Complete Solution Manual: Electrical Engineering Principles and Applications. 7 th Edition. By Allan R. Hambley

Complete Solution Manual: Electrical Engineering Principles and
Applications. 7th Edition. By Allan R. Hambley


APPENDIX A

Exercises

EA.1 Given Z1  2  j 3 and Z2  8  j 6, we have:

Z1  Z2  10  j 3

Z1  Z2  6  j 9

Z1 Z2  16  j 24  j 12  j 218  34  j 12


Z / Z  2  j 3  8  j 6  16  j 12  j 24  j 18  0.02  j 0.36
2

1 2
8 j6 8 j6 100

EA.2 Z1  1545∘  15 cos(45∘ )  j 15 sin(45∘ )  10.6  j 10.6
Z2  10  150∘  10 cos(150∘ )  j 10 sin(150∘ )  8.66  j 5
Z3  590∘  5cos(90∘ )  j 5 sin(90∘ )  j 5

EA.3 Notice that Z1 lies in the first quadrant of the complex plane.
Z1  3  j 4  32  42 arctan(4 /3)  553.13∘

Notice that Z2 lies on the negative imaginary axis.
Z2  j 10  10  90∘

Notice that Z3 lies in the third quadrant of the complex plane.
Z3  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.
Z1  10  j 10  102  102  arctan(10 /10)  14.1445∘  14.14 exp( j 45∘ )

Notice that Z2 lies in the second quadrant of the complex plane.
Z2  10  j 10 
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,102  102 (180 ∘  arctan(10 /10))
 14.14135∘  14.14 exp( j 135∘ )




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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 Z1Z2  (1030∘ )(20135∘ )  (10  20)(30∘  135∘ )  200(165∘ )

Z1 / Z2  (1030∘ ) /(20135∘ )  (10 /20)(30∘  135∘ )  0.5(105∘ )

Z1  Z2  (1030∘ )  (20135∘ )  (8.66  j 5)  (14.14  j 14.14)
 22.8  j 9.14  24.6  21.8∘

Z1  Z2  (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 Z1  2  j 3 and Z2  4  j 3, we have:

Z1  Z2  6  j 0

Z1  Z2  2  j 6

Z1 Z2  8  j 6  j12  j 2 9  17  j 6

Z / Z  2  j 3  4  j 3   1  j 18  0.04  j 0.72
1 2
4  j3 4  j3 25



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

Z1  Z2  3  j1

Z1  Z2  1  j 5

Z1 Z2  2  j 3  j 4  j 2 6  8  j1

1  j2 2  j3  4  j7
Z /Z     0.3077  j 0.5385
1 2
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 Z1  10  j 5 and Z2  20  j 20, we have:

Z1  Z2  30  j 15

Z1  Z2  10  j 25

Z1 Z2  200  j 200  j100  j 2100  300  j100

Z / Z  10  j 5  20  j 20  100  j 300  0.125  j 0.375
1 2
20  j 20 20  j 20 800



PA.4 (a) Za  5  j 5  7.071  45∘  7.071 exp j 45∘ 

(b) Zb  10  j 5  11.18153.43∘  11.18 expj153.43∘ 

(c) Zc  3  j 4  5  126.87∘  5exp j126.87∘ 

(d) Zd   j12  12  90∘  12 exp j 90∘ 



PA.5 (a) Za  545∘  5 expj 45∘   3.536  j 3.536

(b) Zb  10120∘  10 expj120∘   5  j 8.660

(c) Zc  15  90∘  15 exp j 90∘   j15

(d) Zd  1060∘  10 exp j120∘   5  j 8.660



PA.6 (a) Za  5e j30  530∘  4.330  j 2.5
∘




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

∘
(c) Zc  100e j135  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.