 3

get we g simplifyin

0 150

0 I 60 V 50

0 V 250

24 V

b1

11

b1

11

=

− + +

−

=

−

− +

− +

−

But

250

V 24

Ib = − 1

Substituting this into the nodal equation leads to

or V 0 8 100 V 2

24 1− = 1 = 4.165 V

Thus, Ib = (24 – 4.165)/250 = 79.34 mA

Find i0 in the circuit in Fig 3.59

Figure 3.59

Chapter 3, Solution 10

–+

12V

2v 0

+ v 0 –

+ –

v 3

v 3

Find Vo and the power dissipated in all the resistors in the circuit of Fig 3.60

4 Ω

+ _

) 12 ( V 2

0 V 1

− +

Using nodal analysis, determine Vo in the circuit in Fig 3.61

Figure 3.61 For Prob 3.12

Vo

There are two unknown nodes, as shown in the circuit below

0 1

V V 2

0 V 10

30 V

o1

o11

− +

−

(1)

At node o,

0 I 20 V 6 V 5

0 5

0 V I 4 1

V V

xo

1

ox1o

=

− +

−

=

− +

−

−

(2) But Ix = V1/2 Substituting this in (2) leads to

–15V1 + 6Vo = 0 or V1 = 0.4Vo (3)

Substituting (3) into 1,

16(0.4Vo) – 10Vo = 30 or Vo = –8.333 V

Calculate v1 and v2 in the circuit of Fig 3.62 using nodal analysis

Figure 3.62

Chapter 3, Solution 13

At node number 2, [(v2 + 2) – 0]/10 + v2/4 = 3 or v2 = 8 volts

But, I = [(v2 + 2) – 0]/10 = (8 + 2)/10 = 1 amp and v1 = 8x1 = 8volts

Using nodal analysis, find vo in the circuit of Fig 3.63

v

= +

v 5 2

v

+

= +

−

4v1 - 7v0 = -20 (2) Solving (1) and (2), v0 = 27.27 V

Apply nodal analysis to find io and the power dissipated in each resistor in the circuit of Fig 3.64

Figure 3.64

2 2

P35 = ( v − v )2G = ( 2 )23 =

3

Determine voltages v1 through v3 in the circuit of Fig 3.65 using nodal analysis

Using nodal analysis, find current io in the circuit of Fig 3.66

Figure 3.66

v 4

v v 10

v 60 4

v 60

=

− +

− +

Determine the node voltages in the circuit in Fig 3.67 using nodal analysis

v 2

v v 2

−

40 = 2v1 + v3 (2) From Fig (b), - v1 - 10 + v3 = 0 v3 = v1+ 10 (3) Solving (1) to (3), we obtain v1 = 10 V, v2 = 20 V = v3

Figure 3.68

Chapter 3, Solution 19

At node 1,

3 2 1 1

2 1 3

4 8

2 2

2

4 2

V V V

V

V

− +

2 3 1 3

7 2 4 36 0

4 2

Using MATLAB,

V 267 12

V, 933 4 V, 10 267

12

933 4

10

3 2

V B

A

For the circuit in Fig 3.69, find v1, v2, and v3 using nodal analysis

Figure 3.69

Chapter 3, Solution 20

Nodes 1 and 2 form a supernode; so do nodes 1 and 3 Hence

0 4

0 4

5 4 V,

−

V

For the circuit in Fig 3.70, find v1 and v2 using nodal analysis

Figure 3.70

v v 10

v v 4

v

=

− +

−

3v1 - 5v2 - 2v3 = 0 (2) Note that v0 = v2 We now apply KVL in Fig (b)

- v3 - 3v2 + v2 = 0 v3 = - 2v2 (3)

From (1) to (3),

v1 = 1 V, v2 = 3 V

Determine v1 and v2 in the circuit in Fig 3.71

v 2

v

+ +

Use nodal analysis to find Vo in the circuit of Fig 3.72

) V V 2 ( V 2

0 V 1

30

V

1o

1oo

0 V 4

V ) V

V

2

(

o11

o1

_

Vo

Use nodal analysis and MATLAB to find Vo in the circuit in Fig 3.73

4 V 125 0 V 125 1 0 8

4

4 V 25 0 V 75 0 0 4

V V 2

0 V

2 V 75 0 V 25 0 0 2 2

0 V 4

V V

32

32

2 V 125 1 V 125 0 0 1

0 V 8

V V

4 V 125 1 0 0

125 0

0 75

0 25 0 0

0 25 0 75 0 0

125 0 0

0 125

1

Now we can use MATLAB to solve for the unknown node voltages

>> Y=[1.125,0,0,-0.125;0,0.75,-0.25,0;0,-0.25,0.75,0;-0.125,0,0,1.125]

