Circiuts Text Bok

Topics: Voltage source, Electric current, Current source Pages: 22 (4650 words) Published: May 24, 2013
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Chapter 2 Solutions Section 2.1 Introduction 2.1 Current source 2.2 Voltage source 2.3 Resistor 2.4 Capacitor 2.5 Inductor Section 2.2 Charge and Current 2.6 b) The current direction is designated as the direction of the movement of positive charges. 2.7 The relationship of charge and current is t

q(t ) = ! i(t ) + q(t 0 ) dt
t0

so
t

q(t ) = ! 2 sin ( " t ) + q(t0 ) 10 dt
t0

& '2 # q(t )= \$ cos( ( t )! + q(t0 ) 10 " t0 %10(
2.8 The coulomb of one electron is denoted by e and
t

t

q(t ) = ! i (t ) + q(t0 ) dt
t0

So

1 n(t ) = q (t ) / e = ! 12t dt + q(t 0 ) e t0
If t0 = 0 and q(t 0 ) = 0 ,
6 n(t ) = t 2 e

t

2.9

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

q(t ) = ! idt
t

q(t )= ! 5dt
0

q(t ) = 5t
2.10
q(t )= 0 [ t ]= 5(5)! 5(0 ) = 25 Coulombs 5
5

2.11 Using the definition of current-charge relationship, the equation can be rewritten as: i= dq !n = e dt !t

Thus, the current flow within t1 and t2 time interval is,

(5.75 ! 2) " 1019 i= (!1.6 " 10 !19 ) = !3 A 2
The negative sign shows the current flow in the opposite direction with respect to the electric charge. 2.12 Assuming the area of the metal surface is S, The mass of the nickel with depth d = 0.15mm is m=ρ×d×S Meanwhile, using the electro-chemical equivalent, the mass of the nickel can be expressed as m=k×I×t where I = σ × S. Equating the two expressions of the mass, the coating time is found: t = ρ × d / σ = 1.24 × 105 s ≈ 34.4 hour Section 2.3 Voltage 2.13 By the definition of voltage, when a positive charge moves from high voltage to low voltage, its potential energy decreases. So a is “+”, b is “-”. In other words, uab=1V. 2.14 The current i(t) is defined as: &3 0 < t ' 1 # i (t ) = % " \$0 elsewhere!

Therefore, the charge is

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

1

g = ! 3dt =3 C
0

The energy in Joules is given by:
J = V ! C = 5 ! 3 = 15 J

2.15 1 electron = ! 1.6 "10 !19 Coulombs. Therefore, there are 6.25 !1018 electrons in a coulomb. Coulombs of 5 ! 1016 electrons = Therefore, the voltage is V = J 15 = = 1875 V C 8 ! 10 "3
1

5 " 1016 = 8 " 10 !3 6.25 " 1018

2.16
&3 # q(t ) = ' 2 sin \$ ( t !dt %2 " 0

&- 4 , 3 )# = \$ cos* . t '! = 0.4244C 3. + 2 (" 0% J = V ! C = 5 ! 0.4244 = 2.122J 2.17
20 J = 10C 2V dq 10C i= = = 2.5 A dt 4s q=

1

Section 2.4 Respective Direction of Voltage and Current 2.18 True 2.19 2.20 False True

Section 2.5 Kirchoff’s Current Law 2.21 According to KCL