# Solution Sadiku 5th - ch03

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Chapter 3, Solution 1

R1

R2 Ix

12 V

R3

9V

Solution

3 mA

Chapter 3, Solution 2 At node 1, v1 10

v1 5

6

v1

v2 2

60 = - 8v 1 + 5v 2

(1)

At node 2, v2 4

3 6

v1

v2 2

Solving (1) and (2), v 1 = 0 V, v 2 = 12 V

36 = - 2v 1 + 3v 2

(2)

Chapter 3, Solution 3 Applying KCL to the upper node, 8

v0 10

vo 20

vo 30

20

v0 10 v i3 = 0 30

i1 =

v0 = 0 or v 0 = –60 V 60 v0 20 v –2 A, i 4 = 0 60

–6 A , i 2 =

–3 A, 1 A.

Chapter 3, Solution 4 3A

v1

6A

i1

i2

20

10

v2 i3

40

i4 40

At node 1, –6 – 3 + v 1 /(20) + v 1 /(10) = 0 or v 1 = 9(200/30) = 60 V At node 2, 3 – 2 + v 2 /(10) + v 2 /(5) = 0 or v 2 = –1(1600/80) = –20 V i 1 = v 1 /(20) = 3 A, i 2 = v 1 /(10) = 6 A, i 3 = v 2 /(40) = –500 mA, i 4 = v 2 /(40) = –500 mA.

2A

Chapter 3, Solution 5 Apply KCL to the top node. 30 v 0 2k

20 v 0 5k

v0 4k

v 0 = 20 V

Chapter 3, Solution 6. Solve for V 1 using nodal analysis. 10

10 V

– +

5

10 + V1

10

20 V

+

– Figure 3.55 For Prob. 3.6. Step 1. The first thing to do is to select a reference node and to identify all the unknown nodes. We select the bottom of the circuit as the reference node. The only unknown node is the one connecting all the resistors together and we will call that node V 1 . The other two nodes are at the top of each source. Relative to the reference, the one at the top of the 10-volt source is –10 V. At the top of the 20-volt source is +20 V. Step 2. unknown).

Setup the nodal equation (there is only one since there is only one

Step 3.

Simplify and solve.

or V 1 = –2 V. The answer can be checked by calculating all the currents and see if they add up to zero. The top two currents on the left flow right to left and are 0.8 A and 1.6 A respectively. The current flowing up through the 10-ohm resistor is 0.2 A. The current flowing right to left through the 10-ohm resistor is 2.2 A. Summing all the currents flowing out of the node, V 1 , we get, +0.8+1.6 –0.2–2.2 = 0. The answer checks.

Chapter 3, Solution 7 2

Vx 0 10

Vx 0 20

0.2Vx

0

0.35V x = 2 or V x = 5.714 V. Substituting into the original equation for a check we get, 0.5714 + 0.2857 + 1.1428 = 1.9999 checks!

Chapter 3, Solution 8 i1

6

v1

i3

20

i2 + V0

60V 4

+

+ 5V 0 –

20

i1 + i2 + i3 = 0 But

v1 10

( v1 60) 0 20

v1 5 v 0 20

0

2 v1 so that 2v 1 + v 1 – 60 + v 1 – 2v 1 = 0 5 or v 1 = 60/2 = 30 V, therefore v o = 2v 1 /5 = 12 V. v0

Chapter 3, Solution 9

79.34 mA

Chapter 3, Solution 10

v1

v2

v3

At node 1.

[(v 1 –0)/8] + [(v 1 –v 3 )/1] + 4 = 0

At node 2.

–4 + [(v 2 –0)/2] + 2i o = 0

At node 3.

–2i o + [(v 3 –0)/4] + [(v 3 –v 1 )/1] = 0

Finally, we need a constraint equation,

i o = v 1 /8

This produces, 1.125v 1 – v 3 = 4

(1)

0.25v 1 + 0.5v 2 = 4

(2)

–1.25v 1 + 1.25v 3 = 0 or v 1 = v 3

(3)

Substituting (3) into (1) we get (1.125–1)v 1 = 4 or v 1 = 4/0.125 = 32 volts. This leads to, i o = 32/8 = 4 amps.

Chapter 3, Solution 11

Solution

3V 293.9 W 750 mW 121.5 W

Chapter 3, Solution 12 There are two unknown nodes, as shown in the circuit below.

20

10

V1

Vo

Ix

40 V

+ _

20

10

4 Ix

At node 1, V1 40 V1 0 V1 Vo 0 or 20 20 10 (0.05+0.05+.1)V 1 – 0.1V o = 0.2V 1 – 0.1V o = 2

(1)

Vo V1 Vo 0 4I x 0 and I x = V 1 /20 10 10 –0.1V 1 – 0.2V 1 + 0.2V o = –0.3V1 + 0.2Vo = 0 or

(2)

V 1 = (2/3)V o

(3)

At node o,

Substituting (3) into (1), 0.2(2/3)V o – 0.1V o = 0.03333Vo = 2 or V o = 60 V.

Chapter 3, Solution 13 Calculate v 1 and v 2 in the circuit of Fig. 3.62 using nodal analysis. 10 V

15 A

Figure 3.62 For Prob. 3.13. Solution At node number 2, [((v 2 + 10) – 0)/10] + [(v 2 –0)/4] – 15 = 0 or (0.1+0.25)v 2 = 0.35v 2 = –1+15 = 14 or v 2 = 40 volts. Next, I = [(v 2 + 10) – 0]/10 = (40 + 10)/10 = 5 amps and v 1 = 8x5 = 40 volts.

Chapter 3, Solution 14 Using nodal analysis, find v o in the circuit of Fig. 3.63. 12.5 A

8 1

2

+

100 V

50 V

4

vo

Figure 3.63 For Prob. 3.14. 12.5 A

Solution

v0

v1 1

100 V

2

+

+ vo

4

8

50 V

+

At node 1, [(v 1 –100)/1] + [(v 1 –v o )/2] + 12.5 = 0 or 3v 1 – v o = 200–25 = 175 At node o, [(v o –v 1 )/2] – 12.5 + [(v o –0)/4] + [(v o +50)/8] = 0 or –4v 1 + 7v o = 50 (2) Adding 4x(1) to 3x(2) yields,

(1)

4(1) + 3(2) = –4v o + 21v o = 700 + 150 or 17v o = 850 or v o = 50 V. Checking, we get v 1 = (175+v o )/3 = 75 V. At node 1, [(75–100)/1] + [(75–50)/2] + 12.5 = –25 + 12.5 + 12.5 = 0! At node o, [(50–75)/2] + [(50–0)/4] + [(50+50)/8] – 12.5 = –12.5 + 12.5 + 12.5 – 12.5 = 0!

