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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th 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 references 1.
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 example 2). 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
1.
2.
3.
4.
5.
1
2
Identify the principal elements of electrical circuits: nodes, loops, meshes, branches, and voltage
and current sources.
Apply Kirchhoff’s Laws to simple electrical circuits and derive the basic circuit equations.
Apply the passive sign convention and compute power dissipated by circuit elements.
Apply the voltage and current divider laws to calculate unknown variables in simple series,
parallel and series-parallel circuits.
Understand the rules for connecting electrical measuring instruments to electrical circuits for the
measurement of voltage, current, and power.
E. O. Doebelin, Measurement Systems – Application and Design, 4th Edition, McGraw-Hill, New York, 1990.
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, 6th Edition Problem solutions, Chapter 2
Section 2.1: Definitions
Problem 2.1
An isolated free electron is traveling through an electric field from some initial point where its coulombic potential
energy per unit charge (voltage) is 17 kJ/C and velocity = 93 Mm/s to some final point where its coulombic potential
energy per unit charge is 6 kJ/C. Determine the change in velocity of the electron. Neglect gravitational forces.
Solution:
Known quantities:
Initial Coulombic potential energy, Vi = 17kJ /C ; initial velocity, U i = 93M
V f = 6kJ /C .
m
; final Coulombic potential energy,
s
Find:
The change in velocity of the electron.
Assumptions:
∆PEg << ∆PEc
∆KE = −∆PEc
1
me (U 2f − U i2 ) = −q e (V f − Vi )
2
2q
U 2f = U i2 − e (V f − Vi )
me
(
)
2 − 1.6 × 10 −19 C
m

(6kV − 17kV )
=  93M  −
s
9.11 × 10 31 kg

2
2
m2
15 m
= 8.649 × 10 2 − 3.864 × 10 2
s
s
15
m
s
m
m
m
U f −U i = 93 M − 69.17 M
= 23.83 M .
s
s
s
U f = 6.917 ×10 7
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
Problem 2.2
The unit used for voltage is the volt, for current the ampere, and for resistance the ohm. Using the definitions of
voltage, current, and resistance, express each quantity in fundamental MKS units.
Solution:
Known quantities:
MKSQ units.
Find:
Equivalent units of volt, ampere and ohm.
Analysis:
Joule
J
V =
Coulomb
C
Coulomb
C
Current = Ampere =
A=
second
s
Volt
Joule × second J ⋅ s
Resistance = Ohm =
=
= 2
Ampere
Coulomb 2
C
Voltage = Volt =
Conductance = Siemens=
Ampere C 2
=
Volt
J ⋅s
Problem 2.3
A battery can deliver 2.7 · 106 coulombs of charge. Find the capacity and delivered electrons of the battery.
Solution:
Known quantities:
qBattery = 2.7 · 106 C.
Find:
a) The current capacity of the battery in ampere-hours
b) The number of electrons that can be delivered.
Analysis:
a)
There are 3600 seconds in one hour. Amperage is defined as 1 Coulomb per second and is directly
proportional to ampere-hours.
1ℎ𝑟𝑟
2.7 ∙ 106 𝐶𝐶 ∙
= 750 𝐴𝐴ℎ
3600𝑠𝑠
b) The charge of a single electron is -1.602·10-19 C. The negative sign indicates that the particle delivered will
be an electron. Simple division gives the solution:
25
2.7 ∙ 106 𝐶𝐶
�1.602 ∙ 10−19 𝐶𝐶 = 1.685 ∙ 10 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
1 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
Problem 2.4
A three rate charge cycle is used to charge a battery. The current to the battery is held at 30 mA for 6 h and then
switched to 20 mA for the next 3 h.
Solution:
Known quantities:
See Figure P2.4
Find:
a) The total charge transferred to the battery.
b) The energy transferred to the battery.
Analysis:
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶
Current is equal to
, therefore given the current and a duration of that current, the transferred
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆
charge can be calculated by the following equation:
𝐴𝐴 ∙ 𝑡𝑡 = 𝐶𝐶
The two durations should be calculated independently and then added together.
0.030𝐴𝐴 ∙ 21600𝑠𝑠 = 648𝐶𝐶
0.020𝐴𝐴 ∙ 10800𝑠𝑠 = 216𝐶𝐶
648𝐶𝐶 + 216𝐶𝐶 = 𝟖𝟖𝟖𝟖𝟒𝟒𝑪𝑪
b) P=V·I, therefore, an equation for power can be found by multiplying the two graphs together.
First separate the voltage graph into three equations:
−6
0 h → 3 h :𝑉𝑉 = 9.26 · 10 𝑡𝑡 + 0.5
−5
3 h → 6 h :𝑉𝑉 = 5.55 ∙ 10 𝑡𝑡
−4
6 h → 9 h :𝑉𝑉 = 1.11 ∙ 10 𝑡𝑡 − 1.9
Next, multiply the first two equations by 0.03A and the third by 0.02A.
−7
0 h → 3 h :𝑃𝑃 = 2.77 ∙ 10 𝑡𝑡 + 0.015
−6
3 h → 6 h :𝑃𝑃 = 1.66 ∙ 10 𝑡𝑡
−6
6 h → 9 h :𝑃𝑃 = 2.22 ∙ 10 𝑡𝑡 − 0.038
Finally, since Energy is equal to the integral of power, take the integral of each of the equations for their
specified times and add them together.