Y = 1.1250 0 0 -0.1250

0 0.7500 -0.2500 0

0 -0.2500 0.7500 0 -0.1250 0 0 1.1250

>> I=[4,-4,-2,2]'

I =

4 -4 -2

2

>> V=inv(Y)*I

V = 3.8000 -7.0000 -5.0000 2.2000

Vo = V1 – V4 = 3.8 – 2.2 = 1.6 V

Use nodal analysis along with MATLAB to determine the node voltages in Fig 3.74

Figure 3.74 For Prob 3.25

f

b

c b

B = A V V = A-1 B

Using MATLAB leads to

V1 = 25.52 V, V2 = 22.05 V, V3 = 14.842 V, V4 = 15.055 V

Calculate the node voltages v1, v2, and v3 in the circuit of Fig 3.75

Figure 3.75

At node 1,

3 2 1 2

1 3 1

5 10

3

20

15

V V V V

V V V

− +

4

5

3 2 2 2

=

− +

13

233

5

V V 15

V 10 10

0 45 V

V V 11

6

3

3 15

7

2 4

7

32

78 2

19 7 B

A

Thus,

V1 = –7.19V; V2 = –2.78V; V3 = 2.89V

Use nodal analysis to determine voltages v1, v2, and v3 in the circuit in Fig 3.76

Figure 3.76

v v 7 13 4

1 6 1

4 11 7

3 2 1

, 176 7

13 4

1 6 1

4 11 7

1 6 0

4 11 2

, 66 7 4 4

1 0 1

4 2 7

0 6 1

2 11 7

−

−

= Δ

v1 = 0 625 V ,

176

110

1 = = Δ

Δ

v1 = 625 mV, v2 = 375 mV, v3 = 1.625 V

Use MATLAB to find the voltages at nodes a, b, c, and d in the circuit of Fig 3.77

Figure 3.77

At node c,

d c b c

b c

c

d

V V V V

V V

V

V

2 11 5 0 5

4

At node b,

c b a b

b c b

a

V V V V

V V V

V

2 4 45

8 4

a a d

a

V V V V

V V V

V

4 2 7 30 0

8

45 16

d d d

a

V V V V

V V V

V

7 2 5 150 10

20 4

30

− +

=

⎯→

⎯

− +

V V V V

d c b a

7 2

0

5

4 0

2

7

0 2

4

1

2 11

736 1

847 7

14 101

V, 736 1

V, 847 7 V, 14

Use MATLAB to solve for the node voltages in the circuit of Fig 3.78

Figure 3.78

Chapter 3, Solution 29

At node 1,

4 2 1 2

1 1 4

At node 2,

3 2 1 3

2 2

3 3

4 1 4

In matrix form, (1) to (4) become

B AV

V V V

5 1 0 1

1 5 4 0

0 4 7 1

1 0 1 4

4 3 2 1

309 2

209 1

7708 01

B

A

V

i.e

v v 10

v 100 40

v 120 v

v 7 6

9 7

o 1

5 54 49 7

6

9 7

= +

8440 7

720

9 280

280 7

−

= Δ

5

8440

1 = − = − Δ

Δ

Io = –5.6 A

+ v 0 –

1 + 2v0 =

1

v v 1

v 4

31

Solving (2) to (4) leads to,

v1 = 4.97V, v2 = 4.85V, v3 = –0.12V

10 V

We have a supernode as shown in figure (a) It is evident that v2 = 12 V, Applying KVL

to loops 1and 2 in figure (b), we obtain,

-v1 – 10 + 12 = 0 or v1 = 2 and -12 + 20 + v3 = 0 or v3 = -8 V

Thus, v1 = 2 V, v2 = 12 V, v3 = -8V

Which of the circuits in Fig 3.82 is planar? For the planar circuit, redraw the circuits with no crossing branches

Figure 3.82

(a) This is a planar circuit It can be redrawn as shown below

Determine which of the circuits in Fig 3.83 is planar and redraw it with no crossing branches