Chapter 3, Solution 15

5A

v0

v1 1

8

2 4

40 V

20 V

+

+

Nodes 1 and 2 form a supernode so that v 1 = v 2 + 10 At the supernode, 2 + 6v 1 + 5v 2 = 3 (v 3 - v 2 ) At node 3, 2 + 4 = 3 (v 3 - v 2 )

2 + 6v 1 + 8v 2 = 3v 3

v3 = v2 + 2

Substituting (1) and (3) into (2), 2 + 6v 2 + 60 + 8v 2 = 3v 2 + 6 v 1 = v 2 + 10 =

v2 =

54 11

i 0 = 6v i = 29.45 A v2 P 65 = 1 R

v G 56 11

2 2

P 55 = v G P 35 = v L

54 11

2 1

v3

2

G

2

6

144.6 W

2

5

129.6 W

( 2) 2 3

12 W

(1)

56 11

(2) (3)

Chapter 3, Solution 16 2S

v2

v1 i0 2A

+

1S

v0

4S

8S

v3 13 V

+

At the supernode, 2 = v 1 + 2 (v 1 - v 3 ) + 8(v 2 – v 3 ) + 4v 2 , which leads to 2 = 3v 1 + 12v 2 - 10v 3 (1) But v 1 = v 2 + 2v 0 and v 0 = v 2 . Hence v 1 = 3v 2 v 3 = 13V Substituting (2) and (3) with (1) gives, v 1 = 18.858 V, v 2 = 6.286 V, v 3 = 13 V

(2) (3)

Chapter 3, Solution 17

v1 i0

4

2 10

v2

60 V 60 V

3i 0

+

60 v1 v1 v1 v 2 4 8 2 60 v 2 v1 v 2 At node 2, 3i 0 + 10 2 At node 1,

But i 0 =

8

120 = 7v 1 - 4v 2

(1)

0

60 v1 . 4

Hence 3 60 v1 4

60 v 2 10

v1

v2 2

0

1020 = 5v 1 + 12v 2

Solving (1) and (2) gives v 1 = 53.08 V. Hence i 0 =

60 v1 4

1.73 A

(2)

Chapter 3, Solution 18

–+

v2

v1 2

v3 2

15A

8

4

30 V

+

+

v1

v3

(a)

At node 2, in Fig. (a),

v2

v2

v1

At the supernode,

(b)

v1

v2

2

v3 2

v2

2 2v 1 +v 3 = 120

v3 2

–15 = 0 or –0.5v 1 + v 2 – 0.5v 3 = 15

v1 4

(1)

v3 = 0 and (v 1 /4) – 15 + (v 3 /8) = 0 or 8 (2)

From Fig. (b), – v 1 – 30 + v 3 = 0 or v 3 = v 1 + 30 Solving (1) to (3), we obtain, v 1 = 30 V, v 2 = 60 V = v 3

(3)

Chapter 3, Solution 19 At node 1, V1 V3 2 At node 2, 5

V1 V2 8

3

12

V3

V1 V3 8 2 From (1) to (3), 7 1 4

1

16

V2 V3 4

V1 V2 V2 8 2 At node 3, 3

V1 4

4 V1

7

2 V2

2

7 V3

0

V2 V3 4

16 0

V1

7V1 V2

7V2

0

(1)

2V3

36

AV

4V3

(2)

4V1

2V2

7V3 (3)

B

36

Using MATLAB, 10 V

1

A B

4.933 12.267

V1

10 V, V2

4.933 V, V3

12.267 V

Chapter 3, Solution 20 Nodes 1 and 2 form a supernode; so do nodes 1 and 3. Hence V1 V2 V3 0 V1 4V2 V3 0 (1) 4 1 4 . V1

V2

.

4

2

1

4

Between nodes 1 and 3, V1 12 V3 0 V3 V1 12 Similarly, between nodes 1 and 2, V1 V2 2i But i V3 / 4 . Combining this with (2) and (3) gives . V2

V3

6 V1 / 2

(2) (3) (4)

Solving (1), (2), and (4) leads to V1

3V, V2

4.5V, V3

15V

Chapter 3, Solution 21 4k

2k

v1

v3

3v 0 +

3v 0 v2

+ v0

3 mA

1k

+ +

+

v3

v2

(b)

(a) Let v 3 be the voltage between the 2k At node 1, v1 v 2 v1 v 3 3x10 3 4000 2000 At node 2, v1 v 2 4

v1

v3 2

v2 1

resistor and the voltage-controlled voltage source. 12 = 3v 1 - v 2 - 2v 3

3v 1 - 5v 2 - 2v 3 = 0

(1)

(2)

Note that v 0 = v 2 . We now apply KVL in Fig. (b) - v 3 - 3v 2 + v 2 = 0 From (1) to (3), v 1 = 1 V, v 2 = 3 V

v 3 = - 2v 2

(3)

Chapter 3, Solution 22 At node 1,

12 v 0 2

At node 2, 3 +

v1 4

v1

v2 8

3 v2

v1

v0 8

24 = 7v 1 - v 2

(1)

5v 2 1

But, v 1 = 12 - v 1 Hence, 24 + v 1 - v 2 = 8 (v 2 + 60 + 5v 1 ) = 4 V 456 = 41v 1 - 9v 2 Solving (1) and (2), v 1 = - 10.91 V, v 2 = - 100.36 V

(2)

Chapter 3, Solution 23 We apply nodal analysis to the circuit shown below.

1

Vo

+

V1

+

+ _

30 V

2 Vo

4

Vo

2

16

3A

_

At node o, Vo

30 1

Vo 0 2

Vo

(2Vo 4

V1 )

0

1.25Vo

0.25V1

30

(1)

At node 1, (2Vo

V1 ) Vo 4

V1 0 3 16

0

5V1

4Vo

48

From (1), V 1 = 5V o – 120. Substituting this into (2) yields 29V o = 648 or V o = 22.34 V.

(2)

Chapter 3, Solution 24 Consider the circuit below. 8 + Vo _ 4A

V2

V1

2A

4

V3

2

1

V1 0 4 1 V2 4 2 V3 V2 4 V4 2 8

V1

8 V2

0

V3

0

0.75 0.25 0

0 0.25

2

1.125V1 0.125V4 0

V3 0 2 0 2 V1 V4 0 0 1 0

0.125

0

4

1.125 0

V4

V4

1

4

(1)

0.75V2

0.25V3

4

(2)

0.25V2

0.75V3

2

(3)

0.125V1 1.125V4

0.125 0

0.75

0

0

1.125

2

(4)

4 V

4 2 2

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 V o = V 1 – V 4 = 3.8 – 2.2 = 1.6 V.

Chapter 3, Solution 25 Consider the circuit shown below. 20 4 10 1

1

10

2

3 30

8

4

At node 1. V1 V2 V1 V4 4 1 20 At node 2, V1 V2 V2 V2 V3 1 8 10

80

20

21V1 20V2 0

V4

80V1 98V2

(1)

8V3

At node 3, V2 V3 V3 V3 V4 0 2V2 5V3 2V4 10 20 10 At node 4, V1 V4 V3 V4 V4 0 3V1 6V3 11V4 20 10 30 Putting (1) to (4) in matrix form gives:

(3)

(4)

V1

80

21

20

0

0

80

98

8

0

0

0

2

5

2

0

3

0

6

11 V4

B =A V

1

(2)

V2 V3

V = A-1 B

Using MATLAB leads to V 1 = 25.52 V,

V 2 = 22.05 V, V 3 = 14.842 V, V 4 = 15.055 V

Chapter 3, Solution 26 At node 1, V V3 V1 V2 15 V1 3 1 45 20 10 5 At node 2, V1 V2 4 I o V2 V2 V3 5 5 5 V1 V3 But I o . Hence, (2) becomes 10 0 7V1 15V2 3V3 At node 3, V1 V3 10 V3 V2 V3 3 0 10 15 5 Putting (1), (3), and (4) in matrix form produces 7 7 3

4 15

2 V1 3 V2

45 0

6

11

70

V3

AV

7V1

4V2

2V3

(1)

(2)

(3)

70

B

Using MATLAB leads to 7.19 1 V A B 2.78 2.89 Thus, V 1 = –7.19V; V 2 = –2.78V; V 3 = 2.89V.

