Chapter 1, Solution 1
(a) q = 6.482x1017 x [-1.602x10-19 C] = -0.10384 C
(b) q = 1. 24x1018 x [-1.602x10-19 C] = -0.19865 C
(c) q = 2.46x1019 x [-1.602x10-19 C] = -3.941 C
(d) q = 1.628x1020 x [-1.602x10-19 C] = -26.08 C
Chapter 1, Solution 2
(a) i = dq/dt = 3 mA
(b) i = dq/dt = (16t + 4) A
(c) i = dq/dt = (-3e-t + 10e-2t) nA
(d) i=dq/dt = 1200π cos 120π t pA
(e) i =dq/dt = − e −4t (80 cos 50 t + 1000 sin 50 t ) µ A
Chapter 1, Solution 3
(a) q(t) = ∫ i(t)dt + q(0) = (3t + 1) C
(b) q(t) = ∫ (2t + s) dt + q(v) = (t 2 + 5t) mC
(c) q(t) = ∫ 20 cos (10t + π / 6 ) + q(0) = (2sin(10t + π / 6) + 1) µ C
10e -30t
q(t) = ∫ 10e -30t sin 40t + q(0) = ( −30 sin 40t - 40 cos t)
(d) 900 + 1600
= − e - 30t (0.16cos40 t + 0.12 sin 40t) C
Chapter 1, Solution 4
10
−5
q = ∫ idt = ∫ 5sin 6 π t dt = cos 6π t
6π 0
5
= (1 − cos 0.06π ) = 4.698 mC
6π
Chapter 1, Solution 5
2
1
q = ∫ idt = ∫ e dt mC = - e -2t
-2t
2 0
1
= (1 − e 4 ) mC = 490 µC
2
Chapter 1, Solution 6
dq 80
(a) At t = 1ms, i = = = 40 mA
dt 2
dq
(b) At t = 6ms, i = = 0 mA
dt
dq 80
(c) At t = 10ms, i = = = - 20 mA
dt 4
Chapter 1, Solution 7
25A, 0Chapter 1, Solution 8
10 × 1
q = ∫ idt = + 10 × 1 = 15 µC
2
Chapter 1, Solution 9
1
(a) q = ∫ idt = ∫ 10 dt = 10 C
0
3 5 ×1
q = ∫ idt = 10 × 1 + 10 − + 5 ×1
(b) 0
2
= 15 + 10 − 25 = 22.5 C
5
(c) q = ∫ idt = 10 + 10 + 10 = 30 C
0
Chapter 1, Solution 10
q = ixt = 8 x10 3 x15 x10 − 6 = 120 µ C
Chapter 1, Solution 11
q = it = 85 x10-3 x 12 x 60 x 60 = 3,672 C
E = pt = ivt = qv = 3672 x1.2 = 4406.4 J
Chapter 1, Solution 12
For 0 < t < 6s, assuming q(0) = 0,
t t
∫ ∫
q (t ) = idt + q (0 ) = 3tdt + 0 = 1.5t 2
0 0
At t=6, q(6) = 1.5(6)2 = 54
For 6 < t < 10s,
t t
∫ ∫
q (t ) = idt + q (6 ) = 18 dt + 54 = 18 t − 54
6 6
At t=10, q(10) = 180 – 54 = 126
For 10Chapter 1, Solution 13
2 2
w = ∫ vidt = ∫ 1200 cos 2 4 t dt
0 0
2
= 1200 ∫ ( 2 cos 8t - 1)dt (since, cos 2 x = 2 cos 2x - 1)
0
2
2 1
= 1200 sin 8t − t = 1200 sin 16 − 2
8 0 4
= - 2.486 kJ
Chapter 1, Solution 14
q = ∫ idt = ∫ 10(1 - e -0.5t )dt = 10(t + 2e -0.5t )
1 1
(a) 0 0
= 10(1 + 2e -0.5
− 2 ) = 2.131 C
(b) p(t) = v(t)i(t)
p(1) = 5cos2 ⋅ 10(1- e-0.5) = (-2.081)(3.935)
= -8.188 W
Chapter 1, Solution 15
2
2 − 3 2t
q = ∫ idt = ∫ 3e dt =-2t
e
(a) 0 2 0
= −1.5(e − 1) = 1.297 C
-2
5di
v= = −6e 2t ( 5) = −30e -2t
(b) dt
p = vi = − 90 e − 4 t W
3
3 − 90 -4t
(c) w = ∫ pdt = -90∫ e -4t dt = e = − 22.5 J
0 −4 0
Chapter 1, Solution 16
10t V 0< t Chapter 1, Solution 20
Since Σ p = 0
-30×6 + 6×12 + 3V0 + 28 + 28×2 - 3×10 = 0
72 + 84 + 3V0 = 210 or 3V0 = 54
V0 = 18 V
Chapter 1, Solution 21
∆q photon 1 electron
i= = 4 × 1011 ⋅ photon ⋅ 1. 6 × 10 ( C / electron)
19
∆t sec 8
4
= × 1011 × 1. 6 × 10 −19 C/s = 0.8 × 10 -8 C/s = 8 nA
8
Chapter 1, Solution 22
It should be noted that these are only typical answers.
