Principles and Applications of Electrical Engineering 5th Edition Solution Manual

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 1
Chapter 1: Introduction to Electrical Engineering –
Instructor Notes
Chapter 1 is introductory in nature, establishing some rationale for studying electrical engineering methods,
even though the students' primary interest may lie in other areas. The material in this chapter should be included in
every syllabus, and can typically be thoroughly covered in a single-day introductory lecture. Oftentimes, reading of
this material is left up to the discretion of the student.
1.1

G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 1
Problem 1.1
Solution:
A few examples are:
Bathroom
ventilation fan
electric toothbrush
hair dryer
electric shaver
electric heater fan
Kitchen
microwave fan
microwave turntable
mixer
food processor
blender
coffee grinder
garbage disposal
ceiling fan
electric clock
exhaust fan
refrigerator compressor
dish washer
Utility Room
clothes washer
dryer
air conditioner
furnace blower
pump
Family Room
VCR drive
cassette tape drive
reel-to-reel tape drive
record turntable drive
computer fan
treadmill
Miscellaneous
lawn tools
power tools
1.2

G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 1
Problem 1.2
Solution:
Several examples are listed below for each system:
a) A ship
Circuit Analysis
design of the ship's electrical system
Electromagnetics
radar
Solid-State Electronics
radio
sonar
Electric Machines
pump
elevator
Electric Power Systems
lighting
generators
Digital Logic Circuits
elevator control
Computer Systems
navigation
Communication Systems
radio
telephone
Electro-Optics
Morse light
bridge displays
Instrumentation
compass
speed indicator
Control Systems
rudder
HVAC
b) A Commercial Passenger Aircraft
Circuit Analysis
Design of the plane's electrical system
Electromagnetics
radar
microwave oven
Solid-State Electronics
radio
Electric Machines
turbines
fans
Electric Power Systems
lighting
HVAC
Digital Logic Circuits
seat belts
Computer Systems
navigation
1.3

G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 1
Communication Systems
radio
telephone
Electro-Optics
cockpit displays
Instrumentation
compass
air speed indicator
inclinometer
altimeter
Control Systems
rudder
flaps
c) Household
Circuit Analysis
design of the home's electrical system
Electromagnetics
microwave oven
stereo speakers
Solid-State Electronics
television
stereo
VCR
Electric Machines
appliances
power tools
fans
Electric Power Systems
lighting
HVAC
receptacles
Digital Logic Circuits
clocks
timers
Computer Systems
microwave oven
programmable
VCR
Communication Systems
telephone
CB radio
television
radio
Electro-Optics
digital clocks
Instrumentation
electric meter
Control Systems
thermostat
1.4