3V1 6V2 11V3

(4)

Chapter 3, Solution 27 At node 1, 2 = 2v 1 + v 1 – v 2 + (v 1 – v 3 )4 + 3i 0 , i 0 = 4v 2 . Hence, At node 2,

2 = 7v 1 + 11v 2 – 4v 3

(1)

v 1 – v 2 = 4v 2 + v 2 – v 3

0 = – v 1 + 6v 2 – v 3

(2)

At node 3, 2v 3 = 4 + v 2 – v 3 + 12v 2 + 4(v 1 – v 3 ) or

– 4 = 4v 1 + 13v 2 – 7v 3

(3)

In matrix form,

2

7

11

1

6

4

13

v1 =

4 v1

2

1 4

6 13

1 v2 7 v3

0 4

1

176,

1

7

2

1

0

1

11

4

7 4

7

11

0

6

4

4 1

4

2

66,

3

7

110 176

v3 =

0.625V, v 2 =

3

286 176

4 1

13

7

7

11

2

1

6

0

4

13

2

110

286

4

66 176

0.375V

1.625V.

v 1 = 625 mV, v 2 = 375 mV, v 3 = 1.625 V.

Chapter 3, Solution 28 At node c, Vd Vc Vc Vb Vc 0 10 4 5 At node b, Va 90 Vb Vc Vb Vb 8 4 8 At node a, Va 60 Vd Va Va 90 Vb 0 4 16 8 At node d, Va 60 Vd Vd Vd Vc 4 20 10 In matrix form, (1) to (4) become 0 5 11 2 Va 0 1 4 2 0 Vb 90 7

2

0

4 Vc 8 Vd

60

5Vb

11Vc

90

2Vd

Va

4Vb

60

300

5Va

AV

7Va

2Vc

(1)

2Vc (2) 2 Vb

4Vd (3)

8Vd (4)

B

5 0 2 300 We use MATLAB to invert A and obtain 10.56 V

A 1B

20.56 1.389 43.75

Thus, V a = –10.56 V; V b = 20.56 V; V c = 1.389 V; VC d = –43.75 V.

Chapter 3, Solution 29 At node 1, 5 V1 V4 2V1 V1 V2 0 At node 2, V1 V2 2V2 4(V2 V3 ) 0 At node 3, 6 4(V2 V3 ) V3 V4 At node 4, 2 V3 V4 V1 V4 3V4 In matrix form, (1) to (4) become 4 1 0 1 V1 1

7

0 4 1 0 Using MATLAB,

V

A 1B

4 5 1

0

V2

1 V3 5 V4

5 0

6

4V2 2

4V1 V2 V4

(1)

V1

(2)

7V2

4V3

(3)

5V3 V4

V1 V3

5V4

(4)

5 0 6

AV

B

2

0.7708 1.209 2.309 0.7076

i.e. V1

0.7708 V, V 2

1.209 V, V 3

2.309 V, V 4

0.7076 V

Chapter 3, Solution 30 –+ i0 v1

96 V

20

v0

1

10 +

80 V

40

2 4v 0

+ –

2i 0

80

At node 1, [(v 1 –80)/10]+[(v 1 –4v o )/20]+[(v 1 –(v o –96))/40] = 0 or (0.1+0.05+0.025)v 1 – (0.2+0.025)v o = 0.175v 1 – 0.225v o = 8–2.4 = 5.6

(1)

At node 2, –2i o + [((v o –96)–v 1 )/40] + [(v o –0)/80] = 0 and i o = [(v 1 –(v o –96))/40] –2[(v 1 –(v o –96))/40] + [((v o –96)–v 1 )/40] + [(v o –0)/80] = 0 –3[(v 1 –(v o –96))/40] + [(v o –0)/80] = 0 or –0.0.075v 1 + (0.075+0.0125)v o = 7.2 = –0.075v 1 + 0.0875v o = 7.2 (2) Using (1) and (2) we get, 0.175 0.075

0.225 v1 0.0875 vo

5.6 7.2

or

0.0875 0.225 v1 vo

5.6 0.075 0.175 0.0153125 0.016875 7.2

0.0875 0.225 0.075 0.175 0.0015625

5.6 7.2

v 1 = –313.6–1036.8 = –1350.4 v o = –268.8–806.4 = –1.0752 kV and i o = [(v 1 –(v o –96))/40] = [(–1350.4–(–1075.2–96))/40] = –4.48 amps.

Chapter 3, Solution 31 1

+ v0 – v2

v1 1A

2v 0

v3

2

i0 4

1

10 V

4

At the supernode, 1 + 2v 0 =

v1 4

v1

v2 1

v3 1

(1)

But v o = v 1 – v 3 . Hence (1) becomes, 4 = -3v 1 + 4v 2 +4v 3

(2)

At node 3, 2v o + or

v3 4

v1

v3

10 v 3 2

20 = 4v 1 + 0v 2 – v 3

At the supernode, v 2 = v 1 + 4i o . But i o =

(3)

v3 . Hence, 4

v2 = v1 + v3 Solving (2) to (4) leads to, v 1 = 4.97V, v 2 = 4.85V, v 3 = –0.12V.

(4)

+

Chapter 3, Solution 32 5k

v1

v3

v2

+ 10 k

4 mA

v1

10 V

20 V

–+

+–

+

loop 1 12 V

+

loop 2

v3

(b)

(a)

We have a supernode as shown in figure (a). It is evident that v 2 = 12 V, Applying KVL to loops 1and 2 in figure (b), we obtain, -v 1 – 10 + 12 = 0 or v 1 = 2 and -12 + 20 + v 3 = 0 or v 3 = -8 V Thus,

v 1 = 2 V, v 2 = 12 V, v 3 = -8V.

Chapter 3, Solution 33 (a) This is a planar circuit. It can be redrawn as shown below.

5 3

1

2

4

6

2A

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

4

3

12 V

5 +

1

2

Chapter 3, Solution 34 (a)

This is a planar circuit because it can be redrawn as shown below, 7 2

1

3

6 10 V

5

+

4

(b)

This is a non-planar circuit.

Chapter 3, Solution 35

30 V

20 V

+

+

i1 2k

+

i2

v0

5k

4k

Assume that i 1 and i 2 are in mA. We apply mesh analysis. For mesh 1, -30 + 20 + 7i 1 – 5i 2 = 0 or 7i 1 – 5i 2 = 10

(1)

For mesh 2, -20 + 9i 2 – 5i 1 = 0 or -5i 1 + 9i 2 = 20 Solving (1) and (2), we obtain, i 2 = 5. v 0 = 4i 2 = 20 volts.

(2)

Chapter 3, Solution 36 10 V

4 i1

12 V

+–

i2

I1

+

I2

6

i3

2

Applying mesh analysis gives, 10I 1 – 6I 2 = 12 and –6I 1 + 8I 2 = –10 4 3 or

6 5

5 3

3 I1 4

I2

6 5

or

I1 I2

3 5 11

6 5

I 1 = (24–15)/11 = 0.8182 and I 2 = (18–25)/11 = –0.6364 i 1 = –I 1 = –818.2 mA; i 2 = I 1 – I 2 = 0.8182+0.6364 = 1.4546 A; and i 3 = I 2 = –636.4 mA.

