Chapter 3, Solution 1.
v1 40 Ω v2
6A 8Ω
10 A
2Ω
At node 1,
6 = v1/(8) + (v1 - v2)/4 48 = 3v1 - 2v2 (1)
At node 2,
v1 - v2/4 = v2/2 + 10 40 = v1 - 3v2 (2)
Solving (1) and (2),
v1 = 9.143V, v2 = -10.286 V
v1 (9.143)2
2
P8Ω = = = 10.45 W
8 8
(v 1 − v 2 )2
P4Ω = = 94.37 W
4
v1 (= 10.286)2
P2Ω = 2 = = 52.9 W
2 2
Chapter 3, Solution 2
At node 1,
− v1 v1 v − v2
− = 6+ 1 60 = - 8v1 + 5v2 (1)
10 5 2
At node 2,
v2 v − v2
= 3+ 6+ 1 36 = - 2v1 + 3v2 (2)
4 2
Solving (1) and (2),
v1 = 0 V, v2 = 12 V
Chapter 3, Solution 3
Applying KCL to the upper node,
v0 vo vo v
10 = + + +2+ 0 v0 = 40 V
10 20 30 60
v0 v v v
i1 = = 4 A , i2 = 0 = 2 A, i3 = 0 = 1.33 A, i4 = 0 = 67 mA
10 20 30 60
Chapter 3, Solution 4
v1 2A v2
i1 i2 i3 i4
5Ω 10 Ω 10 Ω 5Ω 5A
4A
At node 1,
4 + 2 = v1/(5) + v1/(10) v1 = 20
At node 2,
5 - 2 = v2/(10) + v2/(5) v2 = 10
i1 = v1/(5) = 4 A, i2 = v1/(10) = 2 A, i3 = v2/(10) = 1 A, i4 = v2/(5) = 2 A
Chapter 3, Solution 5
Apply KCL to the top node.
30 − v 0 20 − v 0 v 0
+ = v0 = 20 V
2k 6k 4k
Chapter 3, Solution 6
v 2 − 12 v 0 v 0 − 10
i1 + i2 + i3 = 0 + + =0
4 6 2
or v0 = 8.727 V
Chapter 3, Solution 7
At node a,
10 − Va Va Va − Vb
= + → 10 = 6Va − 3Vb (1)
30 15 10
At node b,
Va − Vb 12 − Vb − 9 − Vb
+ + =0 → 24 = 2Va − 7Vb (2)
10 20 5
Solving (1) and (2) leads to
Va = -0.556 V, Vb = -3.444V
Chapter 3, Solution 8
3Ω i1 v1 i3 5Ω
i2
+ 3V +
V0 2Ω –
+ 4V0
– –
1Ω
v1 v1 − 3 v1 − 4 v 0
i1 + i2 + i3 = 0 + + =0
5 1 5
2 8
But v 0 = v1 so that v1 + 5v1 - 15 + v1 - v1 = 0
5 5
or v1 = 15x5/(27) = 2.778 V, therefore vo = 2v1/5 = 1.1111 V
Chapter 3, Solution 9
3Ω i1 v1 i3 6Ω
+ v0 –
i2
+
12V +
v1 8Ω +
– 2v0
– –
At the non-reference node,
12 − v1 v1 v1 − 2 v 0
= + (1)
3 8 6
But
-12 + v0 + v1 = 0 v0 = 12 - v1 (2)
Substituting (2) into (1),
12 − v1 v1 3v1 − 24
= + v0 = 3.652 V
3 8 6
Chapter 3, Solution 10
At node 1,
v 2 − v1 v
= 4+ 1 32 = -v1 + 8v2 - 8v0 (1)
1 8
1Ω
2i0
4A
v0
v1 v2
8Ω 2Ω 4Ω
i0