Figure 3.83

(a) This is a planar circuit because it can be redrawn as shown below,

Rework Prob 3.5 using mesh analysis

Rework Prob 3.6 using mesh analysis

Chapter 3, Problem 6

Use nodal analysis to obtain v0 in the circuit in Fig 3.55

Figure 3.55

I 4 3

3 5 5

6

, 11 4 3

3 5

3 6

6 5

−

−

= Δ

, 11

Δ

=

i1 = -I1 = -9/11 = -0.8181 A, i2 = I1 – I2 = 10/11 = 1.4545 A

vo = 6i2 = 6x1.4545 = 8.727 V

Rework Prob 3.8 using mesh analysis

Using (1), (2), and (3) we get i1 = -5/9

Therefore, we get v0 = -2i 1 = -2(-5/9) = 1.1111 volts

24 V

+ _ 9 V

1 Ω

4 Ω

2 A

1(I2–I1) + 2(I2–I4) + 9 + 4I2 = 0

I I 1 1 0

6 6 2

1 0 7

432

We can now use MATLAB to solve the problem

>> Z=[7,0,-1;-2,6,6;0,-1,0]

Z =

7 0 -1 -2 6 6

0 -1 0

>> V=[-11,20,4]'

V = -11

20

4

>> I=inv(Z)*V

I = -0.5500 -4.0000 7.1500

Io = I1 – I2 = –2 – 4 = –6 A

Check using the super mesh (equation (3)): 1.1 – 24 + 42.9 = 20!

Determine the mesh currents i1 and i2 in the circuit shown in Fig 3.85

Figure 3.85

Chapter 3, Solution 39

For mesh 1,

0 6 10 2

10 − + 1 − 2 =

But Ix = I1 − I2 Hence,

212

12

A, 8

I

For the bridge network in Fig 3.86, find Io using mesh analysis

0 = -4i1 – 2i2 + 10i3 0 = -2i1 – i2 + 5i3 (3)

Solving (1), (2), and (3), we obtain,

io = i1 = 4.286 mA

Apply mesh analysis to find io in Fig 3.87

Figure 3.87

i i 6 1 0

1 7 2

0 1 6

3 2 1

, 234 6

1 0

1 7 2

0 1 6

1 8 2

0 3 6

Δ

38 2

1 0

8 7 2

3 1 6

At node 0, i + i2 = i3 or i = i3 – i2 =

234

240 38

2 3

−

−

−

= Δ

Δ

− Δ

= 1.188 A

Determine the mesh currents in the circuit of Fig 3.88

Figure 3.88

Chapter 3, Solution 42

For mesh 1,

(1) 2

1 2

2 1

3 1

2 2

40

0

40 100

30

0 30

50

3 2 1

40 0

48 01

B A

I

i.e I1 = 0.48 A, I2 = 0.4 A, I3 = 0.44 A

Use mesh analysis to find vab and io in the circuit in Fig 3.89

Use mesh analysis to obtain io in the circuit of Fig 3.90

Solving (1) to (3), i1 = -3.067, i3 = -1.3333; io = i1 – i3 = -1.7333 A

Find current i in the circuit in Fig 3.91

For the supermesh, 6i3 + 14i4 – 2i1 – 6i2= 0 (3) But i4 – i3 = 4 which leads to i4 = i3 + 4 (4)

Solving (1) to (4) by elimination gives i = i1 = 8.561 A

Calculate the mesh currents i1 and i2 in Fig 3.92

Figure 3.92

Chapter 3, Solution 46

For loop 1,

12 8 11 0

8

77 i2 − i2 = ⎯ ⎯→ i2= A and i1 = i 72 = 1 217 A

Rework Prob 3.19 using mesh analysis

Chapter 3, Problem 3.19

Use nodal analysis to find V1, V2, and V3 in the circuit in Fig 3.68

Figure 3.68

First, transform the current sources as shown below

2

4

2 7

1

4 1

7

3 2 1

Using MATLAB,

8667 1 , 0333 0 , 5 2 8667

1

0333 0

2

3 2

I B

A

I

But

V 10 4 20 4

20

1 1

1 = − V ⎯ ⎯→ V = − I =

I

V 933 4 ) (

8 12 8

12

2 3

3

2 = V − ⎯ ⎯→ V = + I =

Determine the current through the 10-kΩ resistor in the circuit in Fig 3.93 using mesh analysis