Chapter 3, Solution 37 6

60 V +

+ v0

20

4

i1

i2

5v 0

+ –

20

Applying mesh analysis to loops 1 and 2, we get, 30i 1 – 20i 2 + 60 = 0 which leads to i 2 = 1.5i 1 + 3

(1)

–20i 1 + 40i 2 – 60 + 5v 0 = 0

(2)

But, v 0 = –4i 1

(3)

Using (1), (2), and (3) we get –20i 1 + 60i 1 + 120 – 60 – 20i 1 = 0 or 20i 1 = –60 or i 1 = –3 amps and i 2 = 7.5 amps. Therefore, we get, v 0 = –4i 1 = 12 volts.

Chapter 3, Solution 38 Consider the circuit below with the mesh currents. 4

I3

+ _

60 V

3

I4

1

10 A 2

2 Io 1 I1

1

I2

+ _

22.5V

4 5A

I 1 = –5 A

(1)

1(I 2 –I 1 ) + 2(I 2 –I 4 ) + 22.5 + 4I 2 = 0 7I 2 – I 4 = –27.5

(2)

–60 + 4I 3 + 3I 4 + 1I 4 + 2(I 4 –I 2 ) + 2(I 3 – I 1 ) = 0 (super mesh) –2I 2 + 6 I 3 + 6I 4 = +60 – 10 = 50

(3)

But, we need one more equation, so we use the constraint equation –I 3 + I 4 = 10. This now gives us three equations with three unknowns. 7 2 0

0 6 1

1 I2

27.5

6

I3

50

1

I4

10

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=[–27.5,50,10]' V= –27.5 50 10 >> I=inv(Z)*V I= –1.3750 –10.0000 17.8750 I o = I 1 – I 2 = –5 – 1.375 = –6.375 A. Check using the super mesh (equation (3)): –2I 2 + 6 I 3 + 6I 4 = 2.75 – 60 + 107.25 = 50!

Chapter 3, Solution 39

Solution

R1

R2 Ix

12 V

I1

I2 R3

9V

3 mA

Chapter 3, Solution 40 2k

56V

+

6k

6k

i2 2k

i1

i3

4k

4k

Assume all currents are in mA and apply mesh analysis for mesh 1. –56 + 12i 1 – 6i 2 – 4i 3 = 0 or 6i 1 – 3i 2 – 2i 3 = 28

(1)

for mesh 2, –6i 1 + 14i 2 – 2i 3 = 0 or –3i 1 + 7i 2 – i 3

=0

(2)

=0

(3)

for mesh 3, –4i 1 – 2i 2 + 10i 3 = 0 or –2i 1 – i 2 + 5i 3

Solving (1), (2), and (3) using MATLAB, we obtain, i o = i 1 = 8 mA.

Chapter 3, Solution 41 10

i1

6V

2

+– 1

i2 4 8V

5

i3

+

i i2

i3 0

For loop 1, For loop 2,

6 = 12i 1 – 2i 2

3 = 6i 1 – i 2

(1)

-8 = – 2i 1 +7i 2 – i 3

For loop 3,

(2)

-8 + 6 + 6i 3 – i 2 = 0

2 = – i 2 + 6i 3

We put (1), (2), and (3) in matrix form, 6

1 0 i1

3

2

7 1 i2

8

0

1 6 i3

2

6

1 0

2

7 1

0

1 6

6 3 0 234,

2

2 8 1 0 2 6

240

(3)

3

At node 0, i + i 2 = i 3 or i = i 3 – i 2 =

6

1 3

2

7 8

0

1 2

3

2

38

38

240 = 1.188 A 234

Chapter 3, Solution 42 Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Determine the mesh currents in the circuit of Fig. 3.88.

Figure 3.88 Solution For mesh 1, 12 50I 1 30I 2 0 12 50I1 30I 2 For mesh 2, 8 100 I 2 30 I 1 40 I 3 0 8 30 I 1 100 I 2 For mesh 3, 6 50 I 3 40 I 2 0 6 40 I 2 50 I 3 Putting eqs. (1) to (3) in matrix form, we get 50 30

30 100

0

40

I1 0 40 I 2

12 8

50

6

I3

AI

(1) 40 I 3

B

Using Matlab, 0.48 I

1

A B

0.40 0.44 i.e. I 1 = 480 mA, I 2 = 400 mA, I 3 = 440 mA

(2) (3)

Chapter 3, Solution 43 20 a 80 V

+

i1

30

+ i3

30

V ab

20 80 V

+

i2

30 b

20

For loop 1, 80 = 70i 1 – 20i 2 – 30i 3

8 = 7i 1 – 2i 2 – 3i 3

(1)

80 = 70i 2 – 20i 1 – 30i 3

8 = -2i 1 + 7i 2 – 3i 3

(2)

0 = -30i 1 – 30i 2 + 90i 3

0 = i 1 + i 2 – 3i 3

For loop 2,

For loop 3, (3)

Solving (1) to (3), we obtain i 3 = 16/9 I o = i 3 = 16/9 = 1.7778 A V ab = 30i 3 = 53.33 V.

Chapter 3, Solution 44 90 V +

i3

2

4

i2

1 180V +

5

i1 45 A i1

i2

Loop 1 and 2 form a supermesh. For the supermesh, 6i 1 + 4i 2 – 5i 3 + 180 = 0

(1)

For loop 3,

–i 1 – 4i 2 + 7i 3 + 90 = 0

(2)

Also,

i 2 = 45 + i 1

(3)

Solving (1) to (3), i 1 = –46, i 3 = –20;

i o = i 1 – i 3 = –26 A

Chapter 3, Solution 45 4

8

i3

i4

6

2 30V

+

i1

3

i2

1

For loop 1,

30 = 5i 1 – 3i 2 – 2i 3

(1)

For loop 2,

10i 2 - 3i 1 – 6i 4 = 0

(2)

For the supermesh,

6i 3 + 14i 4 – 2i 1 – 6i 2 = 0

(3)

But

i 4 – i 3 = 4 which leads to i 4 = i 3 + 4

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

(4)

Chapter 3, Solution 46 For loop 1, 12 3i1 8(i1 i 2 ) For loop 2, 8(i 2 i1 ) 6i 2 2v o But v o

12 11i1 8i 2 8i1 14i 2

0

2v o

11i1 8i 2

12

(1)

0

3i1 ,

8i1 14i 2 6i1 0 Substituting (2) into (1), 77i 2 8i 2 12

i1

7i 2

i 2 0.1739 A and i1

(2) 7i 2

1.217 A

Chapter 3, Solution 47 First, transform the current sources as shown below. - 6V + 2

V1

8

V2

I3

4

V3

4 8 I1

+ 20V -

2

I2

+ 12V -

For mesh 1, 20 14 I 1 2 I 2 For mesh 2,

10

7 I1

I2

4I 3

(1)

6

I1

7I 2

2I 3

(2)

6 14 I 3 4 I 2 8I 1 0 3 4 I1 Putting (1) to (3) in matrix form, we obtain

2I 2

7I 3

(3)

