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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 10volt source is –10 V. At the top of the 20volt 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 10ohm resistor is 0.2 A. The current flowing right to left through the 10ohm 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 voltagecontrolled 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 = A1 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 nonplanar 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 kiloohms. 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 nonreference 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 nodevoltage 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 meshcurrent 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, 15x103 + 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 kohm 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 = –5000x400x60x109 = –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 100ohm resistor. (v s – v 1 )/200 = v 1 /100 + (v 1 – 103v 0 )/2000
(1)
For the right loop, v 0 = 40i 0 (10,000) = 40(v 1 – 103)10,000/2000, or, v 0 = 200v 1 + 0.2v 0 = 4x103v 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 = 5.298x105 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 = 62 = 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.