COSMOS: Complete Online Solutions Manual Organization System
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
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 2.
(a)
(b)
(c)
(d)
(e)
i = dq/dt = 3 mA
i = dq/dt = (16t + 4) A
i = dq/dt = (-3e-t + 10e-2t) nA
i=dq/dt = 1200π cos 120π t pA
i =dq/dt = − e −4t (80 cos 50 t + 1000 sin 50 t ) µ A
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
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
(d)
10e -30t
( −30 sin 40t - 40 cos t)
900 + 1600
= − e - 30t (0.16cos40 t + 0.12 sin 40t) C
q(t) = ∫ 10e -30t sin 40t + q(0) =
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 4.
q = it = 3.2 x 20 = 64 C
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 5.
10
1
t 2 10
q = ∫ idt = ∫ tdt =
= 25 C
2
4 0
0
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 6.
(a) At t = 1ms, i =
dq 80
=
= 40 A
dt
2
(b) At t = 6ms, i =
dq
= 0A
dt
(c) At t = 10ms, i =
dq 80
=
= –20 A
dt
4
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 7.
25A,
dq 
i=
= - 25A,
dt 
 25A,
0<t<2
2<t<6
6<t<8
which is sketched below:
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 8.
q = ∫ idt =
10 × 1
+ 10 × 1 = 15 µC
2
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 9.
1
(a) q = ∫ idt = ∫ 10 dt = 10 C
0
3
5 ×1

q = ∫ idt = 10 × 1 + 10 −
 + 5 ×1
0
(b)
2 

= 15 + 7.5 + 5 = 22.5C
5
(c) q = ∫ idt = 10 + 10 + 10 = 30 C
0
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 10.
q = it = 8x103x15x10-6 = 120 mC
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
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
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
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 10<t<15s,
t
t
∫
∫
q (t ) = idt + q (10 ) = ( −12)dt + 126 = −12t + 246
10
10
At t=15, q(15) = -12x15 + 246 = 66
For 15<t<20s,
t
∫
q (t ) = 0 dt + q (15) =66
15
Thus,

1.5t 2 C, 0 < t < 6s

 18 t − 54 C, 6 < t < 10s
q (t ) = 
−12t + 246 C, 10 < t < 15s

66 C, 15 < t < 20s

The plot of the charge is shown below.
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
140
120
100
q(t)
80
60
40
20
0
0
5
10
t
15
20
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 13.
dq
= 40π cos 4π t mA
dt
p = vi = 80π cos 2 4π t mW
At t=0.3s,
p = 80π cos 2 (4π x0.3) = 164.5 mW
(a) i =
0.6
0.6
0
0
(b) W = ∫ pdt = 80π ∫ cos 2 4π tdt = 40π ∫ [1 + cos8π t ]dt mJ

0.6 
1
W = 40π 0.6 +
sin 8π t
 = 78.34 mJ
0 
8π

C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 14.
q = ∫ idt = ∫ 10(1 - e -0.5t )dt = 10(t + 2e -0.5t )
1
(a)
(b)
0
= 10(1 + 2e
-0.5
− 2 ) = 2.131 C
1
0
p(t) = v(t)i(t)
p(1) = 5cos2 ⋅ 10(1- e-0.5) = (-2.081)(3.935)
= -8.188 W
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 15.
2
(a)
q = ∫ idt = ∫ 3e
(
0
-2t
)
2
− 3 2t
dt =
e
2
0
= −1.5 e -4 − 1 =
1.4725 C
(b)
5di
= −6e 2t ( 5) = −30e -2t
dt
p = vi = − 90 e − 4 t W
v=
3
(c) w = ∫ pdt = -90∫ e
0
3
-4t
− 90 -4t
e
= − 22.5 J
dt =
−4
0
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 16.
(a)
 30t mA, 0 < t <2
i (t ) = 
120-30t mA, 2 < t<4
5 V, 0 < t <2
v(t ) = 
 -5 V, 2 < t<4
 150t mW, 0 < t <2
p(t ) = 
-600+150t mW, 2 < t<4
which is sketched below.
p(mW)
300
1
2
4
t (s)
-300
(b) From the graph of p,
4
W = ∫ pdt = 0 J
0
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 17.
Σ p=0
→ -205 + 60 + 45 + 30 + p3 = 0
p3 = 205 – 135 = 70 W
Thus element 3 receives 70 W.
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 18.
p1 = 30(-10) = -300 W
p2 = 10(10) = 100 W
p3 = 20(14) = 280 W
p4 = 8(-4) = -32 W
p5 = 12(-4) = -48 W
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 19.
I = 4 –1 = 3 A
Or using power conservation,
9x4 = 1x9 + 3I + 6I = 9 + 9I
4 = 1 + I or I = 3 A
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
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
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 21.
p 60
=
= 0.5 A
v 120
q = it = 0.5x24x60x60 = 43200 C
N e = qx6.24 x1018 = 2.696 x10 23 electrons
p = vi