At node 0,
v0 v
4= + 2I 0 and I 0 = 1 16 = 2v0 + v1 (2)
2 8
At node 2,
v 2 − v1 v 2 v
2I0 = + and I 0 = 1 v2 = v1 (3)
1 4 8
From (1), (2) and (3), v0 = 24 V, but from (2) we get
v
4− o
io = 2 = 2 − 24 = 2 − 6 = - 4 A
2 4
Chapter 3, Solution 11
4Ω i1 v i2 3Ω
i3
10 V +
– 5A
6Ω
Note that i2 = -5A. At the non-reference node
10 − v v
+5= v = 18
4 6
10 − v
i1 = = -2 A, i2 = -5 A
4
Chapter 3, Solution 12
10 Ω v1 20 Ω v2 50 Ω
i3
24 V +
– 5A
40 Ω
24 − v 1 v − v 2 v1 − 0
At node 1, = 1 + 96 = 7v1 - 2v2 (1)
10 20 40
v1 − v 2 v 2
At node 2, 5 + = 500 = -5v1 + 7v2 (2)
20 50
Solving (1) and (2) gives,
v1 = 42.87 V, v2 = 102.05 V
v v
i1 = 1 = 1.072 A, v2 = 2 = 2.041 A
40 50
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
Chapter 3, Solution 14
5A
8Ω
v0
v1
1Ω 2Ω
20 V –
4Ω +
40 V +
–
v1 − v 0 40 − v 0
At node 1, +5= v1 + v0 = 70 (1)
2 1
v1 − v 0 v v + 20
At node 0, +5= 0 + 0 4v1 - 7v0 = -20 (2)
2 4 8
Solving (1) and (2), v0 = 20 V
Chapter 3, Solution 15
5A
8Ω
v0
v1
1Ω 2Ω
20 V –
4Ω +
40 V +
–
Nodes 1 and 2 form a supernode so that v1 = v2 + 10 (1)
At the supernode, 2 + 6v1 + 5v2 = 3 (v3 - v2) 2 + 6v1 + 8v2 = 3v3 (2)
At node 3, 2 + 4 = 3 (v3 - v2) v3 = v2 + 2 (3)
Substituting (1) and (3) into (2),
− 56
2 + 6v2 + 60 + 8v2 = 3v2 + 6 v2 =
11
54
v1 = v2 + 10 =
11
i0 = 6vi = 29.45 A
2 2
v1 54
P65 = = v1 G = 6 = 144.6 W
2
R 11
2
− 56
P55 = v G =
2
2 5 = 129.6 W
11
P35 = (v L − v 3 ) G = (2) 2 3 = 12 W
2
Chapter 3, Solution 16
2S
v1 v2 8S
v3
i0
+ 13 V +
2A 1S v0 4S
–
–
At the supernode,
2 = v1 + 2 (v1 - v3) + 8(v2 – v3) + 4v2, which leads to 2 = 3v1 + 12v2 - 10v3 (1)
But
v1 = v2 + 2v0 and v0 = v2.
Hence
v1 = 3v2 (2)
v3 = 13V (3)
Substituting (2) and (3) with (1) gives,
v1 = 18.858 V, v2 = 6.286 V, v3 = 13 V
Chapter 3, Solution 17
i0
4Ω 2Ω
10 Ω
60 V 8Ω
60 V + 3i0
–
60 − v1 v1 v1 − v 2
At node 1, = + 120 = 7v1 - 4v2 (1)
4 8 2
60 − v 2 v1 − v 2
At node 2, 3i0 + + =0
10 2
60 − v1
But i0 = .