2.77∙10−7 𝑡𝑡 2
10800
+ 0.015𝑡𝑡� |
= 178.2 J
0 h → 3 h :𝐸𝐸 = �
2
0
1.66∙10−6 𝑡𝑡 2 21600
�|
3 h → 6 h :𝐸𝐸 = �
2
10800= 290.43 J
2.22∙10−6 𝑡𝑡 2
32400
+ 0.038𝑡𝑡� |
= 1057.8 J
6 h → 9 h :𝐸𝐸 = �
2
21600
𝑬𝑬𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝑱𝑱
a)
Problem 2.5
Batteries (e.g., lead-acid batteries) store chemical energy and convert it to electric energy on demand. Batteries do
not store electric charge or charge carriers. Charge carriers (electrons) enter one terminal of the battery, acquire
electrical potential energy, and exit from the other terminal at a lower voltage. Remember the electron has a negative
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
charge! It is convenient to think of positive carriers flowing in the opposite direction, that is, conventional current,
and exiting at a higher voltage. All currents in this course, unless otherwise stated, are conventional current.
(Benjamin Franklin caused this mess!) For a battery with a rated voltage = 12 V and a rated capacity = 350 A-h,
determine
a. The rated chemical energy stored in the battery.
b. The total charge that can be supplied at the rated
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)
∆PEc
∆Q
I=
∆Q
∆t
Chemical energy = ∆PEc = ∆V ⋅ ∆Q = ∆V ⋅ (I ⋅ ∆t )
∆V ≡
s
= 15.12 MJ.
hr
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) ∆Q is the total charge passing through the battery and gaining 12 J/C of electrical energy.
= 12 V 350 A − hr 3600
C
s
∆Q = I ⋅ ∆t = 350 Ahr = 350 hr ⋅3600 = 1.26MC.
s
hr
Problem 2.6
What determines the following?
a. How much current is supplied (at a constant voltage) by an ideal voltage source.
b. How much voltage is supplied (at a constant current) by an ideal current source.
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).
V
I = s P = Vs ⋅ I
R
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
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 = Is ⋅ R P = V ⋅ Is
A real source will overheat and, perhaps, burn up if its rated power is exceeded.
Problem 2.7
An automotive battery is rated at 120 A-h. This means that under certain test conditions it can output
1 A at 12 V for 120 h (under other test conditions, the battery may have other ratings).
a. How much total energy is stored in the battery?
b. If the headlights are left on overnight (8 h), how much energy will still be stored in the battery in the
morning? (Assume a 150-W total power rating for both headlights together.)
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 charged
b) Energy stored in the battery after discharging
Analysis:
 60 min  60 sec 
a) Energy = Power × time = (1A)(12V )(120hr )


 hr  min 
w = 5.184 × 10 6 J
b) Assume that 150 W is the combined power rating of both lights; then,
 3600 sec 
6
w used = (150 W )(8hrs )
 = 4.32 × 10 J
 hr 
w stored = w − wused = 864 × 10 3 J
Problem 2.8
A car battery kept in storage in the basement needs recharging. If the voltage and the current provided by the charger
during a charge cycle are shown in FigureP2.8,
a. Find the total charge transferred to the battery.
b. Find the total energy transferred to the battery.
Solution:
Known quantities:
Recharging current and recharging voltage
Find:
a) Total transferred charge
b) Total transferred energy
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
Analysis:
Q = area under the current - time curve = ∫ Idt
a)
=
1
1
1
(4)(30)(60) + 6(30)(60) + (2)(90)(60) + 4(90)(60) + (4)(60)(60) = 48,600 C
2
2
2
Q = 48,600 C
b)
dw
=p
dt
w = ∫ pdt = ∫ vidt
3
v=9+
t V, 0 ≤ t ≤ 10800 s
10800
4
i1 = 10 t A, 0 ≤ t ≤ 1800 s
1800
2
i2 = 6 t A, 1800 ≤ t ≤ 7200 s
5400
4
i 3 = 12 t A, 7200 ≤ t ≤ 10800 s
3600
where i = i1 + i 2 + i3
Therefore,
1800
w=∫
0
vi1 dt + ∫
7200
1800
10800
vi2 dt + ∫
7200

t2
t2
t3
=  90t +
−
−
720 100 4.86 × 10 6

vi3 dt
1800


0
7200

t2
t2
t3

+  54t +
−
−
1200 600 29.16 × 10 6



 1800

t2
t2
t3

+ 108t +
−
−
600 200 9.72 × 10 6

10800


 7200
= 132.9 × 10 3 + 332.8 × 10 3 -94.3 × 10 3 + 648 × 10 3 -566.4 × 10 3
Energy = 453 kJ
Problem 2.9
Suppose the current flowing through a wire is given
by the curve shown in Figure P2.9.
a. Find the amount of charge, q, that flows through
the wire between t1 = 0 and t2 = 1 s.
b. Repeat part a for t2 = 2, 3, 4, 5, 6, 7, 8, 9, and 10 s.
c. Sketch q(t) for 0 ≤ t ≤ 10 s.
Solution:
Known quantities:
Current-time curve
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
Find:
a) Amount of charge during 1st second
b) Amount of charge for 2 to 10 seconds
c) Sketch charge-time curve
Analysis:
a) i =
4×10 −3 t
1
t2
Q1 = ∫ idt = ∫ 4 × 10 tdt = 4 × 10
0
0
2
1
1
−3
1
−3
= 2 × 10 −3 A ⋅ s = 2 × 10 −3 Coulombs
0
b) The charge transferred from t = 1 to t = 2 is the same as from t = 0 to t = 1 .
