Solutions for Electrical Engineering: Principles & Applications, 7th edition by Hambley, All Chapters

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APPENDIX A




R
IF
Exercises




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




D
BR
Z 1  Z 2  10  j 3




AI
Z 1  Z 2  6  j 9




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




BO
2  j 3 8  j 6 16  j 12  j 24  j 2 18
Z1 / Z2   0.02  j 0.36




O
 
8 j6 8 j6 100
Th d co of y th
is is p urs an e
an eir le tro




ST
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




ER
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  )

Notice that Z2 lies in the second quadrant of the complex plane.
Z 2  10  j 10  10 2  10 2 (180   arctan( ))
 14.14 135   14.14 exp( j 135  )




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




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




IE
Z 1  Z 2  (10 30  )  (20135  )  (8.66  j 5)  (14.14  j 14.14)




D
 22.8  j 9.14  24.6  21.8




BR
Z 1  Z 2  (10 30  )  (20135  )  (8.66  j 5)  (14.14  j 14.14)




AI
 5.48  j 19.14  19.9106 




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BO
Problems

PA.1 Given Z 1  2  j 3 and Z 2  4  j 3, we have:




O
Th d co of y th
is is p urs an e
an eir le tro




ST
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




ER
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

1  j2 2  j3  4  j 7
Z1 / Z2     0.3077  j 0.5385
2  j3 2  j3 13




2
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VERIFIEDBRAINBOOSTER
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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




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IF
Z 1  Z 2  30  j 15




IE
Z 1  Z 2  10  j 25




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




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




N
BO
PA.4 (a) Z a  5  j 5  7.071  45  7.071 exp  j 45 




O
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 




ST
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  




ER
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.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





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




3
© 2018 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently
VERIFIEDBRAINBOOSTER
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

, VE
R
(d) Zd  6e j 90  690   j 6





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




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




BR
1045  1045 




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




N
15




BO
(d) Zd   3  90    j 3
590 




O
Th d co of y th
is is p urs an e
an eir le tro




ST
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




ER
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)




4
© 2018 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently
VERIFIEDBRAINBOOSTER
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.