CHAPTER 1 (SOLUTION MANUAL) 1-2.1
R = 30 Ω
V = 500 V
500 I= = 16.66 A 30 1-2.3
(a) W = P × t;
P=
P = I2R = (16.66)2 × 30 = 8.33 kW
V 2 (12) 2 = = 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 = µF C = Q/V= 150 3 4 µF Total capacitance = 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 10 +12 5´12 × 2.73 A = 22
resistance = IT ×
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
i4 =
Þ Þ
∴
20´6 0.2i4 ´6 = 12 + 0.12i4 + 4+6 10 0.88i4 = 12 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 = 495.87 W 6 Ptotal = P4Ω + P20A + P0.2i4 + P6Ω
P6Ω = ∴
=0 1-4.1
I = 2 A; t = 15 s; W = 1000 J Q = It = 30 C V=
1-4.2
1000 = 33.33 V 30
V = 120 V;
I = 0.8 A
120 = 150 Ω 0.8 Q 75 Q 10 (i) (a) I = = 75 A (ii) t = =2s = = 1 5 t I Q 10 = 20 A (b) I = = t 0.5 Q 5 (c) I = = = 2.5 A t 2 1 W = CV2 2 1 0.3 = C(20)2 2 0.3´2 ∴ C= = 1.5 mF 400 R=
1-6.1
1-6.2
1-6.3
The rate of change of voltage across each capacitor and the combination is the same, i.e.,
1-8.1
dV = 4500 volts/second dt
In Fig Q.1 (a),
In Fig Q.1 (b),
50 VAB = VT 50 + 60 50 = 4.55 V = 10 × 110
30 30 +10 + 100 30 = 100 × 140
VAB = VT
1-8.2
VAB = 21.43 V Since each resistor is across the 100 V source, the voltage drop across each source is 100 V. Hence, each meter reads 100 V.
1-8.4
Actual resistance R =
20 = 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
Þ
V2 = VA = 60 V
V3 = –200 V
V4 = VS = 260 V i2 =
Þ
60 =3A 20
i3 =
–200 = – 40 5
i1 = i2 + i3 = –37 A
–V2 = –5 A 12 –35 i5 = i2 = –35 A 3 i4 =
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
The equivalent resistances R1 and R2 are in parallel. R3 = 100 || 100 = 50 Ω The resistance R3 is in series with the 100 Ω resistance. Hence, the total resistance = 150 Ω. 1-9.5
a) Let R = 80 Ω Req = ?
Req = 60 Ω b) If Req = 80 Ω, R = ?
100 ( R + 20) 100 + R + 20 100 R + 2000 80 – 10 = 120 + R 70 (120 + R) = 100 R + 2000 8400 – 2000 = 100 R – 70R 30R = 6400 6400 R= 30 R = 213.33 Ω c) If Req = R; 10 + 100 R + 2000 R= 100 + R + 20 2 120R + R = 10 + 100R + 2000 R2 + 20R – 2010 = 0 80 Ω = 10 +
R=
=
1-9.6
Let
– 20 ± 400 + 4 × 2010 1 440 – 20 2
= 35.93 Ω
RP1 = 2.2 K || 3.3 K || 5.6 K = 1.08 K RP2 = 1 K || 5.6 K || 3.3 K = 0.68 K RP3 = 1 K || 2 K || 5.6 K = 0.61 K RP4 = 1 K || 3.3 K || 2.2 K = 0.57 K
Solution Manual (1) Page No. 5