Q2 = 4 × 10−3 Coulombs
The charge transferred from t = 2 to t = 3 is the same in magnitude and opposite in direction to that from t = 1
to t = 2 .
t=4
Q3 = 2 × 10−3 Coulombs
Q4 = 2 × 10−3 − ∫ 344 × 10−3 dt = 2 × 10−3 − 4 × 10−3 = −2 × 10−3 Coulombs
t = 5, 6, 7
Q5 = −2 × 10−3 + ∫ 45 2 × 10−3dt = 0
Q6 = 0 + ∫ 56 2 × 10−3dt = 2 × 10−3 Coulombs
Q7 = 2 × 10−3 + ∫ 67 2 × 10−3dt = 4 × 10−3 Coulombs
t = 8, 9,10s
Q = 4 × 10−3Coulombs
In Figure P2.10 v1(t) appears twice and should be v2(t) for the second
hour of the voltage graph
Problem 2.10
A two rate charge cycle is used to charge a battery. The current to the battery is held at 70 mA for 1 hour and then
switched to 60 mA for the next hour.
Solution:
Known quantities:
See Figure P2.10
Find:
a) The total charge transferred to the battery.
b) The energy transferred to the battery.
Analysis:
a)
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶
Current is equal to
, therefore given the current and a duration of that current, the transferred
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆
charge can be calculated by the following equation:
𝐴𝐴 ∙ 𝑡𝑡 = 𝐶𝐶
The two durations should be calculated independently and then added together.
0.070𝐴𝐴 ∙ 3600𝑠𝑠 = 252𝐶𝐶
0.060𝐴𝐴 ∙ 3600𝑠𝑠 = 216𝐶𝐶
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
252𝐶𝐶 + 216𝐶𝐶 = 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒
b) P=V·I, therefore, an equation for power can be found by multiplying the two graphs together.
First separate the voltage graph into two equations:
𝑡𝑡�
0 h → 1 h :𝑉𝑉 = 5 + 𝑒𝑒 5194.8 𝑉𝑉
4
4
𝑡𝑡
1 h → 2 h :𝑉𝑉 = �6 − 𝑒𝑒 1−1� + 𝑒𝑒 2−𝑒𝑒 1 ∗ 𝑒𝑒
Next, multiply the first equation by 0.07A and the second by 0.06A.
𝑡𝑡�
0 h → 1 h :𝑃𝑃 = 0.35 + 0.07𝑒𝑒 5194.8
4
4
𝑡𝑡/3600
𝑉𝑉
1 h → 2 h :𝑃𝑃 = 0.06 �6 − 𝑒𝑒 1ℎ−1� + 0.06 𝑒𝑒 2ℎ−𝑒𝑒 1ℎ ∗ 𝑒𝑒
Finally, since Energy is equal to the integral of power, take the integral of each of the equations for their
specified times and add them together.
𝑡𝑡�
3600
0 h → 1 h :𝐸𝐸 = �0.35𝑡𝑡 + 363.64𝑒𝑒 5194.8 � | 0 = 1623.53 J
𝑡𝑡/3600 7200
�|
1 h → 2 h :𝐸𝐸 = �0.22𝑡𝑡 + 184.98𝑒𝑒
3600= 1656 J
𝑬𝑬𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻 = 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑. 𝟓𝟓𝟓𝟓 𝑱𝑱
Problem 2.11
The charging scheme used in Figure P2.11 is an example of a constant-current charge cycle. The
charger voltage is controlled such that the current into the battery is held constant at 40 mA, as shown in
Figure P2.11. The battery is charged for 6 h. Find:
a. The total charge delivered to the battery.
b. The energy transferred to the battery during the
charging cycle.
Hint: Recall that the energy, w, is the integral of power, or P = dw/dt.
Solution:
Known quantities:
Current-time curve and voltage-time curve of battery recharging
Find:
a) Total transferred charge
b) Total transferred energy
Analysis:
a) 40mA = 0.04 A
Q = area under the current - time curve = ∫ Idt = (0.04)(6)(3600) = 864 C
Q = 864 C
b)
dw
=P
dt
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
2
6
0
2
w = ∫ Pdt = ∫ vidt = (3600 )∫ vidt + (3600 )∫ vidt
2
(
)
6
(
)
= (3600 )∫ 1.2 − 0.45e -t/ 0 .4 (0.04 )dt + (3600 )∫ 1.5 − 0.3e - (t- 2 ) / 0 .4 (0.04 )dt
0
2
= 1,167 J
Energy = 1,167 J
Problem 2.12
The charging scheme used in Figure P2.12 is called
a tapered-current charge cycle. The current starts at
the highest level and then decreases with time for the
entire charge cycle, as shown. The battery is charged
for 12 h. Find:
a. The total charge delivered to the battery.
b. The energy transferred to the battery during the
charging cycle.
Hint: Recall that the energy, w, is the integral of power, or
P = dw/dt.
Solution:
Known quantities:
Current-time curve and voltage-time curve of battery recharging
Find:
a) Total transferred charge
b) Total transferred energy
Analysis:
12
a) Q = area under the current - time curve =
-5t/12
∫ Idt = (3600)∫ e dt = 8,582 C
0
Q = 8,582 C
b)
dw
=P
dt
12
(
)(
)
w = ∫ Pdt = ∫ vidt = (3600 )∫ 12 − 3e -5t/12 e -5t/12 dt
0
= 90,022J
Energy = 90,022J
Problem 2.13
Use KCL to determine the unknown currents in Figure P2.13.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
Solution:
Known quantities:
io = 2 A, i2 = -7 A
Find:
a) i1
b) i3
Analysis:
a)
Use KCL at the node between Ro ,R1, and R2.
