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

5
analogy is a valve that does not allow fluid to pass through. This
corresponds to a closed valve. Thus, a closed valve is analogous to an
open switch.
P1.11*      A332 
t
dt
d
dt
t
dq
ti
P1.12 (a) The sine function completes one cycle for each2 radian increase in
the angle. Because the angle is,200
t one cycle is completed for each
time interval of 0.01 s. The sketch is:C0318.0
)200cos()200/10()200sin(10)((b) 005.0
0
005.0
0
005.0
0

 
t
dt
t
dt
ti
Q


C0
)200cos()200/10()200sin(10)((c) 01.0
0
01.0
0
01.0
0

 
t
dt
t
dt
ti
Q



P1.13*coulombs2|22)( 0
00
 




t
t
e
dt
e
dt
ti
Q
P1.14  A633
)(
)( 22
t
t e
e
dt
d
dt
t
dq
ti  
P1.15 The number of electrons passing through a cross section of the wire per
second issecondelectrons/10375.9
106.1
15 19
19 

 
N

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6
The volume of copper containing this number of electrons is310
29
19
m10375.9
10
10375.9
volume 



The cross sectional area of the wire is26
2
m10301.3
4


d
A 
Finally, the average velocity of the electrons ismm/s2840.0
volume 
Au
P1.16* The charge flowing through the battery iscoulombs10432)seconds360024()amperes5( 3

Q
and the stored energy isjoules10184.5)12()10432(Energy 63 
QV
(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 havekm6.17
8.930
10184.5Energy 6




mg
h
(b) Equating kinetic energy to stored energy and solving for velocity, we
havem/s9.587
Energy2 


m
v
(c) The energy density of the battery isJ/kg108.172
30
10184.5 3
6


which is about 0.384% of the energy density of gasoline.
P1.17coulombs40)seconds20()amperes2(timecurrent 
Qjoules200)10(20)(Energy 
QV
Because
iba is positive, if the current were carried by positive charge it
would be entering terminal
b. Electrons enter terminal
a. The energy is
taken from the element.

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7
P1.18joules104.149106.1gainselectronThe 1919  
P1.19*coulombs108.1)seconds000,36()amperes5(timecurrent 5

Qjoules1016.2)12()101.8(Energy 65 
QV
P1.20 If the current is referenced to flow into the positive reference for the
voltage, we say that we have the passive reference configuration. Using
double subscript notation, if the order of the subscripts are the same
for the current and voltage, we have a passive reference configuration.
P1.21* (a)
P -
vaia = 30 W Energy is being absorbed by the element.
(b)
P vbib = 30 W Energy is being absorbed by the element.
(c)
P -
vDEiED = -60 W Energy is being supplied by the element.
P1.22 The amount of energy isJ.30V)(10C)(3 
QV
W Because the
reference polarity is positive at terminal
a and the voltage value is
negative, terminal
b is actually the positive terminal. Because the charge
moves from the negative terminal to the positive terminal, energy is
removed from the device.
P1.23*C50V)(12J)(600 
V
w
Q .
To increase the chemical energy stored in the battery, positive charge
should move from the positive terminal to the negative terminal, in other
words from
a to
b. Electrons move from
b to
a.
P1.24      W20
t
e
titv
tp 


 
0
0 joules20|20)(Energy
t
e
dt
tp
The element absorbs the energy.

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8
P1.25W)200sin(50)(a)(
t
iv
tp ab
ab
J1592.0
)200cos()200/50()200sin(50)((b) 005.0
0
005.0
0
005.0
0

 
t
dt
t
dt
tp
w πππJ0
)200cos()200/50()200sin(50)((c) 01.0
0
01.0
0
01.0
0

 
t
dt
t
dt
tp
w



P1.26*kWh500
$/kWh0.12
$60
Rate
Cost
Energy W694.4
h2430
kWh500
Time
Energy 


PA787.5
120
4.694 
V
P
I%64.8%100
4.694
60
Reduction 
P1.27 (a)
P 60 W delivered to element A.
(b)A.elementfromtakenW60
P
(c)
P 60 W delivered to element
A.
P1.28* (a).elementfromtakenW60
A
P 
(b).elementtodeliveredW60
A
P 
(c).elementfromtakenW60
A
P 
P1.29 The power that can be delivered by the cell isW.12.0
vi
p In 75
hours, the energy delivered iskWhr.0.009Whr9 
pT
W Thus the
unit cost of the energy is$/kWhr56.55)009.0/()50.0( 
Cost which is
463 times the typical cost of energy from electric utilities.