(a) Light bulb 60 W, 100 W
(b) Radio set 4W
(c) TV set 110 W
(d) Refrigerator 700 W
(e) PC 120 W
(f) PC printer 18 W
(g) Microwave oven 1000 W
(h) Blender 350 W
Chapter 1, Solution 23
p 1500
(a) i = = = 12.5 W
v 120
45
(b) w = pt = 1. 5 × 103 × 45 × 60 ⋅ J = 1.5 × kWh = 1.125 kWh
60
(c) Cost = 1.125 × 10 = 11.25 cents
Chapter 1, Solution 24
p = vi = 110 x 8 = 880 W
Chapter 1, Solution 25
4
Cost = 1.2 kW × hr × 30 × 9 cents/kWh = 21.6 cents
6
Chapter 1, Solution 26
0. 8A ⋅ h
(a) i = = 80 mA
10h
(b) p = vi = 6 × 0.08 = 0.48 W
(c) w = pt = 0.48 × 10 Wh = 0.0048 kWh
Chapter 1, Solution 27
(a) Let T = 4h = 4 × 36005
T
q = ∫ idt = ∫ 3dt = 3T = 3 × 4 × 3600 = 43.2 kC
0
T T 0 . 5t
( b) W = ∫ pdt = ∫ vidt = ∫ ( 3) 10 + dt
0 0
3600
4×3600
0. 25t 2
= 310t +
= 3[40 × 3600 + 0. 25 × 16 × 3600]
3600 0
= 475.2 kJ
( c) W = 475.2 kWs, (J = Ws)
475.2
Cost = kWh × 9 cent = 1.188 cents
3600
Chapter 1, Solution 28
P 30
(a) i = = = 0.25 A
V 120
( b) W = pt = 30 × 365 × 24 Wh = 262.8 kWh
Cost = $0.12 × 262.8 = $31.54
Chapter 1, Solution 29
(20 + 40 + 15 + 45) 30
w = pt = 1. 2kW hr + 1.8 kW hr
60 60
= 2.4 + 0.9 = 3.3 kWh
Cost = 12 cents × 3.3 = 39.6 cents
Chapter 1, Solution 30
Energy = (52.75 – 5.23)/0.11 = 432 kWh
Chapter 1, Solution 31
Total energy consumed = 365(4 +8) W
Cost = $0.12 x 365 x 12 = $526.60
Chapter 1, Solution 32
(20 + 40 + 15 + 45) 30
w = pt = 1. 2kW hr + 1.8 kW hr
60 60
= 2.4 + 0.9 = 3.3 kWh
Cost = 12 cents × 3.3 = 39.6 cents
Chapter 1, Solution 33
dq
i= → q = ∫ idt = 2000 × 3 × 10 3 = 6 C
dt
Chapter 1, Solution 34
(b) Energy = ∑ pt = 200 x 6 + 800 x 2 + 200 x 10 + 1200 x 4 + 200 x 2
= 10,000 kWh
(c) Average power = 10,000/24 = 416.67 W
Chapter 1, Solution 35
( a) W = ∫ p( t ) dt = 400 × 6 + 1000 × 2 + 200 × 12 × 1200 × 2 + 400 × 2
= 7200 + 2800 = 10.4 kWh
10.4 kW
( b) = 433.3 W/h
24 h
Chapter 1, Solution 36
160A ⋅ h
(a) i= =4A
40
160Ah 160, 000h
( b) t = = = 6,667 days
0.001A 24h / day
Chapter 1, Solution 37
q = 5 × 10 20 (− 1. 602 × 10 −19 ) = −80. 1 C
W = qv = −80. 1 × 12 = − 901.2 J
Chapter 1, Solution 38
P = 10 hp = 7460 W
W = pt = 7460 × 30 × 60 J = 13.43 × 106 J
Chapter 1, Solution 39
p 2 × 10 3
p = vi → i = = = 16.667 A
v 120
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