12 14 I 2 2 I 1 For mesh 3,

7

1

1

7

4

2

8I 3

4I 3

0

0

I1

10

2 I2 7 I3

6

4

AI

B

3

Using MATLAB, 2 I

But

1

A B

0.0333 1.8667

I1

2.5, I 2

0.0333, I 3

1.8667

I1

20 V 4 V2

V1 2( I 1

20 4 I 1

10 V

I 2 ) 4.933 V

Also, I2

V3 12 8

V3

12 8I 2

12.267 V.

Chapter 3, Solution 48 We apply mesh analysis and let the mesh currents be in mA. 3k I4 4k

2k

5k Io

1k + 6V -

I1

I2 + 4V -

I3

10k

For mesh 1, 6 8 5I1 I 2 4I 4 0 2 5I1 I 2 4I 4 For mesh 2, 4 13I 2 I1 10I 3 2I 4 0 4 I1 13I 2 10I 3 For mesh 3, 3 15I 3 10I 2 5I 4 0 3 10I 2 15I 3 5I 4 For mesh 4, 4 I 1 2 I 2 5 I 3 14 I 4 0 Putting (1) to (4) in matrix form gives 5 1 0 4 I1 2 1 13 10 2 I2 4 AI B 0 10 15 5 I3 3 4 2 5 14 I 4 0 Using MATLAB,

3.608 4.044 I A 1B 0.148 3.896 3 The current through the 10k

3V +

(1) 2I 4

resistor is I o = I 2 – I 3 = 148 mA.

(2)

(3) (4)

Chapter 3, Solution 49 3

i3 2

1

i1

2

27 V

i2

+

2i 0 i1

i2

0 (a)

2

1

2

i1

+

+

v0 or

v0

i2

27V +

(b)

For the supermesh in figure (a), 3i 1 + 2i 2 – 3i 3 + 27 = 0

(1)

At node 0,

i 2 – i 1 = 2i 0 and i 0 = –i 1 which leads to i 2 = –i 1

(2)

For loop 3,

–i 1 –2i 2 + 6i 3 = 0 which leads to 6i 3 = –i 1

(3)

Solving (1) to (3), i 1 = (–54/3)A, i 2 = (54/3)A, i 3 = (27/9)A i 0 = –i 1 = 18 A, from fig. (b), v 0 = i 3 –3i 1 = (27/9) + 54 = 57 V.

Chapter 3, Solution 50

i1

4

2

i3

10 8 35 V

+

i2 3i 0 i2

For loop 1,

i3

16i 1 – 10i 2 – 2i 3 = 0 which leads to 8i 1 – 5i 2 – i 3 = 0

(1)

For the supermesh, –35 + 10i 2 – 10i 1 + 10i 3 – 2i 1 = 0 or

–6i 1 + 5i 2 + 5i 3 = 17.5

Also, 3i 0 = i 3 – i 2 and i 0 = i 1 which leads to 3i 1 = i 3 – i 2 Solving (1), (2), and (3), we obtain i 1 = 1.0098 and i 0 = i 1 = 1.0098 A

(2) (3)

Chapter 3, Solution 51 5A

i1 8 2

i3

1

+

i2 40 V

4

+

v0

20V

+

For loop 1,

i 1 = 5A

(1)

For loop 2,

-40 + 7i 2 – 2i 1 – 4i 3 = 0 which leads to 50 = 7i 2 – 4i 3

(2)

For loop 3,

-20 + 12i 3 – 4i 2 = 0 which leads to 5 = - i 2 + 3 i 3

(3)

Solving with (2) and (3), And,

i 2 = 10 A, i 3 = 5 A

v 0 = 4(i 2 – i 3 ) = 4(10 – 5) = 20 V.

Chapter 3, Solution 52

+ v0 2

i2

VS

+

8

3A

i2

i1

i3 4

i3

+ –

2V 0

For mesh 1, 2(i 1 – i 2 ) + 4(i 1 – i 3 ) – 12 = 0 which leads to 3i 1 – i 2 – 2i 3 = 6

(1)

For the supermesh, 2(i 2 – i 1 ) + 8i 2 + 2v 0 + 4(i 3 – i 1 ) = 0 But v 0 = 2(i 1 – i 2 ) which leads to -i 1 + 3i 2 + 2i 3 = 0 (2) For the independent current source, i3 = 3 + i 2 Solving (1), (2), and (3), we obtain, i 1 = 3.5 A, i 2 = -0.5 A, i 3 = 2.5 A.

(3)

Chapter 3, Solution 53 Applying mesh analysis leads to; –12 + 4kI 1 – 3kI 2 – 1kI 3 = 0 (1) –3kI 1 + 7kI 2 – 4kI 4 = 0 –3kI 1 + 7kI 2 = –12 (2) –1kI 1 + 15kI 3 – 8kI 4 – 6kI 5 = 0 –1kI 1 + 15kI 3 – 6k = –24 (3) I 4 = –3mA (4) –6kI 3 – 8kI 4 + 16kI 5 = 0 –6kI 3 + 16kI 5 = –24 (5) Putting these in matrix form (having substituted I 4 = 3mA in the above), 4

3

1

3

7

0

1

0

15

6

0

6

16

0

I1

0 0

k

I2

12 12

I3

24

I5

24

ZI = V Using MATLAB, >> Z = [4,-3,-1,0;-3,7,0,0;-1,0,15,-6;0,0,-6,16] Z= 4 -3 -1 0 -3 7 0 0 -1 0 15 -6 0 0 -6 16 >> V = [12,-12,-24,-24]' V= 12 -12 -24 -24 We obtain, >> I = inv(Z)*V

I= 1.6196 mA –1.0202 mA –2.461 mA 3 mA –2.423 mA

Chapter 3, Solution 54 Let the mesh currents be in mA. For mesh 1, 12 10 2 I 1 I 2 0 2 2I1 I 2 For mesh 2, 10 3I 2 I 1 I 3 0 10 I1 For mesh 3, 12 2 I 3 I 2 0 12 I 2 2I 3 Putting (1) to (3) in matrix form leads to

2

I1

2

1 I2

10

0 1 2 I3 Using MATLAB,

12

1

1 3

0

AI

(1) 3I 2

I3

(2)

(3)

B

5.25 I

1

A B

8.5

I1

5.25 mA, I 2

8.5 mA, I 3

10.25 mA

10.25

I 1 = 5.25 mA, I 2 = 8.5 mA, and I 3 = 10.25 mA.

Chapter 3, Solution 55

10 V

I2

b

i1 4A

c

+

1A

I2 6

1A I1

i2

I3

12 a

2

i3

I4 4A

I3

d

4

I4

+–

0

8V

It is evident that I 1 = 4 For mesh 4,

12(I 4 – I 1 ) + 4(I 4 – I 3 ) – 8 = 0

For the supermesh At node c,

(1)

6(I 2 – I 1 ) + 10 + 2I 3 + 4(I 3 – I 4 ) = 0 or -3I 1 + 3I 2 + 3I 3 – 2I 4 = -5

I2 = I3 + 1

(2) (3) (4)

Solving (1), (2), (3), and (4) yields, I 1 = 4A, I 2 = 3A, I 3 = 2A, and I 4 = 4A At node b,

i 1 = I 2 – I 1 = -1A

At node a,

i 2 = 4 – I 4 = 0A

At node 0,

i 3 = I 4 – I 3 = 2A

Chapter 3, Solution 56 + v1 – 2

i2

2

12 V

2

2

+

i1

+

2

v2

i3

For loop 1, 12 = 4i 1 – 2i 2 – 2i 3 which leads to 6 = 2i 1 – i 2 – i 3

(1)

For loop 2, 0 = 6i 2 –2i 1 – 2 i 3 which leads to 0 = -i 1 + 3i 2 – i 3

(2)

For loop 3, 0 = 6i 3 – 2i 1 – 2i 2 which leads to 0 = -i 1 – i 2 + 3i 3

(3)

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

1

1

3

1 2 =

1 1

2 3

=

1 1

1 3 1

1 i1 1 i2

1

3

3

0 0

i3

1 1

6

2 8,

2

=

6

1

1 3

1

1 0

24

3

1 6 3 0 1 0

24 , therefore i 2 = i 3 = 24/8 = 3A,

v 1 = 2i 2 = 6 volts, v = 2i 3 = 6 volts

Assume R is in kilo-ohms. V2 4k x15mA 60 V, Current through R is 3 iR i o, V1 i R R 3 R This leads to R = 90/15 =

V1

90 V2 30

.