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2
Chapter 2: Fundamentals of Electrical Circuits – Instructor Notes
Chapter 2 develops the foundations for the first part of the book. Coverage of the entire chapter would be
typical in an introductory course. The first four sections provide the basic definitions (Section 2.1) and cover
Kirchoff’s Laws and the passive sign convention (Sections 2.2, 2.3 And 2.4); A special feature, Focus on
Methodology: The Passive Sign Convention (p. 39) and two examples illustrate this very important topic. A second
feature, that will recur throughout the first six chapters, is presented in the form of sidebars. Make The Connection:
Mechanical Analog of Voltage Sources (p. 24) and Make The Connection: Hydraulic Analog of Current Sources (p.
26) present the concept of analogies between electrical and other physical domains.
Sections 2.5and 2.6 introduce the i-v characteristic and the resistance element. Tables 2.1 and 2.2 on p. 45
summarize the resistivity of common materials and standard resistor values; Table 2.3 on p. 48 provides the
resistance of copper wire for various gauges. The sidebar Make The Connection: Electric Circuit Analog of
Hydraulic Systems – Fluid Resistance (p. 44) continues the electric-hydraulic system analogy.
Finally, Sections 2.7 and 2.8 introduce some basic but important concepts related to ideal and non-ideal
current sources, and measuring instruments. In the context of measuring instruments, Chapter 2 introduces a third
feature of this book, the Focus on Measurements boxes, referenced in the next paragraph.
The Instructor will find that although the material in Chapter 2 is quite basic, it is possible to give an
applied flavor to the subject matter by emphasizing a few selected topics in the examples presented in class. In
particular, a lecture could be devoted to resistance devices, including the resistive displacement transducer of Focus
on Measurements: Resistive throttle position sensor (pp. 56-59), the resistance strain gauges of Focus on
Measurements: Resistance strain gauges (pp. 58-59), and Focus on Measurements: The Wheatstone bridge and
force measurements (pp. 59-60). The instructor wishing to gain a more in-depth understanding of resistance strain
gauges will find a detailed analysis in the references1.
Early motivation for the application of circuit analysis to problems of practical interest to the non-electrical
engineer can be found in the Focus on Measurements: The Wheatstone bridge and force measurements. The
Wheatstone bridge material can also serve as an introduction to a laboratory experiment on strain gauges and the
measurement of force (see, for example2). Finally, the material on practical measuring instruments in Section2.8b
can also motivate a number of useful examples.
The homework problems include a variety of practical examples, with emphasis on instrumentation.
Problem 2.51 illustrates analysis related to fuses; problems 2.65 relates to wire gauges; problem 2.70 discusses the
thermistor; problems 2.71 discusses moving coil meters; problems 2.79 and 2.80 illustrate calculations related to
strain gauge bridges; a variety of problems related to practical measuring devices are also included in the last
section. The 5th Edition of this book includes 26 new problems; some of the 4th Edition problems were removed,
increasing the end-of-chapter problem count from 66 to 80.
It has been the author's experience that providing the students with an early introduction to practical
applications of electrical engineering to their own disciplines can increase the interest level in the course
significantly.
Learning Objectives for Chapter 2
6. Identify the principal elements of electrical circuits: nodes, loops, meshes, branches, and voltage
and current sources.
7. Apply Kirchhoff’s Laws to simple electrical circuits and derive the basic circuit equations.
8. Apply the passive sign convention and compute power dissipated by circuit elements.
9. Apply the voltage and current divider laws to calculate unknown variables in simple series,
parallel and series-parallel circuits.
10. Understand the rules for connecting electrical measuring instruments to electrical circuits for the
measurement of voltage, current, and power.
2.1
1 E. O. Doebelin, Measurement Systems – Application and Design, 4th Edition, McGraw-Hill, New York, 1990.
2 G. Rizzoni, A Practical Introduction to Electronic Instrumentation, 3rd Edition, Kendall-Hunt, 1998.

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2
Section 2.1: Definitions
Problem 2.1
Solution:
Known quantities:
Initial Coulombic potential energy, ; initial velocity,UVi = 17kJ /C i = 93M m
s ; final Coulombic potential energy,
.V f = 6kJ /C
Find:
The change in velocity of the electron.
Assumptions:
∆PE g << ∆PE c
Analysis:
Using the first law of thermodynamics, we obtain the final velocity of the electron:
Q heat −W = ∆KE + ∆PE c + ∆PE g + ...
Heat is not applicable to a single particle. W=0 since no external forces are applied.
∆KE = −∆PE c
1
2 m e (U f
2 −U i2 ) = −Q e (V f − Vi )
U f
2 = U i2 − 2Qe
m e
(V f − Vi )
= 93 M m
s
⎛
⎝
⎜ ⎞
⎠
⎟
2
−
2 −1.6 × 10−19 C( )
9.11×10−37 g 6kV − 17kV( )
= 8.649 ×10 15 m2
s2 − 3.864 ×10 15 m2
s2
U f = 6.917 ×10 7 m
s
U f − U i = 93 M m
s − 69.17 M m
s = 23.83 M m
s .
2.2

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2
Problem 2.2
Solution:
Known quantities:
MKSQ units.
Find:
Equivalent units of volt, ampere and ohm.
Analysis:
Voltage = Volt = Joule
Coulomb V = J
C
Current = Ampere = Coulomb
second a = C
s
Resistance = Ohm = Volt
Ampere = Joule × second
Coulomb 2 Ω = J ⋅ s
C 2
Conductance = Siemens or Mho = Ampere
Volt = C 2
J ⋅ s
Problem 2.3
Solution:
Known quantities:
Battery nominal rate of 100 A-h.
Find:
a) Charge potentially derived from the battery
b) Electrons contained in that charge.
Assumptions:
Battery fully charged.
Analysis:
a)
100A ×1hr = 100 C
s
⎛
⎝
⎜ ⎞
⎠
⎟ 1hr( ) 3600 s
hr
⎛
⎝
⎜ ⎞
⎠
⎟ = 360000 C
b)
charge on electron: −1.602 × 10−19 C
no. of electrons =
360 ×10 3C
1.602 × 10−19C = 224.7 ×10 22
2.3