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9
P1.30 The current supplied to the electronics isA.968.36.12/50/ 
v
p
i
The ampere-hour rating of the battery is the operating time to discharge
the battery multiplied by the current. Thus, the operating time ishours.2.25/100 
i
T
The energy delivered by the battery iskWh.1.26wh1260)2.25(50 
pT
W
Neglecting the cost of
recharging, the cost of energy for 300 discharge cycles is$/kWh.1984.0)26.1300/(75 
Cost
P1.31 A node is a point that joins two or more circuit elements. All points
joined by ideal conductors are electrically equivalent. Thus, there are
four nodes in the circuit at hand:
P1.32 The sum of the currents entering a node equals the sum of the currents
leaving.
P1.33 The currents in series-connected elements are equal.
P1.34 For a proper fluid analogy to electric circuits, the fluid must be
incompressible. Otherwise the fluid flow rate out of an element could be
more or less than the inward flow. Similarly the pipes must be inelastic
so the flow rate is the same at all points along each pipe.
P1.35* Elements
A and
B are in series. Also, elements
E and
F are in series.

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10
P1.36 (a) Elements
C and
D are in series.
(b) Because elements
C and
D are in series, the currents are equal in
magnitude. However, because the reference directions are opposite, the
algebraic signs of the current values are opposite. Thus, we haved
c
i
i  .
(c) At the node joining elements
A,
B, and
C, we can write the KCL
equationA413 
c
a
b
i
i
i . Also we found earlier thatA.1
c
d
i
i
P1.37* At the node joining elements A and B, we have.0
b
a
i
i Thus,A.2
ai
For the node at the top end of element C, we have3
c
b
i
i . Thus,A1
ci
. Finally, at the top right-hand corner node, we have.3
d
e
i
i 
Thus,A4
di . Elements
A and B are in series.
P1.38*findweKCL,ApplyingA.4andA,5A,3A,2givenareWe 
h
d
b
a
i
i
i
iA1
a
b
c
i
i
iA5
h
c
e
i
i
iA3
d
a
f
i
i
iA7
h
f
g
i
i
i
P1.39A.1andA,5A,3A,1givenareWe 
h
g
c
a
i
i
i
i Applying KCL, we findA2
a
c
b
i
i
iA4
h
c
e
i
i
iA7
a
f
d
i
i
iA6
h
g
f
i
i
i
P1.40 If one travels around a closed path adding the voltages for which one
enters the positive reference and subtracting the voltages for which one
enters the negative reference, the total is zero.
P1.41 (a) Elements
A and
B are in parallel.
(b) Because elements
A and
B are in parallel, the voltages are equal in
magnitude. However because the reference polarities are opposite, the
algebraic signs of the voltage values are opposite. Thus, we have.
b
a
v
v 
(c) Writing a KVL equation while going clockwise around the loop
composed of elements
A, C and
D, we obtain.0
c
d
a
v
v
v Solving forcv
and substituting values, we findV.7
cv Also we haveV.2
a
b
v
v

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11
P1.42* Summing voltages for the lower left-hand loop, we have,0105 
av
which yieldsV.5
av Then for the top-most loop, we have,015 
a
c
v
v
which yieldsV.10
cv Finally, writing KCL around the
outside loop, we have,05 
b
c v
v which yieldsV.5
bv
P1.43 We are givenV.6andV,10V,7V,5 
h
f
b
a
v
v
v
v Applying KVL, we
findV12
b
a
d
v
v
vV1
h
f
a
c
v
v
v
vV8
d
c
a
e
v
v
v
vV2
h
e
g
v
v
vV7
e
c
b
v
v
v
P1.44* Applying KCL and KVL, we haveA1
d
a
c
i
i
iA2
a
b
i
iV6
a
d
b
v
v
vV4
d
c
v
v
The power for each element isW20
a
a
A
iv
PW12
b
b
B
iv
PW4
c
c
C
iv
PW4
d
d
D
iv
P0Thus, 
D
C
B
A
P
P
P
P
P1.45 (a) In Figure P1.28, elements
C,
D, and
E are in parallel.
(b) In Figure P1.33, no element is in parallel with another element.
(c) In Figure P1.34, elements
C and
D are in parallel.
P1.46 The points and the voltages specified in the problem statement are:
Applying KVL to the loop
abca, substituting values and solving, we obtain:0
ac
cb
ab
v
v
v0155 
acvV10
acv