90 60 30 V 3 3 R

(15)R

Chapter 3, Solution 58 30

i2 30

10

i1

10

30

i3

+

120 V

For loop 1, 120 + 40i 1 – 10i 2 = 0, which leads to -12 = 4i 1 – i 2

(1)

For loop 2, 50i 2 – 10i 1 – 10i 3 = 0, which leads to -i 1 + 5i 2 – i 3 = 0 For loop 3, -120 – 10i 2 + 40i 3 = 0, which leads to 12 = -i 2 + 4i 3 Solving (1), (2), and (3), we get, i 1 = -3A, i 2 = 0, and i 3 = 3A

(2) (3)

Chapter 3, Solution 59 40

I0

–+ 96 V

i2

10 20 80V

i1

+

4v 0

+

i3

+ –

80

v0

2I 0 i2

i3

For loop 1, -80 + 30i 1 – 20i 2 + 4v 0 = 0, where v 0 = 80i 3 or 4 = 1.5i 1 – i 2 + 16i 3

(1)

For the supermesh, 60i 2 – 20i 1 – 96 + 80i 3 – 4 v 0 = 0, where v 0 = 80i 3 or 4.8 = -i 1 + 3i 2 – 12i 3

(2)

Also, 2I 0 = i 3 – i 2 and I 0 = i 2 , hence, 3i 2 = i 3 (3)

From (1), (2), and (3),

3 =

1 0

2

32

3

12

3

1

3 5,

I0 = i2 =

2

=

2 3

32 12

i1 i2

8 4.8

0

3

1

i3

0

8

1 4.8 0

2/

3 1

0

32 12

3 22.4,

1

= -28/5 = –4.48 A

v 0 = 8i 3 = (-84/5)80 = –1.0752 kvolts

3

=

1 0

2

8

3

4.8

3

0

67.2

Chapter 3, Solution 60 0.5i 0

56 V

4

v1 1

56 V

8

+

v2 2

i0

At node 1, [(v 1 –0)/1] + [(v 1 –56)/4] + 0.5[(v 1 –0)/1] = 0 or 1.75v 1 = 14 or v 1 = 8 V At node 2, [(v 2 –56)/8] – 0.5[8/1] + [(v 2 –0)/2] = 0 or 0.625v 2 = 11 or v 2 = 17.6 V P 1 = (v 1 )2/1 = 64 watts, P 2 = (v 2 )2/2 = 154.88 watts, P 4 = (56 – v 1 )2/4 = 576 watts, P 8 = (56 – v 2 )2/8 = 1.84.32 watts.

Chapter 3, Solution 61 20

v1 is

v2

10 i0

+ v0

30

– + 5v 0

At node 1, i s = (v 1 /30) + ((v 1 – v 2 )/20) which leads to 60i s = 5v 1 – 3v 2 But v 2 = -5v 0 and v 0 = v 1 which leads to v 2 = -5v 1 Hence, 60i s = 5v 1 + 15v 1 = 20v 1 which leads to v 1 = 3i s , v 2 = -15i s i 0 = v 2 /50 = -15i s /50 which leads to i 0 /i s = -15/50 = –0.3

40

(1)

Chapter 3, Solution 62 4k

100V +

A

i1

8k

i2

B

2k

i3

+

40 V

We have a supermesh. Let all R be in k , i in mA, and v in volts. For the supermesh, -100 +4i 1 + 8i 2 + 2i 3 + 40 = 0 or 30 = 2i 1 + 4i 2 + i 3 (1) At node A,

i1 + 4 = i2

(2)

At node B,

i 2 = 2i 1 + i 3

(3)

Solving (1), (2), and (3), we get i 1 = 2 mA, i 2 = 6 mA, and i 3 = 2 mA.

Chapter 3, Solution 63 10

A

5 50 V

+

i1

i2 + –

4i x

For the supermesh, -50 + 10i 1 + 5i 2 + 4i x = 0, but i x = i 1 . Hence, 50 = 14i 1 + 5i 2 At node A, i 1 + 3 + (v x /4) = i 2 , but v x = 2(i 1 – i 2 ), hence, i 1 + 2 = i 2 Solving (1) and (2) gives i 1 = 2.105 A and i 2 = 4.105 A v x = 2(i 1 – i 2 ) = –4 volts and i x = i 2 – 2 = 2.105 amp

(1) (2)

Chapter 3, Solution 64 i1

50

A

i0 i1

i 2 10 + v0

10

i2

+ –

4i 0

i3

40

250V +

5A

0.2V 0

i1 For mesh 2,

B

i3

20i 2 – 10i 1 + 4i 0 = 0

(1)

But at node A, i o = i 1 – i 2 so that (1) becomes i 1 = (16/6)i 2 (2) For the supermesh, –250 + 50i 1 + 10(i 1 – i 2 ) – 4i 0 + 40i 3 = 0 or (3) At node B, But,

28i 1 – 3i 2 + 20i 3 = 125 i 3 + 0.2v 0 = 2 + i 1 v 0 = 10i 2 so that (4) becomes i 3 = 5 + (2/3)i 2

Solving (1) to (5), i 2 = 0.2941 A, v 0 = 10i 2 = 2.941 volts, i 0 = i 1 – i 2 = (5/3)i 2 = 490.2mA.

(4) (5)

Chapter 3, Solution 65 For mesh 1, –12 + 12I 1 – 6I 2 – I 4 = 0 or 12 12 I 1 6 I 2 I 4

(1)

–6I 1 + 16I 2 – 8I 3 – I 4 – I 5 = 0

(2)

–8I 2 + 15I 3 – I 5 – 9 = 0 or 9 = –8I 2 + 15I 3 – I 5

(3)

For mesh 2, For mesh 3, For mesh 4, For mesh 5,

–I 1 – I 2 + 7I 4 – 2I 5 – 6 = 0 or 6 = –I 1 – I 2 + 7I 4 – 2I 5

–I 2 – I 3 – 2I 4 + 8I 5 – 10 = 0 or 10 I 2 I 3 2I 4 8I 5 Casting (1) to (5) in matrix form gives 12 6 0 1 0 I1 12 6 16 8 1 1 I2 0 0 8 15 0 1 I3 9 1 1 0 7 2 I4 6 0

1

1

2

8

I5

(4) (5)

AI

B

10

Using MATLAB we input: Z=[12,-6,0,-1,0;-6,16,-8,-1,-1;0,-8,15,0,-1;-1,-1,0,7,-2;0,-1,-1,-2,8] and V=[12;0;9;6;10] This leads to >> Z=[12,-6,0,-1,0;-6,16,-8,-1,-1;0,-8,15,0,-1;-1,-1,0,7,-2;0,-1,-1,-2,8] Z= 12 -6 0 -1 0

-6 0 -1 0 16 -8 -1 -1 -8 15 0 -1 -1 0 7 -2 -1 -1 -2 8

>> V=[12;0;9;6;10] V= 12

0 9 6 10 >> I=inv(Z)*V I= 2.1701 1.9912 1.8119 2.0942 2.2489 Thus, I = [2.17, 1.9912, 1.8119, 2.094, 2.249] A.

