Solution Manual for Electrical Engineering: Principles & Applications, 7th Edition by Allan R. Hambley |9780137562855| All Chapters| LATEST 2026

SOLUTION MANUAL
ELECTRIACAL ENGINEERING PRINCIPLES &
APPLICATIONS

,CONTENTS

Chapter 1........................................................................................................................ 1

Chapter 2 ....................................................................................................................... 24

Chapter 3 ....................................................................................................................... 84

Chapter 4 ....................................................................................................................... 121

Chapter 5 ....................................................................................................................... 174

Chapter 6 ....................................................................................................................... 221

Chapter 7 ....................................................................................................................... 329

Chapter 8 ....................................................................................................................... 286

Chapter 9 ....................................................................................................................... 347

Chapter 10 ..................................................................................................................... 359

Chapter 11 ...................................................................................................................... 407

Chapter 12 ..................................................................................................................... 458

Chapter 13 ..................................................................................................................... 487

Chapter 14 ..................................................................................................................... 520

Chapter 15 ......................................................................................................................572

Chapter 16 ..................................................................................................................... 608

Chapter 17 ..................................................................................................................... 646

Appendix A .................................................................................................................... 685

Appendix C..................................................................................................................... 689

, CHAPTER 1

Exercises

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

Dq (T ) D
E1.2 I (T )   (0.01sin(200t)  0.01  200Cos(200t )  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 Direction To The Reference. Thus Positive Charge Moves
Upward In Element E.

E1.4 Energy = Charge  Voltage = (2 C)  (20 V) = 40 J

Because Vab Is Positive, The Positive Terminal Is A And The Negative
Terminal Is B. Thus The Charge Moves From The Negative Terminal To
The Positive Terminal, And Energy Is Removed From The Circuit
Element.

E1.5 Iab Enters Terminal A. Furthermore, Vab Is Positive At Terminal A.
Thus The Current Enters The Positive Reference, And We Have The
Passive Reference Configuration.

E1.6 Pa (T )  Va (T )Ia (T )  20t 2
(A
)
10 10
2 20t 3 10 20t 3
Wa   Pa (T   20t dt    6667 J
0 3 0 3
)Dt
0

(B) Notice That The References Are Opposite To The Passive Sign
Convention. Thus We Have:

Pb (T )  Vb (T )Ib (T )  20Wt  200

10 (T )Dt P 10

1

, (20t  200)Dt 2
1010
t0
200

 Jt 
100
B  B  0
0 0




2

,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

E1.10 Elements E And F Are In Parallel; Elements A And B Are In Series.

E1.11 The Resistance Of A Wire Is Given By R . Using A  D 2 / And
Ρ
 4
A
Substituting Values, We Have:

1.12  106  L
9.6   L = 17.2 M
 (1.6  103 )

E1.12 P V 2 R  R V 2 / P  144   I V /R  120 /144  0.833 A

E1.13 P V 2 R  V  PR  0.25  1000  15.8 V
I V / R  15.8 /1000  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
PR  V2i2  (25)  (1)  25 W And Ps  V1i1  (25)  (1)  25 W.

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

3

,Passive Sign Configuration). Also We Have PR  Vrir  160 W.




4

, Problems

P1.1 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.
4. To Be Able To Communicate Effectively With Electrical Engineers.

P1.2 Broadly, The Two Objectives Of Electrical Systems Are:
1. To Gather, Store, Process, Transport, And Display Information.
2. To Distribute, Store, And Convert Energy Between Various Forms.

P1.3 Eight Subdivisions Of EE Are:

1. Communication Systems.
2. Computer Systems.
3. Control Systems.
4. Electromagnetics.
5. Electronics.
6. Photonics.
7. Power Systems.
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.
(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 Units Of Volts, Which Are Equivalent To Joules Per
Coulomb.
(C) The Current Through An Open Switch Is Zero. The Voltage Across
The Switch Can Be Any Value Depending On The Circuit.



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, (d) The Voltage Across A Closed Switch Is Zero. The Current Through
The Switch Can Be Any Value Depending Of The Circuit.
(e) Direct Current Is Constant In Magnitude And Direction With Respect
To Time.
(f) Alternating Current Varies Either In Magnitude Or Direction With Time.

P1.6 (A) A Conductor Is Analogous To A Frictionless Pipe.
(B) An Open Switch Is Analogous To A Closed Valve.
(C) A Resistance Is Analogous To A Constriction In A Pipe Or To A Pipe
With Friction.
(D) A Battery Is Analogous To A Pump.

P1.7* The Reference Direction For Iab Points From A To B. Because Iab
Has A Negative Value, The Current Is Equivalent To Positive
Charge Moving Opposite To The Reference Direction. Finally, Since
Electrons Have
Negative Charge, They Are Moving In The Reference Direction (I.E., From A
To B).
For A Constant (Dc) Current, Charge Equals Current Times The Time
Interval. Thus, Q  (3 A)  (3 S)  9 C.

2 Coulomb/S
P1.8 Electrons Per Second   12.5  1018
1.60  1019 Coulomb/Electron
Dq T  D
P1.9* I T    2t T 2   2  2t A
Dt Dt

P1.10 The Positive Reference For V Is At The Head Of The Arrow, Which
Is Terminal B. The Positive Reference For Vba Is Terminal B. Thus,
We Have
Vba  V  10 Also, I Is The Current Entering Terminal A, And Iba Is The
V.
Current Leaving Terminal A. Thus, We   3 A. The True
Have I Iba
Polarity Is Positive At Terminal A, And The True Current Direction Is
Entering Terminal A. Thus, Current Enters The Positive Reference And
Energy Is Being Delivered To The Device.

P1.11 To Cause Current To Flow, We Make Contact Between The Conducting

6

,Parts Of The Switch, And We Say That The Switch Is Closed. The
Corresponding Fluid Analogy Is A Valve That Allows Fluid To Pass
Through. This Corresponds To An Open Valve. Thus, An Open Valve Is
Analogous To A Closed




7

, Switch.

 

P1.12* Q   I (T  2e  2e T |  2 Coulombs
0
)Dt dt
T
0
0



P1.13 (A) The Sine Function Completes One Cycle For Each 2 Radian Increase In
The Angle. Because The Angle Is One Cycle Is Completed For Each
200T , Time Interval Of 0.01 S. The
Sketch Is:




(B) Q )0.01
Dt (T  0.01 )Dt
T  ( ) Cos(200T ) 0.01
I
0
 10 Sin(200
0
0

0 C

(B) Q 0.015
)Dt (T  0.015 )Dt
T  ( ) Cos(200T ) 0.015
I
0
 10 Sin(200
0
0

 0.0318 C


P1.14* The Charge Flowing Through The Battery Is
Q  (5 Amperes)  (24  3600 Seconds)  432  103 Coulombs
The Stored Energy Is
Energy  QV  (432  103 )  (12)  5.184  106 Joules
(a) Equating Gravitational Potential Energy, Which Is Mass Times Height
Times The Acceleration Due To Gravity, To The Energy Stored In The
Battery And Solving For The Height, We Have
Energy 5.184  106
H   17.6 Km
Mg 30  9.8
(b) Equating Kinetic Energy To Stored Energy And Solving For Velocity,
We Have

8