Solution Manual For Circuits and Networks Analysis and Synthesis 6th Edition A Sudhakar , Shyammohan S Palli

CHAPTER 1 (SOLUTION MANUAL)

1-2.1 R = 30 Ω V = 500 V
500
I= = 16.66 A P = I2R = (16.66)2 × 30 = 8.33 kW
30
V 2 (12) 2
1-2.3 (a) W = P × t; P= = = 14.4 watts
R 10
∴ W = 14.4 × 2 × 60 = 1728 J
(b) Decrease

1-2.4 Q = 50 × 10–6 C; V = 150 V
50´10 –6 1
C = Q/V= = µF
150 3
4
Total capacitance = µF
15
1 1 4
W = CV2 = ´ × 10–6 × (150)2 = 0.3 × 10–2 J
2 2 15

1-2.5 In the circuit shown in Fig Q.5 , the 20 Ω resistance is parallel with the 30 Ω resistance.
The equivalent resistance R = 20 || 30 = 12 Ω.
The 12 Ω resistance is parallel with the 10 Ω resistance. Hence, the current through the 5 Ω
12
resistance = IT ×
10 +12
5´12
= × 2.73 A
22
Power consumed by the 5 Ω resistance = I2R
= 52 × 5 = 125 watts

1-2.6




Power absorbed by the 5 Ω resistance
5 + 5v = I3 + I5
v = 2 × I3
5 + 10I3 = I3 + I5
5 + 9I3 = I5
I3 = V : I 5 = V
3 5
9V V
5+ =
3 5
V
5 + 3V – = 0
5
25 + (15V – V) = 0
14V = – 25
25
V =–
14

Solution Manual (1) Page No. 1

, = –1.79V
V 2 1.792
P5Ω = = = 638mW
R 5

1-2.7 The circuit is redrawn as shown below.




0.1 k
1
= 0.25
4
0.025 × 103
= 25 Ω



Current flowing through the 25 Ω resistance due to the 5 A source
100 × 5
I 25 =
100 + 25
100 × 5
=
125
Iy = I25 = 4 A
Current flowing through the 25 Ω resistance due to the dependent source
100 × 0.8 ix
I25d =
100 + 25
4
= × 0.8 ix
5
Total current flowing through the 25 Ω resistance
Ix = Iy + I25d
= 4 + 0.64 iy
1-2.8




20´6 0.2i4 ´6
i4 = + = 12 + 0.12i4
4+6 10
Þ 0.88i4 = 12
Þ i4 = 13.63 A
∴ P4Ω = (13.63)2 × 4 = 743.8 W
V = 13.63 × 4 = 54.54 V
P20A = –1090.9 W
P0.2i4 = (54.54)(0.2 × 13.63) = –148.76 W



Solution Manual (1) Page No. 2

, 54.54
P6Ω = = 495.87 W
6
∴ Ptotal = P4Ω + P20A + P0.2i4 + P6Ω
=0

1-4.1 I = 2 A; t = 15 s; W = 1000 J
Q = It = 30 C
1000
V= = 33.33 V
30
1-4.2 V = 120 V; I = 0.8 A
120
R= = 150 Ω
0.8
Q 75 Q 10
1-6.1 (i) (a) I = = = 75 A (ii) t = = =2s
t 1 I 5
Q 10
(b) I = = = 20 A
t 0.5
Q 5
(c) I = = = 2.5 A
t 2
1
1-6.2 W = CV2
2
1
0.3 = C(20)2
2
0.3´2
∴ C= = 1.5 mF
400
1-6.3 The rate of change of voltage across each capacitor and the combination is the same,
dV
i.e., = 4500 volts/second
dt

1-8.1 In Fig Q.1 (a), In Fig Q.1 (b),
50 30
VAB = VT VAB = VT
50 + 60 30 +10 + 100
50 30
= 10 × = 4.55 V = 100 ×
110 140
VAB = 21.43 V
1-8.2 Since each resistor is across the 100 V source, the voltage drop across each source is 100 V.
Hence, each meter reads 100 V.
20
1-8.4 Actual resistance R = = 0.2 kΩ
100 mA
Voltage applied V = 0.2 × 103 × 150 × 10–3 = 30 V
∴ the change in voltage = 10 V
1-8.5




Total voltage V = 5.5 + 7.2 + 12.3 = 25 V

Solution Manual (1) Page No. 3

, If R4 is added to the circuit, remaining voltage appearing across the resistor
= 30 – 25 V = 5 V
1-9.1




V1 = –60 V V3 = VA – VB = 60 – 260
V2 = VA = 60 V Þ V3 = –200 V

V4 = VS = 260 V

60 –200
i2 = =3A i3 = = – 40
20 5
Þ i1 = i2 + i3 = –37 A
–V2
i4 = = –5 A
12
–35
i5 = i2 = –35 A
3

1-9.3 Find Req.:




= Req = RAB = 22.5Ω

1-9.4 In the circuit shown in Fig Q.4, the 40 Ω resistor is parallel with the 60 Ω resistor. The equivalent
is in series with the 76 Ω resistor.
The equivalent resistance R1 = {[40 || 60] + 76} = 100 Ω
In the other branch, the 82 Ω resistor is in series with the 18 Ω resistor.
The equivalent resistance R2 = 82 + 18 = 100 Ω

Solution Manual (1) Page No. 4