4
Hence
3(60 − v1 ) 60 − v 2 v1 − v 2
+ + =0 1020 = 5v1 - 12v2 (2)
4 10 2
60 − v1
Solving (1) and (2) gives v1 = 53.08 V. Hence i0 = = 1.73 A
4
Chapter 3, Solution 18
–+
v2
v1 v3
10 V
2Ω 2Ω + +
5A
v1 v3
4Ω 8Ω – –
(a) (b)
v 2 − v1 v 2 − v3
At node 2, in Fig. (a), 5 = + 10 = - v1 + 2v2 - v3 (1)
2 2
v 2 − v1 v 2 − v 3 v1 v 3
At the supernode, + = + 40 = 2v1 + v3 (2)
2 2 4 8
From Fig. (b), - v1 - 10 + v3 = 0 v3 = v1 + 10 (3)
Solving (1) to (3), we obtain v1 = 10 V, v2 = 20 V = v3
Chapter 3, Solution 19
At node 1,
V1 − V3 V1 − V2 V1
5 = 3+ + +
→ 16 = 7V1 − V2 − 4V3 (1)
2 8 4
At node 2,
V1 − V2 V2 V2 − V3
= +
→ 0 = −V1 + 7V2 − 2V3 (2)
8 2 4
At node 3,
12 − V3 V1 − V3 V2 − V3
3+ + + =0
→ − 36 = 4V1 + 2V2 − 7V3 (3)
8 2 4
From (1) to (3),
7 − 1 − 4 V1 16
− 1 7 − 2 V2 = 0
→ AV = B
4 2 − 7 V3 − 36
Using MATLAB,
10
V = A −1 B = 4.933
→ V1 = 10 V, V2 = 4.933 V, V3 = 12.267 V
12.267
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 2Ω V3
4Ω 1Ω 4Ω
Between nodes 1 and 3,
− V1 + 12 + V3 = 0
→ V3 = V1 − 12 (2)
Similarly, between nodes 1 and 2,
V1 = V2 + 2i (3)
But i = V3 / 4 . Combining this with (2) and (3) gives
. V2 = 6 + V1 / 2 (4)
Solving (1), (2), and (4) leads to
V1 = −3V, V2 = 4.5V, V3 = −15V
Chapter 3, Solution 21
4 kΩ
2 kΩ 3v0
v3 3v0
v1 v2
+ +
+ + +
v0 v3 v2
3 mA 1 kΩ – –
–
(b)
(a)
Let v3 be the voltage between the 2kΩ resistor and the voltage-controlled voltage source.
At node 1,
v − v 2 v1 − v 3
3x10 −3 = 1 + 12 = 3v1 - v2 - 2v3 (1)
4000 2000
At node 2,
v1 − v 2 v1 − v 3 v 2
+ = 3v1 - 5v2 - 2v3 = 0 (2)
4 2 1
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
Chapter 3, Solution 22
12 − v 0 v1 v − v0
At node 1, = +3+ 1 24 = 7v1 - v2 (1)
2 4 8
v1 − v 2 v 2 + 5v 2
At node 2, 3 + =
8 1
But, v1 = 12 - v1
Hence, 24 + v1 - v2 = 8 (v2 + 60 + 5v1) = 4 V
456 = 41v1 - 9v2 (2)
Solving (1) and (2),
v1 = - 10.91 V, v2 = - 100.36 V
Chapter 3, Solution 23
v1 v 2
At the supernode, 5 + 2 = + 70 = v1 + 2v2 (1)
10 5
Considering Fig. (b), - v1 - 8 + v2 = 0 v2 = v1 + 8 (2)
Solving (1) and (2),
v1 = 18 V, v2 = 26 V
v1 v2
–+
5A 2A 8V
+ +
10 Ω 5Ω
v1 v2
– –
(a) (b)
Chapter 3, Solution 24
6mA
1 kΩ 2 kΩ 3 kΩ
V1 V2
+ io -
30V 15V
- 4 kΩ 5 kΩ +
At node 1,
30 −V 1 V V − V2
=6+ 1 + 1 → 96 = 7V1 − 2V2 (1)
1 4 2
At node 2,
(−15 −V 2) V2 V2 − V1
6+ = + → 30 = −15V1 + 31V2 (2)
3 5 2
Solving (1) and (2) gives V1=16.24. Hence