→
i=
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 22.
q = it = 30 x103 x 2 x10−3 = 60 C
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 23.
W = pt = 1.8x(15/60) x30 kWh = 13.5kWh
C = 10cents x13.5 = $1.35
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 24.
W = pt = 40 x24 Wh = 0.96 kWh
C = 8.5 cents x0.96 = 8.16 cents
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 25.
Cost = 1.2 kW ×
4
hr × 30 × 9 cents/kWh = 21.6 cents
60
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 26.
0. 8A ⋅ h
= 80 mA
10h
(b) p = vi = 6 × 0.08 = 0.48 W
(c) w = pt = 0.48 × 10 Wh = 0.0048 kWh
(a) i =
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 27.
(a) Let T = 4h = 4 × 3600
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 

= 310t +

3600  0

= 475.2 kJ
( c)
= 3[40 × 3600 + 0.25 × 16 × 3600]
W = 475.2 kWs, (J = Ws)
475.2
Cost =
kWh × 9 cent = 1.188 cents
3600
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 28.
(a) i =
P 30
=
= 0.25 A
V 120
( b) W = pt = 30 × 365 × 24 Wh = 262.8 kWh
Cost = $0.12 × 262.8 = $31.54
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 29.
(20 + 40 + 15 + 45)
 30 
hr + 1.8 kW  hr
60
 60 
= 2.4 + 0.9 = 3.3 kWh
Cost = 12 cents × 3.3 = 39.6 cents
w = pt = 1. 2kW
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 30.
Monthly charge = $6
First 250 kWh @ $0.02/kWh = $5
Remaining 968 kWh @ $0.07/kWh= $67.76
Total = $78.76
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 31.
Total energy consumed = 365(120x4 + 60x8) W
Cost = $0.12x365x960/1000 = $42.05
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 32.
i = 20 µA
q = 15 C
t = q/i = 15/(20x10-6) = 750x103 hrs
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 33.
i=
dq
→ q = ∫ idt = 2000 × 3 × 10 − 3 = 6 C
dt
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 34.
(a)
Energy =
∑ pt
= 200 x 6 + 800 x 2 + 200 x 10 + 1200 x 4 + 200 x 2
= 10 kWh
(b)
Average power = 10,000/24 = 416.7 W
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 35.
energy = (5x5 + 4x5 + 3x5 + 8x5 + 4x10)/60 = 2.333 MWhr
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 36.
160A ⋅ h
=4A
40
160Ah 160, 000h
( b) t =
=
= 6,667 days
0.001A 24h / day
(a)
i=
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 37.
W = pt = vit = 12x 40x 60x60 = 1.728 MJ
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 38.
P = 10 hp = 7460 W
W = pt = 7460 × 30 × 60 J = 13.43 × 106 J
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 1, Solution 39.
W = pt = 600x4 = 2.4 kWh
C = 10cents x2.4 = 24 cents
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e, © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.