SOLUTION MANUAL FOR Fundamentals of Electric Circuits 5th Edition by Charles Alexander Matthew Sadik||COMPLETE GUIDE|| REVISED

SOLUTION MANUAL FOR
Fundamentals of Electric Circuits 5th Edition
by Charles Alexander Matthew Sadik||COMPLETE GUIDE|| REVISED

5TH EDITION

,(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 = 1200 Cos120 T Pa
(e) I =Dq/Dt = − E−4t (80 Cos50t + 1000 Sin 50t )  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 +  / 6) + Q(0) = (2sin(10t +  / 6) +1)C

10e -30t
Q(T) = -30t
Sin 40t + Q(0) = (−30 Sin 40t - 40
(d) Cos T) 900 + 1600
10e
= − E -30t (0.16cos40 T + 0.12 Sin 40t) C



Chapter 1, Solution 4

−5
10

Q =  Idt = 5sin 6 Π T Dt = Cos 6π T
6 0


5
= (1 − Cos 0.06 ) = 4.698 Mc
6

,Chapter 1, Solution 5
2

Q =  Idt = e- dt Mc = - 1 E-2t
2t

2

0


1
= (1 − E4 ) 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

Dq 25A, 0T2
I= = - 2T6
25A,
Dt 
25A, 6 T8

Which Is Sketched Below:

, Chapter 1, Solution 8
10 1
Q = Idt = + 10 1 = 15 µc
 2


Chapter 1, Solution 9
1

 0
(a) Q = Idt = 10 Dt = 10 C

 5 1 
Q =  Idt = 10 1 + 10 −
3
+ 5 1
(b)  
0
 2 
= 15 + 10 − 25 = 22.5 C

5
(c) Q = 0 Idt = 10 + 10 + 10 = 30 C




Chapter 1, Solution 10


Q = Ixt = 8 X103 X15x10 −6 = 120  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,