Figure 3.93

We apply mesh analysis and let the mesh currents be in mA

3 1

4 1 − 2 − 3 + 4 =

Putting (1) to (4) in matrix form gives

B AI

I I I

14 5 2

4

5 15 10

0

2 10 13

1

4 0 1

5

4 3 2 1

087 8

217 7

Find vo and io in the circuit of Fig 3.94

Figure 3.94

At node 0, i2 – i1 = 2i0 and i0 = -i1 which leads to i2 = -i1 (2)

For loop 3, -i1 –2i2 + 6i3 = 0 which leads to 6i3 = -i1 (3) Solving (1) to (3), i1 = (-32/3)A, i2 = (32/3)A, i3 = (16/9)A

i0 = -i1 = 10.667 A, from fig (b), v0 = i3-3i1 = (16/9) + 32 = 33.78 V

Use mesh analysis to find the current io in the circuit in Fig 3.95

For loop 1, 16i1 – 10i2 – 2i3 = 0 which leads to 8i1 – 5i2 – i3 = 0 (1)

For the supermesh, -60 + 10i2 – 10i1 + 10i3 – 2i1 = 0

Also, 3i0 = i3 – i2 and i0 = i1 which leads to 3i1 = i3 – i2 (3) Solving (1), (2), and (3), we obtain i1 = 1.731 and i0 = i1 = 1.731 A

Apply mesh analysis to find vo in the circuit in Fig 3.96

Use mesh analysis to find i1, i2, and i3 in the circuit of Fig 3.97

Figure 3.97

+

V S

+ – 2V 0

i1 = 3.5 A, i2 = -0.5 A, i3 = 2.5 A

Find the mesh currents in the circuit of Fig 3.98 using MATLAB

2 kΩ

+ _

Applying mesh analysis leads to;

–12 + 4kI1 – 3kI2 – 1kI3 = 0 (1)

–3kI1 + 7kI2 – 4kI4 = 0

I I I

I k 16 6 0 0

6 15 0 1

0 0 7 3

0 1 3 4

5321

3 mA –2.423 mA

Find the mesh currents i1, i2, and i3 in the circuit in Fig 3.99

Figure 3.99

Chapter 3, Solution 54

Let the mesh currents be in mA For mesh 1,

2 1 2

2 1 3

Using MATLAB,

mA 25 10 ,

mA 5 8 , mA 25 5 25

10

5 8

25 5

3 2

I B

A

I

In the circuit of Fig 3.100, solve for i1, i2, and i3

Figure 3.100

For mesh 4, 12(I4 – I1) + 4(I4 – I3) – 8 = 0 (2)

For the supermesh 6(I2 – I1) + 10 + 2I3 + 4(I3 – I4) = 0

Determine v1 and v2 in the circuit of Fig 3.101

Figure 3.101

For loop 1, 12 = 4i1 – 2i2 – 2i3 which leads to 6 = 2i1 – i2 – i3 (1)

For loop 2, 0 = 6i2 –2i1 – 2 i3 which leads to 0 = -i1 + 3i2 – i3 (2)

For loop 3, 0 = 6i3 – 2i1 – 2i2 which leads to 0 = -i1 – i2 + 3i3 (3)

In matrix form (1), (2), and (3) become,

1 3 1

1 1 2

3 2 1

3 1 1

1 3 1

1 1 2

1 3 1

1 6 2

0 3 1

6 1 2

In the circuit in Fig 3.102, find the values of R, V1, and V2 given that io = 18 mA

Figure 3.102

Chapter 3, Solution 57

Assume R is in kilo-ohms

V V

V V

mA x k

V2 = 4 Ω 18 = 72 , 1 = 100 − 2 = 100 − 72 = 28

Current through R is

R R R

i V i

R

3

3 28 3

Find i1, i2, and i3 the circuit in Fig 3.103

For loop 1, 120 + 40i1 – 10i2 = 0, which leads to -12 = 4i1 – i2 (1)

For loop 2, 50i2 – 10i1 – 10i3 = 0, which leads to -i1 + 5i2 – i3 = 0 (2)

For loop 3, -120 – 10i2 + 40i3 = 0, which leads to 12 = -i2 + 4i3 (3)

