Electrical Engineering: Principles & Applications, 7th Edition - Solutions Manual by Allan R. Hambley

Passvibes

, APPENDIX A

Exercises

EA.1 Given Z 1  2  j 3 and Z 2  8  j 6, we have:

Z 1  Z 2  10  j 3

Z 1  Z 2  6  j 9

Z1 Z 2  16  j 24  j 12  j 2 18  34  j 12

2  j 3 8  j 6 16  j 12  j 24  j 2 18
Z1 / Z2     0.02  j 0.36
8 j6 8 j6 100
Th d co of y th
is is p urs an e
an eir le tro

w ro es y p int
th sa es


or v
or ill d




k ide an art egr
is
w




pr d s as f th y o




EA.2 Z 1  1545   15 cos( 45  )  j 15 sin(45  )  10.6  j 10.6
ot ole se is f t
ec ly s w he
te fo sin or w




Z 2  10  150   10 cos( 150  )  j 10 sin(150  )  8.66  j 5
d
d o it


by r th g s (in ork
U e u tud clu an
ni s en d d




Z 3  590   5 cos(90  )  j 5 sin(90  )  j 5
te e
d of t le ng is n
St in ar on ot
at st ni t p
es ru ng he er
k




co cto . D W mit
py rs is or ted




EA.3 Notice that Z1 lies in the first quadrant of the complex plane.
rig in se ld .
i




ht te min Wi
la ach at de




Z 1  3  j 4  32  42  arctan( )  553.13
w
s ing ion We




Notice that Z2 lies on the negative imaginary axis.
Z 2   j 10  10  90 
b)




Notice that Z3 lies in the third quadrant of the complex plane.
Z 3  5  j 5  52  52 (180   arctan( 5 / 5))  7.07 225   7.07   135 

EA.4 Notice that Z1 lies in the first quadrant of the complex plane.
Z 1  10  j 10  10 2  10 2  arctan()  14.1445   14.14 exp( j 45  )
Pa




Notice that Z2 lies in the second quadrant of the complex plane.
ss




Z 2  10  j 10  10 2  10 2 (180   arctan( ))
 14.14 135   14.14 exp( j 135  )
vi
be




1
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s




Passvibes
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

,EA.5 Z 1Z 2  (1030  )(20135  )  (10  20)(30   135  )  200 (165  )

Z1 / Z 2  (1030  ) /(20135  )  ()(30   135  )  0.5(105  )

Z 1  Z 2  (10 30  )  (20135  )  (8.66  j 5)  (14.14  j 14.14)
 22.8  j 9.14  24.6  21.8

Z 1  Z 2  (10 30  )  (20135  )  (8.66  j 5)  (14.14  j 14.14)
 5.48  j 19.14  19.9106 

Problems

PA.1 Given Z 1  2  j 3 and Z 2  4  j 3, we have:
Th d co of y th
is is p urs an e
an eir le tro

w ro es y p int
th sa es


or v




Z1  Z2  6  j 0
or ill d




k ide an art egr
is
w




pr d s as f th y o
ot ole se is f t
ec ly s w he
te fo sin or w




Z 1  Z 2  2  j 6
d
d o it


by r th g s (in ork
U e u tud clu an
ni s en d d
te e
d of t le ng is n
St in ar on ot
at st ni t p




Z 1 Z 2  8  j 6  j 12  j 2 9  17  j 6
es ru ng he er
k




co cto . D W mit
py rs is or ted
rig in se ld .
i




ht te min Wi




2  j 3 4  j 3  1  j 18
la ach at de
w




Z1 / Z2     0.04  j 0.72
s ing ion We




4  j3 4  j3 25
b)




PA.2 Given that Z 1  1  j 2 and Z 2  2  j 3, we have:

Z1  Z2  3  j 1

Z 1  Z 2  1  j 5

Z1 Z2  2  j 3  j 4  j 2 6  8  j 1
Pa




1  j2 2  j3  4  j 7
Z1 / Z2     0.3077  j 0.5385
ss




2  j3 2  j3 13
vi
be




2
© 2018 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently
s




Passvibes
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

,PA.3 Given that Z 1  10  j 5 and Z 2  20  j 20, we have:

Z 1  Z 2  30  j 15

Z 1  Z 2  10  j 25

Z1 Z 2  200  j 200  j 100  j 2 100  300  j 100

10  j 5 20  j 20 100  j 300
Z1 / Z2     0.125  j 0.375
20  j 20 20  j 20 800



PA.4 (a) Z a  5  j 5  7.071  45  7.071 exp  j 45 
Th d co of y th
is is p urs an e
an eir le tro




Z b  10  j 5  11.18153 .43  11.18 exp j 153 .43 
w ro es y p int
th sa es




(b)
or v
or ill d




k ide an art egr
is
w




pr d s as f th y o
ot ole se is f t
ec ly s w he




Zc  3  j 4  5  126 .87   5 exp  j 126 .87  
te fo sin or w




(c)
d
d o it


by r th g s (in ork
U e u tud clu an
ni s en d d
te e
d of t le ng is n




Zd   j 12  12  90   12 exp  j 90  
St in ar on ot




(d)
at st ni t p
es ru ng he er
k




co cto . D W mit
py rs is or ted
rig in se ld .
i




ht te min Wi
la ach at de
w
s ing ion We




PA.5 (a) Z a  545  5 exp j 45   3.536  j 3.536

(b) Z b  10120   10 exp j 120    5  j 8.660
b)




(c) Zc  15  90   15 exp  j 90     j 15

(d) Zd  1060   10 exp  j 120    5  j 8.660
Pa




PA.6 (a) Z a  5e j 30  530   4.330  j 2.5





(b) Z b  10e  j 45  10  45  7.071  j 7.071

ss




(c) Zc  100e j 135  100 135   70.71  j 70.71

vi
be




3
© 2018 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently
s




Passvibes
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

, (d) Zd  6e j 90  690   j 6





PA.7 (a) Z a  5  j 5  1030   13.66  j 10

(b) Z b  545  j 10  3.536  j 6.464

1045  1045 
(c) Zc   2  8.13  1.980  j 0.283
3  j4 553.13

15
(d) Zd   3  90    j 3
590 
Th d co of y th
is is p urs an e
an eir le tro

w ro es y p int
th sa es


or v
or ill d




k ide an art egr
is
w




pr d s as f th y o
ot ole se is f t
ec ly s w he
te fo sin or w
d
d o it


by r th g s (in ork
U e u tud clu an
ni s en d d
te e
d of t le ng is n
St in ar on ot
at st ni t p
es ru ng he er
k




co cto . D W mit
py rs is or ted
rig in se ld .
i




ht te min Wi
la ach at de
w
s ing ion We
b)
Pa
ss
vi
be




4
© 2018 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently
s




Passvibes
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

, CHAPTER 1

Exercises

E1.1 Charge = Current  Time = (2 A)  (10 s) = 20 C

dq (t ) d
E1.2 i (t )   (0.01sin(20 0t)  0.01  200cos(200 t )  2cos(200t ) A
dt dt

E1.3 Because i2 has a positive value, positive charge moves in the same
direction as the reference. Thus, positive charge moves downward in
element C.

Because i3 has a negative value, positive charge moves in the opposite
Th d co of y th
is is p urs an e
an eir le tro

w ro es y p int
th sa es




direction to the reference. Thus positive charge moves upward in
or v
or ill d




k ide an art egr
is
w




element E.
pr d s as f th y o
ot ole se is f t
ec ly s w he
te fo sin or w
d
d o it


by r th g s (in ork
U e u tud clu an




E1.4 Energy = Charge  Voltage = (2 C)  (20 V) = 40 J
ni s en d d
te e
d of t le ng is n
St in ar on ot
at st ni t p
es ru ng he er
k




Because vab is positive, the positive terminal is a and the negative
co cto . D W mit
py rs is or ted
rig in se ld .
i




terminal is b. Thus the charge moves from the negative terminal to the
ht te min Wi
la ach at de
w




positive terminal, and energy is removed from the circuit element.
s ing ion We




E1.5 iab enters terminal a. Furthermore, vab is positive at terminal a. Thus
b)




the current enters the positive reference, and we have the passive
reference configuration.