𝑖𝑖0 − 𝑖𝑖1 + 𝑖𝑖2 = 0
𝑖𝑖1 = 𝑖𝑖0 + 𝑖𝑖2
𝒊𝒊𝟏𝟏 = −𝟓𝟓𝟓𝟓
b) Use KCL at the node between R2 ,R3, and the current source.
6𝐴𝐴 + 𝑖𝑖3 − 𝑖𝑖2 = 0
𝑖𝑖3 = 𝑖𝑖2 − 6𝐴𝐴
𝒊𝒊𝟑𝟑 = −𝟏𝟏𝟏𝟏𝟏𝟏
Problem 2.14
Use KCL to determine the unknown currents in Figure P2.14.
Solution:
Known quantities:
ia = 3 A, ib = -2 A, ic = 1 A, id = 6 A, ie = -4 A
Find:
a) i1
b) i2
Analysis:
a)
Use KCL at Node A.
𝑖𝑖1 + 𝑖𝑖𝑏𝑏 − 𝑖𝑖𝑎𝑎 − 𝑖𝑖𝑐𝑐 = 0
𝑖𝑖1 = 𝑖𝑖𝑎𝑎 − 𝑖𝑖𝑏𝑏 + 𝑖𝑖𝑐𝑐
𝒊𝒊𝟏𝟏 = 𝟔𝟔𝟔𝟔
b) Use KCL at Node B.
𝑖𝑖𝑒𝑒 + 𝑖𝑖𝑑𝑑 − 𝑖𝑖1 − 𝑖𝑖2 = 0
𝑖𝑖2 = 𝑖𝑖𝑒𝑒 + 𝑖𝑖𝑑𝑑 − 𝑖𝑖1
2.11
𝒊𝒊𝟐𝟐 = −𝟒𝟒𝟒𝟒
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Problem 2.15
Use KCL to determine the unknown currents in Figure P2.15.
Solution:
Known quantities:
ia = 2 mA,
ib = 7 mA, ic = 4 mA
Find:
a) i1
b) i2
c) i3
Analysis:
a)
Use KCL at Node A.
𝑖𝑖𝑏𝑏 − 𝑖𝑖𝑎𝑎 − 𝑖𝑖1 = 0
𝑖𝑖1 = 𝑖𝑖𝑏𝑏 − 𝑖𝑖𝑎𝑎
𝒊𝒊𝟏𝟏 = 𝟓𝟓𝟓𝟓𝟓𝟓
b) Use KCL at Node C.
𝑖𝑖𝑐𝑐 − 𝑖𝑖2 − 𝑖𝑖𝑏𝑏 = 0
𝑖𝑖2 = 𝑖𝑖𝑐𝑐 − 𝑖𝑖𝑏𝑏
c)
𝒊𝒊𝟐𝟐 = −𝟑𝟑𝟑𝟑𝟑𝟑
Use KCL at Node D.
𝑖𝑖3 + 𝑖𝑖𝑎𝑎 − 𝑖𝑖𝑐𝑐 = 0
𝑖𝑖3 = 𝑖𝑖𝑐𝑐 − 𝑖𝑖𝑎𝑎
𝒊𝒊𝟑𝟑 = 𝟐𝟐𝟐𝟐𝟐𝟐
Problem 2.16
Use KVL to determine the unknown voltages in Figure P2.16.
Solution:
Known quantities:
Va = 2 V, Vb = 4 V, Vc = 5 V
Find:
a)
V1
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b) V2
c) V3
Analysis:
a)
Use KVL at the third loop.
𝑉𝑉3 − 𝑉𝑉𝑐𝑐 = 0
𝑉𝑉3 = 𝑉𝑉𝑐𝑐
𝑽𝑽𝟑𝟑 = 𝟓𝟓𝟓𝟓
b) Use KVL at the second loop.
𝑉𝑉2 −𝑉𝑉3 − 𝑉𝑉𝑏𝑏 = 0
𝑉𝑉2 = 𝑉𝑉3 + 𝑉𝑉𝑏𝑏
c)
Use KCL at the first loop.
𝑽𝑽𝟐𝟐 = 𝟗𝟗𝟗𝟗
𝑉𝑉𝑎𝑎 −𝑉𝑉1 − 𝑉𝑉2 = 0
𝑉𝑉1 = 𝑉𝑉𝑎𝑎 − 𝑉𝑉2
𝑽𝑽𝟏𝟏 = −𝟕𝟕𝟕𝟕
Problem 2.17
Use KCL to determine the unknown currents in Figure P2.17.
Solution:
Known quantities:
ia = -2 A,
ib = 6 A, ic = 1 A, id = -4 A
Find:
a)
b)
c)
d)
i1
i2
i3
i4
Analysis:
a)
Use KCL at Node A.
𝑖𝑖1 − 𝑖𝑖𝑎𝑎 − 𝑖𝑖𝑐𝑐 = 0
𝑖𝑖1 = 𝑖𝑖𝑎𝑎 + 𝑖𝑖𝑐𝑐
𝒊𝒊𝟏𝟏 = −𝟏𝟏𝟏𝟏
b) Use KCL at Node B.
𝑖𝑖2 − 𝑖𝑖1 − 𝑖𝑖𝑏𝑏 = 0
𝑖𝑖2 = 𝑖𝑖1 + 𝑖𝑖𝑏𝑏
2.13
𝒊𝒊𝟐𝟐 = 𝟓𝟓𝟓𝟓
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c)
Use KCL at Node C.
𝑖𝑖3 − 𝑖𝑖2 − 𝑖𝑖𝑑𝑑 = 0
𝑖𝑖3 = 𝑖𝑖2 + 𝑖𝑖𝑑𝑑
𝒊𝒊𝟑𝟑 = 𝟏𝟏𝟏𝟏
d) Use KCL at Node D.