Solving (1), (2), and (3), we get, i1 = -3A, i2 = 0, and i3 = 3A

Rework Prob 3.30 using mesh analysis

Chapter 3, Problem 30

Using nodal analysis, find vo and io in the circuit of Fig 3.79

Figure 3.79

12 3

1

32 2 3

i i

3 2 1

1 3

0

12 3

1

32 2

12 6

1

32 10 3

6 3 1

10 2 3

Calculate the power dissipated in each resistor in the circuit in Fig 3.104

At node 1, (v1/1) + (0.5v1/1) = (10 – v1)/4, which leads to v1 = 10/7

At node 2, (0.5v1/1) + ((10 – v2)/8) = v2/2 which leads to v2 = 22/7

P1Ω = (v1)2/1 = 2.041 watts, P2Ω = (v2)2/2 = 4.939 watts

P4Ω = (10 – v1)2/4 = 18.38 watts, P8Ω = (10 – v2)2/8 = 5.88 watts

Calculate the current gain io/is in the circuit of Fig 3.105

At node 1, is = (v1/30) + ((v1 – v2)/20) which leads to 60is = 5v1 – 3v2 (1)

But v2 = -5v0 and v0 = v1 which leads to v2 = -5v1

Hence, 60is = 5v1 + 15v1 = 20v1 which leads to v1 = 3is, v2 = -15is

i0 = v2/50 = -15is/50 which leads to i0/is = -15/50 = –0.3

Find the mesh currents i1, i2, and i3 in the network of Fig 3.106

We have a supermesh Let all R be in kΩ, i in mA, and v in volts

For the supermesh, -100 +4i1 + 8i2 + 2i3 + 40 = 0 or 30 = 2i1 + 4i2 + i3 (1)

Solving (1), (2), and (3), we get i1 = 2 mA, i2 = 6 mA, and i3 = 2 mA

Find vx, and ix in the circuit shown in Fig 3.107

At node A, i1 + 3 + (vx/4) = i2, but vx = 2(i1 – i2), hence, i1 + 2 = i2 (2)

Solving (1) and (2) gives i1 = 2.105 A and i2 = 4.105 A

vx = 2(i1 – i2) = –4 volts and ix = i2 – 2 = 2.105 amp

Find vo, and io in the circuit of Fig 3.108

Figure 3.108

For mesh 2, 20i2 – 10i1 + 4i0 = 0 (1)

But at node A, io = i1 – i2 so that (1) becomes i1 = (16/6)i2 (2) For the supermesh, -100 + 50i1 + 10(i1 – i2) – 4i0 + 40i3 = 0

Use MATLAB to solve for the mesh currents in the circuit of Fig 3.109

1 6 12

–I1 – I2 + 7I4 – 2I5 – 6 = 0 or

6 = –I1 – I2 + 7I4 – 2I5 (4) For mesh 5,

–I2 – I3 – 2I4 + 8I5 – 10 = 0 or

5 4 3

10 = − I − I − I + I (5)

B AI 10

6 9 0 12

I I I I I

8 2 1 1

0

2 7 0

1

1

1 0 15

8

0

1 1 8 16

6

0 1 0

6

12

54321

Write a set of mesh equations for the circuit in Fig 3.110 Use MATLAB to determine the mesh currents

+ _

+ _

8 Ω

10 Ω

+ _

I5

32 V

8 Ω

18 4 0

6

0

4 12 4 2

2

0 4 18

0

6

6 2 0

30

4

0 2 6 4

Obtain the node-voltage equations for the circuit in Fig 3.111 by inspection Then solve

V 3 2 V 5 0 5 0 0

5 0 95 0 25 0

0 25 0 35

equation

Vo = V2 – V3 Substituting this back into the matrix equation, the first equation becomes,

2 V 5 0 5 0 0

5 0 95 0 25 0

3 25 3 35 0

Now we can use MATLAB to solve for V

>> Y=[0.35,-3.25,3;-0.25,0.95,-0.5;0,-0.5,0.5]

Y = 0.3500 -3.2500 3.0000 -0.2500 0.9500 -0.5000

0 -0.5000 0.5000

>> I=[-2,0,6]'

I = -2

0

6

>> V=inv(Y)*I

V = -164.2105 -77.8947 -65.8947

Vo = V2 – V3 = –77.89 + 65.89 = –12 V

Let us now do a quick check at node 1

–3(–12) + 0.1(–164.21) + 0.25(–164.21+77.89) + 2 =

+36 – 16.421 – 21.58 + 2 = –0.001; answer checks!