Solutions manual for fundamentals of electric circuits 5th edition by alexander 2019 0723 25597 16grxc5 PDF

Preview

Summary

solutions...

Description

FiFt h Edition

Fundamentals of

Electric Circuits INSTRUCTOR SOLUTIONS MANUAL

Charles K. Alexander | Matthew n. o. 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 cos120 t pA i =dq/dt =  e 4t (80 cos50t  1000 sin50t )  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

(d)

10e -30t ( 30 sin 40 t - 40 cos t) 900  1600   e -30t (0.16cos40 t  0.12 sin 40t) C

q(t)   10e -30t sin 40 t  q(0) 

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

Chapter 1, Solution 5 10 t 2 10 1 q   idt   tdt   25 C 4 0 0 2

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

dq 30   15 A 2 dt

(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   - 25A, i 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

Chapter 1, Solution 10 q = it = 10x103x15x10-6 = 150 mC

Chapter 1, Solution 11 q= it = 90 x10-3 x 12 x 60 x 60 = 3.888 kC E = pt = ivt = qv = 3888 x1.5 = 5.832 kJ

Chapter 1, Solution 12 For 0 < t < 6s, assuming q(0) = 0, t

t





0

0

q( t)  idt  q(0)  3 tdt  0  1.5 t 2

At t=6, q(6) = 1.5(6)2 = 54 For 6 < t < 10s, t

t





6

6

q( t)  idt  q(6 )  18 dt  54  18 t  54

At t=10, q(10) = 180 – 54 = 126 For 10...