E1.6 (a) pa (t )  v a (t )ia (t )  20t 2
10 10 10
20t 3 20t 3
w a   pa (t )dt   20t dt  2
  6667 J
0 0
3 0
3
(b) Notice that the references are opposite to the passive sign
Pa




convention. Thus we have:

pb (t )  v b (t )ib (t )  20t  200
ss




10 10
10
w b   pb (t )dt   (20t  200 )dt  10t 2  200t  1000 J
vi




0
0 0
be




1
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s




Passvibes
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

,E1.7 (a) Sum of currents leaving = Sum of currents entering
ia = 1 + 3 = 4 A

(b) 2 = 1 + 3 + ib  ib = -2 A

(c) 0 = 1 + ic + 4 + 3  ic = -8 A

E1.8 Elements A and B are in series. Also, elements E, F, and G are in series.

E1.9 Go clockwise around the loop consisting of elements A, B, and C:
-3 - 5 +vc = 0  vc = 8 V

Then go clockwise around the loop composed of elements C, D and E:
- vc - (-10) + ve = 0  ve = -2 V
Th d co of y th
is is p urs an e
an eir le tro

w ro es y p int
th sa es


or v




E1.10 Elements E and F are in parallel; elements A and B are in series.
or ill d




k ide an art egr
is
w




pr d s as f th y o
ot ole se is f t
ec ly s w he
te fo sin or w




ρL
d
d o it


by r th g s (in ork




E1.11 The resistance of a wire is given by R  . Using A  d and
U e u tud clu an




A
ni s en d d
te e
d of t le ng is n
St in ar on ot




substituting values, we have:
at st ni t p
es ru ng he er
k




co cto . D W mit
py rs is or ted
rig in se ld .
i




1.12  10 6  L
ht te min Wi
la ach at de




9.6   L = 17.2 m
w
s ing ion We




 (1.6  10  3 )

E1.12 P V 2 R  R V 2 / P  144   I  V / R   0.833 A
b)




E1.13 P V 2 R  V  PR  0.25  1000  15.8 V
I  V / R  15.  15.8 mA

E1.14 Using KCL at the top node of the circuit, we have i1 = i2. Then, using KVL
going clockwise, we have -v1 - v2 = 0; but v1 = 25 V, so we have v2 = -25 V.
Next we have i1 = i2 = v2/R = -1 A. Finally, we have
Pa




PR  v 2i2  (25)  (1)  25 W and Ps  v 1i1  (25)  (1)  25 W.
ss




E1.15 At the top node we have iR = is = 2A. By Ohm’s law we have vR = RiR = 80
V. By KVL we have vs = vR = 80 V. Then ps = -vsis = -160 W (the minus sign
is due to the fact that the references for vs and is are opposite to the
vi




passive sign configuration). Also we have PR  v R iR  160 W.
be




2
© 2018 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently
s




Passvibes
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

, Problems

P1.1 Broadly, the two objectives of electrical systems are:
1. To gather, store, process, transport and present information.
2. To distribute, store, and convert energy between various forms.

P1.2 Four reasons that non-electrical engineering majors need to learn the
fundamentals of EE are:

1. To pass the Fundamentals of Engineering Exam.
2. To be able to lead in the design of systems that contain
electrical/electronic elements.
3. To be able to operate and maintain systems that contain
electrical/electronic functional blocks.
Th d co of y th




4. To be able to communicate effectively with electrical engineers.
is is p urs an e
an eir le tro

w ro es y p int
th sa es


or v
or ill d




k ide an art egr
is
w




pr d s as f th y o




P1.3 Eight subdivisions of EE are:
ot ole se is f t
ec ly s w he
te fo sin or w
d
d o it


by r th g s (in ork
U e u tud clu an




1. Communication systems.
ni s en d d
te e
d of t le ng is n
St in ar on ot




2. Computer systems.
at st ni t p
es ru ng he er
k




co cto . D W mit




3. Control systems.
py rs is or ted
rig in se ld .
i




4. Electromagnetics.
ht te min Wi
la ach at de
w
s ing ion We




5. Electronics.
6. Photonics.
7. Power systems.
b)




8. Signal Processing.

P1.4 Responses to this question are varied.

P1.5 (a) Electrical current is the time rate of flow of net charge through a
conductor or circuit element. Its units are amperes, which are equivalent
to coulombs per second.
Pa




(b) The voltage between two points in a circuit is the amount of energy
transferred per unit of charge moving between the points. Voltage has
ss




units of volts, which are equivalent to joules per coulomb.
vi
be




3
© 2018 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently
s




Passvibes
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.