io = V1/4 = 4.06 mA
Chapter 3, Solution 25
i0 v0
20V + 2Ω
10V +
– –
4Ω
40V +
1Ω – 2Ω
Using nodal analysis,
20 − v 0 40 − v 0 10 − v 0 v −0
+ + = 0 v0 = 20V
1 2 2 4
20 − v 0
i0 = = 0A
1
Chapter 3, Solution 26
At node 1,
15 − V1 V − V3 V1 − V2
= 3+ 1 + → − 45 = 7V1 − 4V2 − 2V3 (1)
20 10 5
At node 2,
V1 − V2 4 I o − V2 V2 − V3
+ = (2)
5 5 5
V − V3
But I o = 1 . Hence, (2) becomes
10
0 = 7V1 − 15V2 + 3V3 (3)
At node 3,
V − V3 − 10 − V3 V2 − V3
3+ 1 + + =0 → − 10 = V1 + 2V2 − 5V3 (4)
10 5 5
Putting (1), (3), and (4) in matrix form produces
7 − 4 − 2 V1 − 45
7 − 15 3 V2 = 0
→ AV = B
1 2 − 5 V3 − 10
Using MATLAB leads to
− 9.835
−1
V = A B = − 4.982
− 1.96
Thus,
V1 = −9.835 V, V2 = −4.982 V, V3 = −1.95 V
Chapter 3, Solution 27
At node 1,
2 = 2v1 + v1 – v2 + (v1 – v3)4 + 3i0, i0 = 4v2. Hence,
2 = 7v1 + 11v2 – 4v3 (1)
At node 2,
v1 – v2 = 4v2 + v2 – v3 0 = – v1 + 6v2 – v3 (2)
At node 3,
2v3 = 4 + v2 – v3 + 12v2 + 4(v1 – v3)
or – 4 = 4v1 + 13v2 – 7v3 (3)
In matrix form,
7 11 − 4 v 1 2
1 − 6 1 v = 0
2
4 13 − 7 v 3 − 4
7 11 −4 2 11 −4
∆ = 1 −6 1 = 176, ∆ 1 = 0 −6 1 = 110
4 13 −7 −4 13 −7
7 2 −4 7 11 2
∆2 = 1 0 1 = 66, ∆ 3 = 1 − 6 0 = 286
4 −4 −7 4 13 − 4
∆ 1 110 ∆ 66
v1 = = = 0.625V, v2 = 2 = = 0.375V
∆ 176 ∆ 176
∆3 286
v3 = = = 1.625V.
∆ 176
v1 = 625 mV, v2 = 375 mV, v3 = 1.625 V.
Chapter 3, Solution 28
At node c,
Vd − Vc Vc − Vb Vc
= + → 0 = −5Vb + 11Vc − 2Vd (1)
10 4 5
At node b,
Va + 45 − Vb Vc − Vb Vb
+ = → − 45 = Va − 4Vb + 2Vc (2)
8 4 8
At node a,
Va − 30 − Vd Va Va + 45 − Vb
+ + =0
→ 30 = 7Va − 2Vb − 4Vd (3)
4 16 8
At node d,
Va − 30 − Vd Vd Vd − Vc
= + → 150 = 5Va + 2Vc − 7Vd (4)
4 20 10
In matrix form, (1) to (4) become
0 − 5 11 − 2 Va 0
1 − 4 2 0 Vb − 45
7 − 2 0 − 4 V = 30 → AV = B
c
5 0 2 − 7 V 150
d
We use MATLAB to invert A and obtain
− 10.14
−1 7.847
V = A B=
− 1.736
− 29.17
Thus,
Va = −10.14 V, Vb = 7.847 V, Vc = −1.736 V, Vd = −29.17 V
Chapter 3, Solution 29
At node 1,
5 + V1 − V4 + 2V1 + V1 − V2 = 0
→ − 5 = 4V1 − V2 − V4 (1)
At node 2,
V1 − V2 = 2V2 + 4(V2 − V3 ) = 0
→ 0 = −V1 + 7V2 − 4V3 (2)
At node 3,
6 + 4(V2 − V3 ) = V3 − V4 → 6 = −4V2 + 5V3 − V4 (3)
At node 4,
2 + V3 − V4 + V1 − V4 = 3V4
→ 2 = −V1 − V3 + 5V4 (4)
In matrix form, (1) to (4) become
4 − 1 0 − 1 V1 − 5
− 1 7 − 4 0 V2 0
0 − 4 5 − 1 V = 6 → AV = B
3
− 1 0 − 1 5 V 2
4
Using MATLAB,
− 0.7708
−1 1.209
V = A B=
2.309
0.7076
i.e.