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12
Similiarly, applying KVL to the loop
abcda, substituting values and solving,
we obtain:0
da
cd
cb
ab
v
v
v
v010155 
cdvV20
cdv
P1.47 (a) The voltage between any two points of an ideal conductor is zero
regardless of the current flowing.
(b) An ideal voltage source maintains a specified voltage across its
terminals.
(c) An ideal current source maintains a specified current through itself.
P1.48 Four types of controlled sources and the units for their gain constants
are:
1. Voltage-controlled voltage sources. V/V or unitless.
2. Voltage-controlled current sources. A/V or siemens.
3. Current-controlled voltage sources. V/A or ohms.
4. Current-controlled current sources. A/A or unitless.
P1.49 Provided that the current reference points into the positive voltage
reference, the voltage across a resistance equals the current through
the resistance times the resistance. On the other hand, if the current
reference points into the negative voltage reference, the voltage equals
the negative of the product of the current and the resistance.
P1.50*

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13
P1.51
P1.52 The resistance of the copper wire is given byA
L
R Cu
Cu
 , and the
resistance of the tungsten wire isA
L
R W
W
 . Taking the ratios of the
respective sides of these equations yieldsCu
W
Cu
W R
R

 . Solving forWR
and substituting values, we haveΩ1.58
)1072.1()10(5.44(0.5) 88-




Cu
W
Cu
W
R
R ρρ
P1.53
P1.54
P1.55*   100
100
1002
1
2
1
P
V
R  powerinreduction19%aforW81
100
902
2
2
2

R
V
P

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14
P1.56 The power delivered to the resistor is)4exp(5.2/)()( 2
t
R
t
v
tp 
and the energy delivered isJ625.0
4
5.2
)4exp(
4
5.2
)4exp(5.2)(
000



 




t
dt
t
dt
tp
w
P1.57 The power delivered to the resistor is)4cos(25.125.1)2(sin5.2/)()( 22
t
t
R
t
v
tp

 
and the energy delivered is  J5.12)4sin(
4
25.1
25.1)4cos(25.125.1)(
10
0
10
0
10
0



  
t
t
dt
t
dt
tp
w



P1.58 Equation 1.10 gives the resistance asA
L
R ρ

(a) Thus, if the length of the wire is doubled, the resistance doubles to.1 
(b) If the diameter of the wire is doubled, the cross sectional area
A is
increased by a factor of four. Thus, the resistance is decreased by
a factor of four to 0.125.
P1.59 (a) The voltage across the voltage source is 10 V independent of the
current. Thus, we have
v  10 which plots as a vertical line in the
v
i
plane.
(b) The current source has
i  2 independent of
v, which plots as a
horizontal line in the
v
i plane.
(c) Ohm's law gives
i 
v/5.
(d) Applying Ohm's law and KVL, we obtain105 
i
v which is equivalent
to22.0 
v
i .
(e) Applying KCL and Ohm's law, we obtain obtain105 
i
v which is
equivalent to22.0 
v
i .

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15
The plots for all five parts are shown. (Parts d and e have the same plot.)
P1.60* (a) Not contradictory.
(b) A 2-A current source in series with a 3-A current source is
contradictory because the currents in series elements must be equal.
(c) Not contradictory.
(d) A 2-A current source in series with an open circuit is contradictory
because the current through a short circuit is zero by definition and
currents in series elements must be equal.
(e) A 5-V voltage source in parallel with a short circuit is contradictory
because the voltages across parallel elements must be equal and the
voltage across a short circuit is zero by definiton.
P1.61 The power for each element is 20 W. The current source delivers power
and the voltage source absorbs it.

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Electrical Engineering & Electronics

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

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