Solution Manual for Electrical Engineering: Principles and Applications, 7th Edition

Name: Solution Manual for Electrical Engineering: Principles and Applications, 7th Edition
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1
CHAPTER 1
Exercises
E1.1 Charge = Current  Time = (2 A)  (10 s) = 20 C
E1.2A)2cos(200)200cos(2000.010t)0.01sin(20(
)(
)(
t
t
dt
d
dt
t
dq
ti 
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 (a)2
20)()()(
t
tit
v
t
p
a
a
a J6667
3
20
3
20
20)(
310
0
310
0
10
0
2   
t
t
dt
t
dt
t
p
w a
a
(b) Notice that the references are opposite to the passive sign
convention. Thus we have:20020)()()( 
t
tit
v
t
p
b
b
bJ100020010)20020()( 10
0
2
10
0
10
0
  
t
t
dt
t
dt
t
p
w b
b

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 byA
L
R ρ
 . Using4/2
d
A  and
substituting values, we have:4/)106.1(
1012.1
6.9 23
6






L

L = 17.2 m
E1.12R
V
P 2
  144/2
P
V
R A833.0144/120/ 
R
V
I
E1.13R
V
P 2
 V8.15100025.0 
PR
VmA8.151000/8.15/ 
R
V
I
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 haveW25)1()25(22 
iv
PR
andW.25)1()25(11 
iv
Ps
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
passive sign configuration). Also we haveW.160
R
R
R
iv
P

3
Problems
P1.1 Broadly, the two objectives of electrical systems are:
1. To gather, store, process, transport and present information.
2. To distribute, store, and convert energy between various forms.
P1.2 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.
P1.3 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.

4
(c) The current through an open switch is zero. The voltage across the
switch can be any value depending on the circuit.
(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 anagolous to a frictionless pipe.
(b) An open switch is anagolous to a closed valve.
(c) A resistance is anagolous to a constriction in a pipe or to a pipe with
friction.
(d) A battery is analogous to a pump.
P1.718
19 1025.6
ectroncoulomb/el1060.1
coulomb/s1
secondperElectrons 

 
P1.8* The reference direction forabi points from
a to
b. Becauseabi 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,C.15s)(3A)5( 
Q
P1.9 The positive reference for
v is at the head of the arrow, which is
terminal
a. The positive reference for
vba is terminal
b. Thus, we haveV.12
v
vba
Also,
i is the current entering terminal
a, and
iba is the
current leaving terminal
a. Thus, we haveA.2
bai
i Thus, current
enters the positive reference and energy is being delivered to the device.
P1.10 To stop current flow, we break contact between the conducting parts of
the switch, and we say that the switch is open. The corresponding fluid

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Solution Manual for Electrical Engineering: Principles and Applications, 7th Edition

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