Solutions Manual Fundamentals of Electric Circuits 5th Edition Alexander Sadiku

Instant download and all chapters Solutions Manual Fundamentals of Electric Circuits 5th Edition Alexander, Sadiku https://testbankdata.com/download/solutions-manual-fundamentals-electric-circuits-5th-editionalexander-sadiku/ Chapter 1, Solution 1 (a) q = 6.482x1017 x [-1.602x10-19 C] = –103.84 mC (b) q = 1. 24x1018 x [-1.602x10-19 C] = –198.65 mC (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) (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 = 1200 cos 120 t pA i =dq/dt =  e 4t (80 cos 50 t  1000 sin 50 t )  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)   10e sin 40t  q(0)  ( 30 sin 40t - 40 cos t) (d) 900  1600   e - 30t (0.16cos40 t  0.12 sin 40t) C -30t

Chapter 1, Solution 4 q = it = 7.4 x 20 = 148 C

Chapter 1, Solution 5 10

1 t 2 10 q   idt   tdt   25 C 2 4 0 0

Chapter 1, Solution 6 (a) At t = 1ms, i 

dq 30   15 A dt 2

(b) At t = 6ms, i 

dq  0A dt

(c) At t = 10ms, i 

dq  30   –7.5 A dt 4

Chapter 1, Solution 7 25A, dq  i  - 25A, dt   25A,

0t2 2t6 6t8

which is sketched below:

Chapter 1, Solution 8

q   idt 

10  1  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 0 (b) 2    15  7.5  5  22.5C 5

(c) q   idt  10  10  10  30 C 0