Solutions Manual — Electrical Engineering: Principles & Applications, 7th Edition (Hambley, 2017), Chapters 1-16 | All Chapters Covered

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Electrical Engineering: Principles &
UV Applications – 7th Edition
IA_
SOLUTIONS
AP
PR
MANUAL
O VE
D?
Allan R. Hambley



Comprehensive Solutions Manual for

Instructors and Students

© Allan R. Hambley

All rights reserved. Reproduction or distribution without permission is prohibited.




Created by MedConnoisseur ©2025/2026

,ST
TABLE OF CONTENTS
Electrical Engineering: Principles & Applications –
UV
7th Edition
Allan R. Hambley
IA_
Chapter 1. Introduction: Circuits, Currents, Voltages, Power and Energy,
AP
Kirchhoff’s Laws, Circuit Elements, and Circuit Basics
Chapter 2. Resistive Circuits
Chapter 3. Inductance and Capacitance
PR
Chapter 4. Transients (First and Second Order Circuits, RC, RL, etc.)
Chapter 5. Steady-State Sinusoidal (AC) Analysis
Chapter 6. Frequency Response, Bode Plots, Resonance, Filters
O
Chapter 7. Logic Circuits
Chapter 8. Computers, Microcontrollers, and Computer-Based Instrumentation
Systems
Chapter 9. Diodes
VE
Chapter 10. Amplifiers: Specifications and External Characteristics
Chapter 11. Field-Effect Transistors
D?
Chapter 12. Bipolar Junction Transistors
Chapter 13. Operational Amplifiers
Chapter 14. Magnetic Circuits and Transformers
Chapter 15. DC Machines
Chapter 16. AC Machines




Created by MedConnoisseur ©2025/2026

,ST APPENDIX A

UV Exercises

EA.1 Given Z 1  2  j 3 and Z 2  8  j 6, we have:
IA_ Z 1  Z 2  10  j 3

Z 1  Z 2  6  j 9
AP
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
PR
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




OV
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




ED
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  )




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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.

, ST
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  )
UV Z 1  Z 2  (10 30  )  (20135  )  (8.66  j 5)  (14.14  j 14.14)
IA_  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 
AP Problems

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




OV
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




ED
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
© 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.