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2
Problem 2.4
Solution:
Known quantities:
Two-rate change charge cycle shown in Figure P2.4.
Find:
a) The charge transferred to the battery
b) The energy transferred to the battery.
Analysis:
a) To find the charge delivered to the battery during the charge
cycle, we examine the charge-current relationship:
i = dq
dt or dq = i ⋅ dt
thus:
Q = i(t)dtt0
t1∫
Q = 50mAdt0
5hrs
∫ + 20mAdt5hrs
10hrs
∫
= 0.05dt0
18000s
∫ + 0.02dt18000
36000s
∫
= 900 + 360
= 1260C
b) To find the energy transferred to the battery, we examine the energy relationship
p = dw
dt or dw = p(t)dt
w = p(t)dtt0
t1∫ = v(t)i(t)dtt0
t1∫
observing that the energy delivered to the battery is the integral of the power over the charge cycle. Thus,
w = 0.05(1 + 0.75t
18000) dt
0
18000
∫ + 0.02(1 + 0.25t
18000) dt
18000
36000
∫
= (0.05t + 0.75
36000 t 2 ) 0
18000 + (0.02t + 0.25
36000 t 2 ) 18000
36000
J1732.5=w
2.4

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2
Problem 2.5
Solution:
Known quantities:
Rated voltage of the battery; rated capacity of the battery.
Find:
a) The rated chemical energy stored in the battery
b) The total charge that can be supplied at the rated voltage.
Analysis:
a)
∆V ≡ ∆PE c
∆Q I = ∆Q
∆t
Chemical energy = ∆PE c = ∆V ⋅ ∆Q = ∆V ⋅ I ⋅ ∆t( )
= 12 V 350 A − hr 3600 s
hr = 15.12 MJ.
As the battery discharges, the voltage will decrease below the rated voltage. The remaining chemical energy stored
in the battery is less useful or not useful.
b) is the total charge passing through the battery and gaining 12 J/C of electrical energy.∆Q
∆Q = I ⋅ ∆t = 350 a hr = 350 C
s hr ⋅ 3600 s
hr = 1.26 MC.
Problem 2.6
Solution:
Known quantities:
Resistance of external circuit.
Find:
a) Current supplied by an ideal voltage source
b) Voltage supplied by an ideal current source.
Assumptions:
Ideal voltage and current sources.
Analysis:
a) An ideal voltage source produces a constant voltage at or below its rated current. Current is determined by the
power required by the external circuit (modeled as R).
IVP
R
V
I s
s ⋅==
b) An ideal current source produces a constant current at or below its rated voltage. Voltage is determined by the
power demanded by the external circuit (modeled as R).
V = I s ⋅ R P = V ⋅ I s
A real source will overheat and, perhaps, burn up if its rated power is exceeded.
2.5

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2
Problem 2.7
Solution:
Known quantities:
Rated discharge current of the battery; rated voltage of the battery; rated discharge time of the battery.
Find:
a) Energy stored in the battery when fully recharging
b) Energy stored in the battery after discharging
Analysis:
a) Energy = Power × time = 1A( ) 12V( ) 120hr( ) 60 min
hr
⎛
⎝
⎜ ⎞
⎠
⎟ 60 sec
min
⎛
⎝
⎜ ⎞
⎠
⎟
J105.184=w 6
×
b) Assume that 150 W is the combined power rating of both lights; then,
J104.32
sec3600
)8)(150(w 6
used ×=⎟
⎠
⎞
⎜
⎝
⎛
= hr
hrsW
J10864 3
×=−= usedstored www
Problem 2.8
Solution:
Known quantities:
Recharging current and recharging voltage
Find:
a) Total transferred charge
b) Total transferred energy
Analysis:
a)
C48,600)(4)(60)(60
2
1
4(90)(60))(2)(90)(60
2
1
6(30)(60))(4)(30)(60
2
1
curvetime-currentunder theareaQ
=++++=
∫== Idt
C48,600=Q
b) p
dt
dw = so ∫ ∫== vidtpdtw
s10800t0027A,t
3600
4
-12=i
s0027t1800A,t
5400
2
-6=i
s1800t0A,t
1800
4
-10=i
s10800t0V,t
10800
3
+9=v
3
2
1
≤≤
≤≤
≤≤
≤≤
2.6

