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

Electrical Engineering: Principles &
ST
Applications – 7th Edition
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SOLUTIONS
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MANUAL
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Allan R. Hambley
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Comprehensive Solutions Manual for
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Instructors and Students
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© Allan R. Hambley
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All rights reserved. Reproduction or distribution without permission is prohibited.




©STUDYSTREAM

, APPENDIX A
ST
Exercises

EA.1 Given Z 1  2  j 3 and Z 2  8  j 6, we have:
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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
IA
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
_A
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
PP
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
RO
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.
VE
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  )
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Notice that Z2 lies in the second quadrant of the complex plane.
Z 2  10  j 10  10 2  10 2 (180   arctan( ))
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 14.14 135   14.14 exp( j 135  )




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© 2018 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently
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  )
ST
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
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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 
IA
Problems

PA.1 Given Z 1  2  j 3 and Z 2  4  j 3, we have:
Th d co of y th
_A
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
PP
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
RO
b)




PA.2 Given that Z 1  1  j 2 and Z 2  2  j 3, we have:
VE
Z1  Z2  3  j 1

Z 1  Z 2  1  j 5
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Z1 Z2  2  j 3  j 4  j 2 6  8  j 1

1  j2 2  j3  4  j 7
Z1 / Z2     0.3077  j 0.5385
2  j3 2  j3 13
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© 2018 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently
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
ST
Z 1  Z 2  10  j 25
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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
IA
PA.4 (a) Z a  5  j 5  7.071  45  7.071 exp  j 45 
Th d co of y th
_A
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
PP
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
RO
(b) Z b  10120   10 exp j 120    5  j 8.660
b)




(c) Zc  15  90   15 exp  j 90     j 15
VE
(d) Zd  1060   10 exp  j 120    5  j 8.660
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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

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(c) Zc  100e j 135  100 135   70.71  j 70.71





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© 2018 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.