𝑖𝑖𝑐𝑐 + 𝑖𝑖4 − 𝑖𝑖3 = 0
𝑖𝑖4 = 𝑖𝑖3 − 𝑖𝑖𝑐𝑐
𝒊𝒊𝟒𝟒 = 𝟎𝟎𝟎𝟎
Section 2.4 Sign Convention
Problem 2.18
In the circuits of Figure P2.18, the directions of
current and polarities of voltage have already been
defined. Find the actual values of the indicated currents
and voltages.
Solution:
Known quantities:
Circuit shown in Figure P2.18.
Find:
Voltages and currents in every figure.
Analysis:
15
(clockwise current) : i1 = −0.3 A;i2 = 0.3 A;v1 = 6V )
30+20
20
(b) The voltage across the 20 Ω resistor is
= 5V ; since the current flows from
4
(a) Using I =
top to bottom, the polarity of this voltage is positive on top.
Then it follows that
5
v1 = 5V and i 2 =
= -0.167A
30
(the negative sign follows from the direction of i2 in the drawing).
(c) Since -0.5A pointing upward is the same current as 0.5A pointing downward,
the voltage across the 30 Ω resistor is
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V30Ω = 15V (positive on top); and i 2 =
15
= 0.75A ,
20
since V30Ω is also the voltage across the 20 Ω resistor. Finally,
i1 = -(i2 + 0.5 ) = -1.25 A ,and v1 = -30 i1 + 15 = 52.5V
Problem 2.19
In the circuits of Figure P2.19, find the power delivered by each source.
Solution:
Known quantities:
Circuit shown in Figure P2.19.
a) Power delivered by the 3A Current Source
b) Power delivered by the -9V Voltage Source
Analysis:
a)
Follow the counterclockwise current:
𝑃𝑃 = +𝑣𝑣𝑣𝑣 = (+3𝐴𝐴) ∙ (+10𝑉𝑉)
b) Follow the counterclockwise current:
𝑷𝑷 = +𝟑𝟑𝟑𝟑𝟑𝟑delivered
𝑃𝑃 = −𝑣𝑣𝑣𝑣 = −(+5𝐴𝐴) ∙ (−9𝑉𝑉)
𝑷𝑷 = +𝟒𝟒𝟒𝟒𝟒𝟒delivered
Problem 2.20
Determine which elements in the circuit of Figure
P2.20 are supplying power and which are dissipating
power. Also determine the amount of power dissipated
and supplied.
Solution:
Known quantities:
Circuit shown in Figure P2.20.
Find:
Determine power dissipated or supplied for each power source.
Analysis:
Element A:
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P = -vi = -(-12V)( 25 A) = 300W (dissipating)
Element B:
P = vi = ( 15V)( 25 A) = 375W (dissipating)
Element C:
P = −vi = −( 27V)( 25 A) = −675W (supplying)
Problem 2.21
In the circuits of Figure P2.21, find the power absorbed by R4 and the power delivered by the current source.
Solution:
Known quantities:
Circuit shown in Figure P2.21.
Find:
a) Power absorbed by R4
b) Power delivered by the current source
Analysis:
a)
Follow the current in the rightmost loop through R4:
𝑃𝑃 = (2𝐴𝐴) ∙ (15𝑉𝑉)
𝑷𝑷 = +𝟑𝟑𝟑𝟑𝟑𝟑dissipated
b) Use KVL at the leftmost loop to find V3:
10𝑉𝑉 − 2𝑉𝑉 − 1𝑉𝑉 − 𝑣𝑣3 = 0
Use KVL at the rightmost loop to find V5:
𝑣𝑣3 = 7𝑉𝑉
7𝑉𝑉 − 15𝑉𝑉 − 𝑣𝑣5 = 0
𝑣𝑣5 = −8𝑉𝑉
The current source has a -8V drop across it. Use this to calculate the power dissipated using the proper sign
convention.
𝑷𝑷 = −𝒗𝒗𝒗𝒗 = −(+𝟐𝟐𝟐𝟐) ∙ (−𝟖𝟖𝟖𝟖) = +𝟏𝟏𝟏𝟏𝟏𝟏dissipated
Problem 2.22
For the circuit shown in Figure P2.22:
a. Determine which components are absorbing power
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
and which are delivering power.
b. Is conservation of power satisfied? Explain your
answer.
Solution:
Known quantities:
Circuit shown in Figure P2.22.
Find:
a) Determine power absorbed or power delivered
b) Testify power conservation
Analysis:
By KCL, the current through element B is 5A, to the right.
By KVL on the outer loop, −v a − 3 + 10 + 5 = 0 .
Therefore, the voltage across element A is
va = 12V (positive at the top).
A supplies (12V )(5 A) = 60W
B supplies (3V )(5 A) = 15W
C absorbs (5V )(5 A) = 25W
D absorbs (10V )(3 A) = 30W
E absorbs (10V )(2 A) = 20W
Total power supplied = 60W + 15W = 75W
Total power absorbed = 25W + 30W + 20W = 75W
Tot. power supplied = Tot. power absorbed
∴conservation of power is satisfied.
Problem 2.23
For the circuit shown in Figure P2.23, determine
the power absorbed by the 5 Ω resistor.
Solution:
Known quantities:
Circuit shown in Figure P2.23.
Find:
Power absorbed by the 5Ω resistance.