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2
where 321 iiii ++=
Therefore,
33333
10800
7200
6
322
7200
1800
6
322
1800
0
6
322
10800
7200 3
7200
1800 2
1800
0 1
10456610648104105108380109132
10729200600
108
1016296001080
60
10864100720
90
××+××+×=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
−−++
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
−−++
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
−−+=
∫+∫+∫=
.-.-..
.
ttt
t
.
ttt
t
.
ttt
t
dtvidtvidtviw
kJ.Energy = 9489
Problem 2.9
Solution:
Known quantities:
Current-time curve
Find:
a) Amount of charge during 1st second
b) Amount of charge for 2 to 10 seconds
c) Sketch charge-time curve
Analysis:
a) 1
104 3 ti −
×=
Coulombs102
sec
amp
102
2
104104
0 33
1
0
2
31
0 31
1 −−−− ×=×=×=∫ ×=∫= t
tdtidtQ
b) The charge transferred from to t is the same as fromt = 1 = 2 t = 0 to t = 1. Q2 = 4 × 10−3 Coulombs
The charge transferred from to t is the same in magnitude at = 2 = 3 nd opposite in direction to that from t = 1
to t = 2 . Q3 = 2 ×10−3 Coulombs
t = 4
Q4 = 2 × 10−3 − 4 ×10−3dt3
4
∫ = 2 ×10−3 − 4 ×10−3 = −2 ×10−3 Coulombs
t = 5,6, 7
Q5 = −2 ×10−3 + 2 × 10−3
4
5
∫ dt = 0
Q6 = 0 + 2 ×10−3
5
6
∫ dt = 2 ×10−3 Coulombs
Q7 = 2 × 10−3 + 2 ×10−3
6
7
∫ dt = 4 × 10−3 Coulombs
t = 8,9,10s
Q = 4 × 10−3 Coulombs
2.7

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2
2.8
Problem 2.10
Known quantities:
urve and voltage-time curve of battery recharging
Total transferred energy
Ana
a)
Solution:
Current-time c
Find:
a) Total transferred charge
b)
lysis:
AmA 1.0100 =
- time curve = Idt∫ = 0.1( ) 2( ) 3600( ) + 3600( ) 0.1e - t-2( )/0.82 dt
2
6
∫ = 1,011 CQ = area under the current
Q = 1, 011 C
dw
dt = P so
( )
b)
( ) ( ) ( )( ) ( ) ( )( ) 8,114J0.1e936001.01.89e6.81360036003600 4
2
/0.822-t-
2
0
t/0.82
4
2
2
0
=∫+∫ +=∫+∫=∫ ∫= vidtPdtw = dtdtdtvidtvi
Energy = 8,114J
roblem 2.11
Known quantities:
urve and voltage-time curve of battery recharging
Total transferred energy
Ana
a)
P
Solution:
Current-time c
Find:
c) Total transferred charge
d)
lysis:
AmA 04.0=40
Q = area under the current - time curve = Idt∫ = 0.04( ) 6( ) 3600( ) = 864 C
C648=Q
b) P
dt
dw = so
./t--.-t/
1671
04030513600040450213600
360000
4
2
402
2
0
40
4
2
2
0
=
∫ −+∫ −=
∫+∫= (dtPdtw 36=∫ ∫= ) ( )
( ) ( )( ) ( ) ( )( )( )
J,
dt.e..dt.e..
dtvidtvivi
J,Energy 1671=

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2
2.9
Problem 2.12
Solution:
Known quantities:
Current-time curve and voltage-time curve of battery recharging
Find:
ed e
ed y
a
0
5t/12 =∫=∫= Idt
e) Total transferr charg
f) Total transferr energ
Analysis:
a)
( )12 -
areQ = C8,564dte3600curvetime-currentunder the
C8,564=Q
dw
dt = Pb) so
( ) ( )( )
8,986J
ee3123600 2
0
5t/12-5t/12-
=
∫ −=∫ ∫== dtvidtPdtw
Energy = 8,986J

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2
Section 2.2, 2.3 KCL, KVL
Problem 2.13
Solution:
Known quantities:
Circuit shown in Figure P2.13 with currents I 0 = −2 A,
A, A, and voltage sourceI 1 = −4 I S = 8 VS = 12V.
Find:
The unknown currents.
Analysis:
Applying KCL to node (a) and node (b):
I0 + I1 + I2 = 0
I0 + I S + I1 - I3 = 0
⎧
⎨
⎩ ⇒
I2 = −(I0 + I1 ) = 6 A
I3 = I0 + I S + I1 = 2 A
⎧
⎨
⎪
⎩
⎪
Problem 2.14
Solution:
Known quantities:
Circuit shown in Figure P2.14.
Find:
The unknown currents.
Analysis:
Applying KCL at the node: 0562 =++ --i
thus i = 3 A which means that a 3-A current is leaving the node.
Problem 2.15
Solution:
Known quantities:
Circuit shown in Figure P2.15.
Find:
The unknown currents.
Analysis:
Applying KCL at the node: 0256 =++ -i
thus i = -3 A which means that a 3-A current is leaving the node.
2.10

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