Copyright 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the

Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Solution 1.1

(a) q = 6.482x10 17 x [-1.602x10 -19 C] = –103.84 mC

(b) q = 1. 24x10 18 x [-1.602x10 -19 C] = –198.65 mC

(c) q = 2.46x10 19 x [-1.602x10 -19

C] = –3.941 C

(d) q = 1.628x10 20 x [-1.602x10 -19

C] = –26.08 C

(Fundamentals of Electrical Circuits, 6e Charles Alexander, Matthew Sadiku) (Solution Manual, For Complete File, Download link at the end of this File) 1 / 4

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Solution 1.2

Determine the current flowing through an element if the charge flow is given by (a) ()() mC 3tq= (b) () C 4)-20t(4ttq 2 += (c) ()( ) 2e15etq 18t-3t − −= nC (d) q(t) = 5t 2 (3t 3

• 4) pC

(e) q(t) = 2e -3t sin(20πt) µC

(a) i = dq/dt = 0 mA (b) i = dq/dt = (8t + 20) A (c) i = dq/dt = (–45e -3t

• 36e

-18t

• nA

(d) i=dq/dt = (75t 4

• 40t) pA

(e) i =dq/dt = {- 6e -3t sin(20πt) + 40πe -3t cos(20πt)} µA

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Solution 1.3

(a) C 1)(3t+=+=∫ q(0)i(t)dt q(t) (b) mC 5t)(t 2

+=++=∫

q(v)dt s)(2tq(t) (c) ( )q(t) 20 cos 10t / 6 q(0) (2sin(10 / 6) 1) Ctππµ= + += + +∫

(d) C 40t) si n 0.12t(0.16cos40e 30t- +−= − + =+= ∫ t)cos 40-t40sin30(

1600900

e10 q(0)t40sin10eq(t) -30t 30t-

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Solution 1.4

Since i is equal to Δq/Δt then i = 300/30 = 10 amps.

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