Analysis:
The current flowing clockwise in the series circuit is
The voltage across the 5 Ω resistor, positive on the left, is
i=
20V
= 1A
20Ω
v5Ω = (1A)(5Ω ) = 5V
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Therefore, P5Ω = (5V )(1A) = 5 W
Problem 2.24
For the circuit shown in Figure P2.24, determine
which components are supplying power and which are
dissipating power. Also determine the amount of
power dissipated and supplied.
Solution:
Known quantities:
Circuit shown in Figure P2.24.
Find:
Determine power absorbed or power delivered and corresponding
amount.
Analysis:
If current direction is out of power source, then power source is supplying, otherwise it is absorbing.
A supplies (100V )(4 A) = 400 W
B absorbs (10V )(4 A) = 40 W
C supplies (100V )(1A) = 100W
D supplies (− 10V )(1A) = − 10W , i.e absorbs 10W
E absorbs (90V )(5 A) = 450W
Problem 2.25
For the circuit shown in Figure P2.25.determine
which components are supplying power and which are
dissipating power. Also determine the amount of
power dissipated and supplied.
Solution:
Known quantities:
Circuit shown in Figure P2.25.
Find:
Determine power absorbed or power delivered and corresponding amount.
Analysis:
If current direction is out of the positive terminal, then this is a power source which is supplying, otherwise it is
absorbing.
A absorbs (5V )(4 A) = 20W
B supplies (2V )(6 A) = 12W
D supplies (3V )(4 A) = 12W
Since conservation of power is satisfied, Tot. power supplied = Tot. power absorbed
Total power supplied = 12W + 12W = 24W
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
∴Cabsorbs 24W − 20W = 4W
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
Problem 2.26
If an electric heater requires 23 A at 110 V, determine
a. The power it dissipates as heat or other losses.
b. The energy dissipated by the heater in a 24-hperiod.
c. The cost of the energy if the power company charges at the rate 6 cents/kWh.
Solution:
Known quantities:
Current absorbed by the heater; voltage at which the current is supplied; cost of the energy.
Find:
a) Power consumption
b) Energy dissipated in 24 hr.
c) Cost of the Energy
Assumptions:
The heater works for 24 hours continuously.
Analysis:
a)
b)
c)
P = VI = 110V (23 A) = 2.53 × 10 3 W = 2.53KW
s
J
W = Pt = 2.53 × 10 3 × 24hr × 3600 = 218.6MJ = 60.72KW - hr
hr
s
$
(2.53kW )(24hr ) = $3.64
Cost = (Rate) × W = 0.06
kW − hr
Problem 2.27
In the circuit shown in Figure P2.27, determine the
terminal voltage of the source, the power supplied to
the circuit (or load), and the efficiency of the circuit.
Assume that the only loss is due to the internal
resistance of the source. Efficiency is defined as the
ratio of load power to source power.
Solution:
Known quantities:
Circuit shown in Figure P2.27 with voltage source,
resistance of the load,
v s = 12V ; internal resistance of the source, Rs = 5kΩ ; and
R0 = 7 kΩ .
Find:
The terminal voltage of the source; the power supplied to the circuit, the efficiency of the circuit.
Assumptions:
Assume that the only loss is due to the internal resistance of the source.
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Analysis:
KVL :− v S + iT RS + vT = 0
OL :vT = iT R0
∴ iT =
vT
RL
− vS +
vT
R S + vT = 0
R0
vT =
vS
12V
=
= 7V or
RS
5kΩ
1+
1+
7kΩ
R0
VD :vT =
PL =
v S R0
12V ⋅7kΩ
=
= 7V .
RS + R0 5kΩ + 7kΩ
vT2
=
R0
(7V )2
= 7mW
V
7 × 10
A
P
i2R
PL
7kΩ
η = out =
= 2 T 02 =
= 0.5833 or 58.33% .
Pin
PRS + PL iT RS + iT R0 5kΩ + 7kΩ
3
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
Problem 2.28
A 24-volt automotive battery is connected to two
headlights, such that the two loads are in parallel; each
of the headlights is intended to be a 75-W load,
however, a 100-W headlight is mistakenly installed.
What is the resistance of each headlight, and what is
the total resistance seen by the battery?
Solution:
Known quantities:
Headlights connected in parallel to a 24-V automotive battery; power absorbed by each headlight.
Find:
Resistance of each headlight; total resistance seen by the battery.
Analysis:
Headlight no. 1:
P = v × i = 100 W =
R=
v2
R
or
v 2 576
= 5.76 Ω
=
100 100
Headlight no. 2:
P = v × i = 75 W =
R=
v2
R
or
v 2 576
=
= 7.68 Ω
75 75
The total resistance is given by the parallel combination:
1
1
1
or RTOTAL = 3.29 Ω
+
=
RTOTAL 5.76 Ω 7.68 Ω
Problem 2.29
What is the equivalent resistance seen by the
battery of Problem 2.28 if two 15-W taillights are
added (in parallel) to the two 75-W (each)
headlights?
Solution:
Known quantities:
Headlights and 24-V automotive battery of problem 2.13 with 2 15-W taillights added in parallel; power absorbed
by each headlight; power absorbed by each taillight.
Find:
Equivalent resistance seen by the battery.
Analysis:
The resistance corresponding to a 75-W headlight is:
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R75W =
v 2 576
= 7.68 Ω
=
75
75
For each 15-W tail light we compute the resistance:
v 2 576
= 38.4 Ω
=
R15W =
15
15
Therefore, the total resistance is computed as:
1
1
1
1
1
or RTOTAL = 3.2 Ω
=
+
+
+
RTOTAL 7.68Ω 7.68Ω 38.4Ω 38.4Ω
Problem 2.30
Using the circuit in Figure P2.30, determine the power
absorbed by the potentiometer and plot it as a function of R.
