FiFt h Edition
Fundamentals of
Electric Circuits
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) 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 cos 120 t pA
(e) 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)
-30t
( 30 sin 40t - 40 cos t)
(d) 900 1600
e - 30t (0.16cos40 t 0.12 sin 40t) C
Fundamentals of
Electric Circuits
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) 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 cos 120 t pA
(e) 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)
-30t
( 30 sin 40t - 40 cos t)
(d) 900 1600
e - 30t (0.16cos40 t 0.12 sin 40t) C