V1 = −0.7708 V, V2 = 1.209 V, V3 = 2.309 V, V4 = 0.7076 V
Chapter 3, Solution 30
v2
–+
40 Ω
I0
120 V
v1 20 Ω v0
10 Ω 1 2
+
100 V
– 4v0 + 2I0
– 80 Ω
At node 1,
v 1 − v 2 100 − v 1 4 v o − v 1
= + (1)
40 10 20
But, vo = 120 + v2 v2 = vo – 120. Hence (1) becomes
7v1 – 9vo = 280 (2)
At node 2,
vo − 0
Io + 2Io =
80
v + 120 − v o v o
3 1 =
40 80
or 6v1 – 7vo = -720 (3)
7 − 9 v 1 280
from (2) and (3), 6 − 7 v = − 720
o
7 −9
∆= = −49 + 54 = 5
6 −7
280 − 9 7 280
∆1 = = −8440 , ∆2 = = −6720
− 720 − 7 6 − 720
∆1 − 8440 ∆ − 6720
v1 = = = −1688, vo = 2 = − 1344 V
∆ 5 ∆ 5
Io = -5.6 A
Chapter 3, Solution 31
1Ω
+ v0 –
v1 v2 2v0 v3 2Ω
i0
1A 10 V +
4Ω 1Ω 4Ω –
At the supernode,
v1 v 2 v1 − v 3
1 + 2v0 = + + (1)
4 1 1
But vo = v1 – v3. Hence (1) becomes,
4 = -3v1 + 4v2 +4v3 (2)
At node 3,
v2 10 − v 3
2vo + = v1 − v 3 +
4 2
or 20 = 4v1 + v2 – 2v3 (3)
v3
At the supernode, v2 = v1 + 4io. But io = . Hence,
4
v2 = v1 + v3 (4)
Solving (2) to (4) leads to,
v1 = 4 V, v2 = 4 V, v3 = 0 V.
Chapter 3, Solution 32
5 kΩ v3
10 V 20 V
–+ +–
v1
v2 +
+ loop 1 v3
v1 loop 2
10 kΩ 12 V + –
– –
4 mA
(b)
(a)
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.
Chapter 3, Solution 33
(a) This is a non-planar circuit because there is no way of redrawing the circuit
with no crossing branches.
(b) This is a planar circuit. It can be redrawn as shown below.
4Ω
3Ω
5Ω
12 V + 2Ω
–
1Ω
Chapter 3, Solution 34
(a) This is a planar circuit because it can be redrawn as shown below,
7Ω
2Ω
1Ω 3Ω
6Ω
10 V + 5Ω
–
4Ω
(b) This is a non-planar circuit.
Chapter 3, Solution 35
30 V + 20 V +
– –
+
i1 i2 v0 4 kΩ
2 kΩ 5 kΩ –
Assume that i1 and i2 are in mA. We apply mesh analysis. For mesh 1,
-30 + 20 + 7i1 – 5i2 = 0 or 7i1 – 5i2 = 10 (1)
For mesh 2,
-20 + 9i2 – 5i1 = 0 or -5i1 + 9i2 = 20 (2)
Solving (1) and (2), we obtain, i2 = 5.
v0 = 4i2 = 20 volts.
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