Solution:
Known quantities:
vs=15V, Rs=10 Ohms, and the circuit in Figure P2.30.
Find:
R
Analysis:
Use ohms law to find an equation for P as a function of R:
𝑃𝑃𝑅𝑅 = 𝑉𝑉𝑅𝑅 ∗ 𝐼𝐼𝑅𝑅
The voltage across R is equal to the source voltage minus the voltage across Rs:
𝑉𝑉𝑅𝑅 = 15𝑉𝑉 − 𝑉𝑉𝑅𝑅𝑅𝑅
VRS is determined by the current through the loop which can be found by adding the resistors in series:
15𝑉𝑉
𝐼𝐼𝑅𝑅 =
(𝑅𝑅𝑠𝑠 + 𝑅𝑅)
𝑉𝑉𝑅𝑅𝑅𝑅 = 10Ω ∗ 𝐼𝐼𝑅𝑅
Simplify:
15𝑉𝑉
15𝑉𝑉
𝑃𝑃𝑅𝑅 = �15𝑉𝑉 − �10Ω ∗
�� ∗ �
�
10Ω + R
10Ω + R
Plot:
PR
R
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Problem 2.31
Using the circuit in Figure P2.27 to determine the following
values relating to the power supply and consumption of the circuit.
Solution:
Known quantities:
vs=15V, Rs=100 Ohms, iT= 0, 10, 20, 30, 80, 100 mA.
The circuit in Figure P2.27.
Find:
a)
b)
c)
d)
The total power supplied by the ideal source
The power dissipated within the non-ideal source
How much power is supplied to the load resistor
Plot vT and power supplied to R0 as a function of iT.
Analysis:
a)
The power supplied by the ideal source is equal to the current through the loop times the 15V of the supply.
From current lowest to highest the power supplied would be:
0𝑊𝑊
0.15𝑊𝑊
0.3𝑊𝑊
0.45𝑊𝑊
1.2𝑊𝑊
1.5𝑊𝑊
b) The power dissipated within the non-ideal source is the power dissipated by Rs which can be found using
P=i2*r. From current lowest to highest the power dissipated would be:
0𝑊𝑊
0.01𝑊𝑊
0.04𝑊𝑊
0.09𝑊𝑊
0.64𝑊𝑊
1𝑊𝑊
c) The power supplied to the load resistor is equal to the total power supplied minus the power dissipated by
the non-ideal source. From current lowest to highest the power supplied would be:
0𝑊𝑊
0.14𝑊𝑊
0.26𝑊𝑊
0.36𝑊𝑊
0.56𝑊𝑊
0.5𝑊𝑊
d) For the vT plot Ohm’s Law can be used to find the voltage drop across Rs which is vT = 15 – 100iT. For the
power plot, the data from part c can be used directly.
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Vs across R2 in figure P2.32 should be replaced with V2
Problem 2.32
In the circuit in Figure P2.32, assume v2 = vs/6 and the power
delivered by the source is 150 mW. Find R, vs, v2, and i.
Solution:
Known quantities:
R1=8kΩ, R2=10kΩ, R3=12kΩ, Ps = 150 mW and the circuit in Figure P2.32.
Find:
R,vs, v2, and i.
Analysis:
Use ohms law to find vs and i:
𝑣𝑣2 =
Also:
So:
𝟐𝟐
Since we know i, vs can easily be found:
𝑣𝑣𝑠𝑠
= 𝑅𝑅2 ∗ 𝑖𝑖
6
𝑣𝑣𝑠𝑠 ∗ 𝑖𝑖 = 150𝑚𝑚𝑚𝑚
𝒊𝒊 = �
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
= 𝟏𝟏. 𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓
𝑹𝑹𝟐𝟐 ∗ 𝟔𝟔
𝑣𝑣𝑠𝑠 ∗ 𝑖𝑖 = 150𝑚𝑚𝑚𝑚
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𝒗𝒗𝒔𝒔 = 𝟗𝟗𝟗𝟗. 𝟗𝟗𝟗𝟗𝟗𝟗
Use Ohm’s law to find Req:
𝒗𝒗𝟐𝟐 =
𝒗𝒗𝒔𝒔
= 𝟏𝟏𝟏𝟏. 𝟖𝟖𝟖𝟖𝟖𝟖
𝟔𝟔
𝑣𝑣𝑠𝑠
= 𝑅𝑅𝑒𝑒𝑒𝑒 = 60𝑘𝑘Ω
𝑖𝑖
Use Req to find R:
𝑅𝑅𝑒𝑒𝑒𝑒 = 𝑅𝑅1 + 𝑅𝑅 + 𝑅𝑅2 + 𝑅𝑅3
𝑅𝑅 = 𝑅𝑅𝑒𝑒𝑒𝑒 − (𝑅𝑅1 + 𝑅𝑅2 + 𝑅𝑅3 ) = 30𝑘𝑘Ω
Problem 2.33
A GE SoftWhite Longlife lightbulb is rated as follows:
PR = rated power = 60 W
POR = rated optical power = 820 lumens (lm) (average)
1 lumen = 1/680W
Operating life = 1,500 h (average)
VR = rated operating voltage = 115 V
The resistance of the filament of the bulb, measured with a standard multimeter, is 16.7 Ω. When the bulb is connected into a
circuit and is operating at the rated values given above, determine
a. The resistance of the filament.
b. The efficiency of the bulb.
Solution:
Known quantities:
Rated power; rated optical power; operating life; rated operating voltage; open-circuit resistance of the filament.