Chapter 3, Solution 66 The mesh equations are obtained as follows. 12 24 30I1 4I2

6I3

2I4

0

or 30I 1 – 4I 2 – 6I 3 – 2I 4 = –12 24 40 4I1 30I2 2I4 6I5 0 or –4I 1 + 30I 2 – 2I 4 – 6I 5 = –16

(2)

–6I 1 + 18I 3 – 4I 4 = 30

(3)

–2I 1 – 2I 2 – 4I 3 + 12I 4 –4I 5 = 0

(4)

–6I 2 – 4I 4 + 18I 5 = –32

(5)

Putting (1) to (5) in matrix form 30

4

4

30

0

6 2

0 2

18 4

6

0

0

(1)

6

2 2 4 12 4

0 6 0 I 4 18

ZI = V Using MATLAB, >> Z = [30,-4,-6,-2,0; -4,30,0,-2,-6; -6,0,18,-4,0; -2,-2,-4,12,-4; 0,-6,0,-4,18] Z= 30 -4 -6 -2 0 -4 30 0 -2 -6 -6 0 18 -4 0 -2 -2 -4 12 -4

12 16 30 0 32

0

-6

0

-4

18

>> V = [-12,-16,30,0,-32]' V= -12 -16 30 0 -32 >> I = inv(Z)*V I= -0.2779 A -1.0488 A 1.4682 A -0.4761 A -2.2332 A

Chapter 3, Solution 67 Consider the circuit below.

5A

V1

4

V2

2

V3

+ Vo 3 Vo

5

10

0.35 0.25 0

0.25 0.95 0.5

0 0.5 V 0.5

10 A

5 3Vo 0 15

Since we actually have four unknowns and only three equations, we need a constraint equation. Vo = V2 – V3 Substituting this back into the matrix equation, the first equation becomes, 0.35V 1 – 3.25V 2 + 3V 3 = –5 This now results in the following matrix equation,

0.35 0.25 0

3.25 0.95 0.5

3 0.5 V 0.5

5 0 15

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=[–5,0,15]' I= –5 0 15 >> V=inv(Y)*I V= –410.5262 –194.7368 –164.7368 V o = V 2 – V 3 = –77.89 + 65.89 = –30 V. Let us now do a quick check at node 1. –3(–30) + 0.1(–410.5) + 0.25(–410.5+194.74) + 5 = 90–41.05–102.62+48.68+5 = 0.01; essentially zero considering the accuracy we are using. The answer checks.

Chapter 3, Solution 68 Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find the voltage V o in the circuit of Fig. 3.112. 3A

10

25 +

4A

Vo

40

20

+ _

24 V

+ _

24 V

_

Figure 3.112 For Prob. 3.68. Solution Consider the circuit below. There are two non-reference nodes. 3A

V1

10

Vo

25

+ 4A

40

Vo _

20

0.125

0.1

0.1

0.19

V

4 3 3 24 / 25

7 2.04

Using MATLAB, we get, >> Y=[0.125,-0.1;-0.1,0.19] Y= 0.1250 -0.1000 -0.1000 0.1900 >> I=[7,-2.04]' I= 7.0000 -2.0400 >> V=inv(Y)*I V= 81.8909 32.3636 Thus, V o = 32.36 V. We can perform a simple check at node V o , 3 + 0.1(32.36–81.89) + 0.05(32.36) + 0.04(32.36–24) = 3 – 4.953 + 1.618 + 0.3344 = – 0.0004; answer checks!

Chapter 3, Solution 69 Assume that all conductances are in mS, all currents are in mA, and all voltages are in volts. G 11 = (1/2) + (1/4) + (1/1) = 1.75, G 22 = (1/4) + (1/4) + (1/2) = 1, G 33 = (1/1) + (1/4) = 1.25, G 12 = -1/4 = -0.25, G 13 = -1/1 = -1, G 21 = -0.25, G 23 = -1/4 = -0.25, G 31 = -1, G 32 = -0.25 i 1 = 20, i 2 = 5, and i 3 = 10 – 5 = 5 The node-voltage equations are: 1.75 0.25 1

0.25 1 0.25

v1

20

0.25 v 2 1.25 v 3

5 5

1

Chapter 3, Solution 70

3 0 0 5

4I x

V

20

4I x

7

With two equations and three unknowns, we need a constraint equation, I x = 2V 1 , thus the matrix equation becomes, 5 0 8

5

V

20 7

This results in V 1 = 20/(–5) = –4 V and V 2 = [–8(–4) – 7]/5 = [32 – 7]/5 = 5 V.

Chapter 3, Solution 71

9 4 5

4

5

7 1

1 I 9

30 15 0

We can now use MATLAB solve for our currents. >> R=[9,–4,–5;–4,7,–1;–5,–1,9] R= 9 –4 –5 –4 7 –1 –5 –1 9 >> V=[30,–15,0]' V= 30 –15 0 >> I=inv(R)*V I= 6.255 A 1.9599 A 3.694 A

Chapter 3, Solution 72 R 11 = 5 + 2 = 7, R 22 = 2 + 4 = 6, R 33 = 1 + 4 = 5, R 44 = 1 + 4 = 5, R 12 = -2, R 13 = 0 = R 14 , R 21 = -2, R 23 = -4, R 24 = 0, R 31 = 0, R 32 = -4, R 34 = -1, R 41 = 0 = R 42 , R 43 = -1, we note that R ij = R ji for all i not equal to j. v 1 = 8, v 2 = 4, v 3 = -10, and v 4 = -4 Hence the mesh-current equations are:

7 2

2 6

0 4

0

4

5

0

0

1

0 0

i1

8

i2

4

1 i3 5

i4

10 4

Chapter 3, Solution 73 R 11 = 2 + 3 +4 = 9, R 22 = 3 + 5 = 8, R 33 = 1+1 + 4 = 6, R 44 = 1 + 1 = 2, R 12 = -3, R 13 = -4, R 14 = 0, R 23 = 0, R 24 = 0, R 34 = -1 v 1 = 6, v 2 = 4, v 3 = 2, and v 4 = -3 Hence, 9

3

4

0

i1

6

3

8

0

0

i2

4

4

0

6

1 i3 2 i4

2

0

0

1

3

Chapter 3, Solution 74 R 11 = R 1 + R 4 + R 6 , R 22 = R 2 + R 4 + R 5 , R 33 = R 6 + R 7 + R 8 , R 44 = R 3 + R 5 + R 8 , R 12 = -R 4 , R 13 = -R 6 , R 14 = 0, R 23 = 0, R 24 = -R 5 , R 34 = -R 8 , again, we note that R ij = R ji for all i not equal to j.

