Solution of Fundamental of Electric Circuits

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Chapter 1, Problem 1 How many coulombs are represented by these amounts of electrons: (a) 6.482 × 1017 (b) 1.24 × 1018 (c) 2.46 × 1019 (d) 1.628 × 10 20

Chapter 1, Solution 1 (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, Problem 2. Determine the current flowing through an element if the charge flow is given by (a) q(t ) = (3t + 8) mC (b) q(t ) = ( 8t 2 + 4t-2) C (c) q (t ) = 3e -t − 5e −2 t nC (d) q(t ) = 10 sin 120π t pC (e) q(t ) = 20e −4 t cos 50t μC

(

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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 = 1200π cos 120π t pA i =dq/dt = − e −4t (80 cos 50 t + 1000 sin 50 t ) μ A

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Chapter 1, Problem 3. Find the charge q(t) flowing through a device if the current is: (a) i (t ) = 3A, q(0) = 1C (b) i ( t ) = ( 2t + 5) mA, q(0) = 0 (c) i ( t ) = 20 cos(10t + π / 6) μA, q(0) = 2 μ C (d) i (t ) = 10e −30t sin 40tA, q(0) = 0

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 q(t) = ∫ 10e -30t sin 40t + q(0) = (c) q(t) = ∫ 20 cos (10t + π / 6 ) + q(0) = (2sin(10t + π / 6) + 1) μ C (d) 10e -30t ( −30 sin 40 t - 40 cos t) 900 + 1600 = − e - 30t (0.16cos40 t + 0.12 sin 40t) C

Chapter 1, Problem 4. A current of 3.2 A flows through a conductor. Calculate how much charge passes through any cross-section of the conductor in 20 seconds.

Chapter 1, Solution 4 q = it = 3.2 x 20 = 64 C

Chapter 1, Problem 5. Determine the total charge transferred over the time interval of 0 ≤ t ≤ 10s when 1 i (t ) = t A. 2 Chapter 1, Solution 5 1 t 2 10 q = ∫ idt = ∫ tdt = = 25 C 2 4 0 0
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Chapter 1, Problem 6. The charge entering a certain element is shown in Fig. 1.23. Find the current at: (a) t = 1 ms (b) t = 6 ms (c) t = 10 ms

Figure 1.23

Chapter 1, Solution 6 (a) At t = 1ms, i = (b) At t = 6ms, i = dq 80 = = 40 A dt 2

dq = 0A dt dq 80 = = –20 A dt 4

(c) At t = 10ms, i =

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 1, Problem 7. The charge flowing in a wire is plotted in Fig. 1.24. Sketch the corresponding current.

Figure 1.24

Chapter 1, Solution 7

⎡ 25A, dq ⎢ i= = - 25A, dt ⎢ ⎢ 25A, ⎣

0< t> I = inv(Z)*V I=
1.6196 mA –1.0202 mA –2.461 mA 3 mA –2.423 mA

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