V1 The input voltage vector is =

V2 V3 V4

R1

R4 R4

R6

R4 R2

R4

R6

0

0

R5

R6 R5 R6

0 R7 R8

R8 R3

i1

0 R5

i2

R8

i3

R5

R8

i4

V1 V2 V3 V4

Chapter 3, Solution 75

i3

i1

i2

Clearly, i 1 = –3 amps, i 2 = 0 amps, and i 3 = 3 amps, which agrees with the answers in Problem 3.44.

Chapter 3, Solution 76

Clearly, v 1 = 625 mVolts, v 2 = 375 mVolts, and v 3 = 1.625 volts, which agrees with the solution obtained in Problem 3.27.

1 Chapter 3, Solution 77

As a check we can write the nodal equations, 1.7 1.2

0.2 1.2

V

5 2

Solving this leads to V 1 = 3.111 V and V 2 = 1.4444 V. The answer checks!

Chapter 3, Solution 78 The schematic is shown below. When the circuit is saved and simulated the node voltages are displayed on the pseudocomponents as shown. Thus, V1

.

3V, V2

4.5V, V3

15V,

Chapter 3, Solution 79 The schematic is shown below. When the circuit is saved and simulated, we obtain the node voltages as displayed. Thus, V a = –10.556 volts; V b = 20.56 volts; V c = 1.3889 volts; and V d = –43.75 volts. 1.3889 V R3 10

–43.75 V R6 4

R4 5

R5 4

R1

R2

20

8

20.56 V

R7 16

R8 8 V1

V2 60Vdc

90Vdc

–10.556 V

Chapter 3, Solution 80

Clearly, v 1 = 84 volts, v 2 = 4 volts, v 3 = 20 volts, and v 4 = -5.333 volts

Chapter 3, Solution 81

Clearly, v 1 = 26.67 volts, v 2 = 6.667 volts, v 3 = 173.33 volts, and v 4 = –46.67 volts which agrees with the results of Example 3.4. This is the netlist for this circuit.

Chapter 3, Solution 82

2i 0

+ v0 – 3k

1

2

2k

+

3v 0

3

6k

4

4A 4k

8k

0

This network corresponds to the Netlist.

100V +

Chapter 3, Solution 83 The circuit is shown below. 1

20 V

+

20

70

2

2A

50

3

30

0

When the circuit is saved and simulated, we obtain v 2 = –12.5 volts

Chapter 3, Solution 84 From the output loop, v 0 = 50i 0 x20x103 = 106i 0

(1)

From the input loop, 15x10-3 + 4000i 0 – v 0 /100 = 0

(2)

From (1) and (2) we get, i 0 = 2.5 A and v 0 = 2.5 volt.

Chapter 3, Solution 85 The amplifier acts as a source. Rs + Vs -

RL

For maximum power transfer, RL

Rs

9

Chapter 3, Solution 86 Let v 1 be the potential across the 2 k-ohm resistor with plus being on top. Then, Since i = [(0.047–v 1 )/1k] [(v 1 –0.047)/1k] – 400[(0.047–v 1 )/1k] + [(v 1 –0)/2k] = 0 or 401[(v 1 –0.047)] + 0.5v 1 = 0 or 401.5v 1 = 401x0.047 or v 1 = 0.04694 volts and i = (0.047–0.04694)/1k = 60 nA Thus,

v 0 = –5000x400x60x10-9 = –120 mV.

Chapter 3, Solution 87 v 1 = 500(v s )/(500 + 2000) = v s /5 v 0 = -400(60v 1 )/(400 + 2000) = -40v 1 = -40(v s /5) = -8v s , Therefore, v 0 /v s = –8

Chapter 3, Solution 88 Let v 1 be the potential at the top end of the 100-ohm resistor. (v s – v 1 )/200 = v 1 /100 + (v 1 – 10-3v 0 )/2000

(1)

For the right loop, v 0 = -40i 0 (10,000) = -40(v 1 – 10-3)10,000/2000, or, v 0 = -200v 1 + 0.2v 0 = -4x10-3v 0

(2)

Substituting (2) into (1) gives, (v s + 0.004v 1 )/2 = -0.004v 0 + (-0.004v 1 – 0.001v 0 )/20 This leads to 0.125v 0 = 10v s or (v 0 /v s ) = 10/0.125 = –80

Chapter 3, Solution 89

22.5 µA

12.75 V

Chapter 3, Solution 90 1k

IB

10 k

vs

+

-

i1

i2

+ V CE + V BE – –

500

IE

+

18V

+

-

V0

For loop 1, -v s + 10k(I B ) + V BE + I E (500) = 0 = -v s + 0.7 + 10,000I B + 500(1 + )I B which leads to v s + 0.7 = 10,000I B + 500(151)I B = 85,500I B But, v 0 = 500I E = 500x151I B = 4 which leads to I B = 5.298x10-5 Therefore, v s = 0.7 + 85,500I B = 5.23 volts

Chapter 3, Solution 91 We first determine the Thevenin equivalent for the input circuit. R Th = 6||2 = 6x2/8 = 1.5 k

and V Th = 2(3)/(2+6) = 0.75 volts

5k

IC

IB

1.5 k

0.75 V

+

-

i1

i2

+ V CE + V BE – –

400

IE

+

9V

+

-

V0

For loop 1, -0.75 + 1.5kI B + V BE + 400I E = 0 = -0.75 + 0.7 + 1500I B + 400(1 + )I B I B = 0.05/81,900 = 0.61 A v 0 = 400I E = 400(1 + )I B = 49 mV For loop 2, -400I E – V CE – 5kI C + 9 = 0, but, I C = I B and I E = (1 + )I B V CE = 9 – 5k I B – 400(1 + )I B = 9 – 0.659 = 8.641 volts

Chapter 3, Solution 92 Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find I B and V C for the circuit in Fig. 3.128. Let

= 100, V BE = 0.7V.

Figure 3.128 Solution I1

5k

10 k

VC IC

IB

+ V CE + V BE – – 4k

IE

+ V0

I 1 = I B + I C = (1 + )I B and I E = I B + I C = I 1

12V

+

-

Applying KVL around the outer loop, 4kI E + V BE + 10kI B + 5kI 1 = 12 12 – 0.7 = 5k(1 + )I B + 10kI B + 4k(1 + )I B = 919kI B I B = 11.3/919k = 12.296 A Also, 12 = 5kI 1 + V C which leads to V C = 12 – 5k(101)I B = 5.791 volts

Chapter 3, Solution 93 1

v1

4

i1 24V

+

3v 0

i 2

+

v2 i3

i

i2

2

8

3v 0

2

4

+

+

+

+

v0

v1

v2

(a)

(b)

From (b), -v 1 + 2i – 3v 0 + v 2 = 0 which leads to i = (v 1 + 3v 0 – v 2 )/2 At node 1 in (a), ((24 – v 1 )/4) = (v 1 /2) + ((v 1 +3v 0 – v 2 )/2) + ((v 1 – v 2 )/1), where v 0 = v 2 or 24 = 9v 1 which leads to v 1 = 2.667 volts At node 2, ((v 1 – v 2 )/1) + ((v 1 + 3v 0 – v 2 )/2) = (v 2 /8) + v 2 /4, v 0 = v 2 v 2 = 4v 1 = 10.66 volts Now we can solve for the currents, i 1 = v 1 /2 = 1.333 A, i 2 = 1.333 A, and i 3 = 2.6667 A. ## Related documents

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