Find:
a) The resistance of the filament in operation
b) The efficiency of the bulb.
Analysis:
a)
P = VI∴ I =
OL: R =
PR 60W
=
= 521.7mA
V R 115V
V VR
115 V
=
=
= 220.4 Ω
I
I
521.7 mA
b)
Efficiency is defined as the ratio of the useful power dissipated by or supplied by the load to the total power supplied
by the source. In this case, the useful power supplied by the load is the optical power.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
Po ,out = Optical Power Out = 820 lum
η = efficiency =
Po ,out
PR
=
W
= 1.206 W
680 lum
1.206W
= 0.0201 = 2.01% .
60W
Problem 2.34
An incandescent lightbulb rated at 100 W will dissipate 100 W as heat and light when connected across a 110-V ideal voltage
source. If three of these bulbs are connected in series across the same source, determine the power each bulb will dissipate.
Solution:
Known quantities:
Rated power; rated voltage of a light bulb.
Find:
The power dissipated by a series of three light bulbs connected to the nominal voltage.
Assumptions:
The resistance of each bulb doesn’t vary when connected in series.
Analysis:
When connected in series, the voltage of the source will divide equally across the three bulbs. The voltage across
each bulb will be 1/3 what it was when the bulbs were connected individually across the source. Power dissipated in
a resistance is a function of the voltage squared, so the power dissipated in each bulb when connected in series will
be 1/9 what it was when the bulbs were connected individually, or 11.11 W:
VB2
P
=
IV
=
I
R
=
Ohm’s Law:
B
B
RB
2
V 2 (110V )
RB = B =
= 121Ω
P
100W
2
VB = VS = 110 V
Connected in series and assuming the resistance of each bulb remains the same as when connected individually:
OL: −VS + IRB3 + IRB2 + IRB1 = 0
KVL: −VS + VB1 + VB2 + VB3 = 0
I=
VS
110V
=
= 303 mA
R B1 + R B 2 + R B 3 (121 + 121 + 121)Ω
1
2
PB1 = I 2 RB1 = (303 mA ) (121 Ω ) = 11.11W = 100W.
9
Problem 2.35
An incandescent lightbulb rated at 60 W will
dissipate 60 W as heat and light when connected
across a 100-V ideal voltage source. A 100-W bulb
will dissipate 100 W when connected across the same
source. If the bulbs are connected in series across the
same source, determine the power that either one of the
two bulbs will dissipate.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
Solution:
Known quantities:
Rated power and rated voltage of the two light bulbs.
Find:
The power dissipated by the series of the two light bulbs.
Assumptions:
The resistance of each bulb doesn’t vary when connected in series.
Analysis:
2
Ohm’s Law: P = IVB = I RB =
VB = VS = 110 V
VB2
RB
VB2 (110V )
R60 =
=
= 201.7Ω
P60
60 W
2
VB2 (110V )
=
= 121Ω
P100
100 W
2
R100 =
When connected in series and assuming the resistance of each bulb remains the same as when connected
individually:
KVL: −VS + VB60 + VB100 = 0
OL: −VS + IRB60 + IRB100 = 0
110 V
VS
=
= 340.9 mA
I=
RB60 + RB100 201.7 + 121 V
A
PB 60 = I 2 R B 60 = (340.9mA) (201.7 Ω ) = 23.44W
2
PB100 = I 2 R B100 = (340.9mA) (121 Ω ) = 14.06W
2
Notes: 1.It’s strange but it’s true that a 60 W bulb connected in series with a 100 W bulb will dissipate more power
than the 100 W bulb. 2. If the power dissipated by the filament in a bulb decreases, the temperature at which the
filament operates and therefore its resistance will decrease. This made the assumption about the resistance necessary.
Problem 2.36
For the circuit shown in Figure P2.36, find the voltage
vab and the power dissipated in R2.
Solution:
Known quantities:
Circuit of Figure P2.36. R1=5Ω, R2=3Ω, R3=4Ω,
R4=5Ω, vs=12V.
Find:
vab and the power dissipated in R2.
Analysis:
Find the current through each resistor pair by combining them in series:
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
𝑅𝑅1 + 𝑅𝑅2 = 8Ω
𝑅𝑅3 + 𝑅𝑅4 = 9Ω
Since there is 12V across both of them Ohm’s law is used to find the currents:
12𝑉𝑉 = 8Ω ∗ 𝐼𝐼𝑎𝑎
𝐼𝐼𝑎𝑎 = 1.5𝐴𝐴
12𝑉𝑉 = 9Ω ∗ 𝐼𝐼𝑏𝑏
𝐼𝐼𝑏𝑏 = 1.33𝐴𝐴
Now find the voltage drop across R1 and R3:
𝑉𝑉𝑎𝑎 = 12𝑉𝑉 − (𝐼𝐼𝑎𝑎 ∗ 𝑅𝑅1 ) = 4.5𝑉𝑉
𝑉𝑉𝑏𝑏 = 12𝑉𝑉 − (𝐼𝐼𝑏𝑏 ∗ 𝑅𝑅3 ) = 6.67𝑉𝑉
𝑽𝑽𝒂𝒂𝒂𝒂 = −𝟐𝟐. 𝟏𝟏𝟏𝟏𝟏𝟏
PR2 is equal to Ib times Va:
𝑷𝑷𝑹𝑹𝑹𝑹 = 𝟒𝟒. 𝟓𝟓𝟓𝟓 ∗ 𝟏𝟏. 𝟓𝟓𝟓𝟓 = 𝟔𝟔. 𝟕𝟕𝟕𝟕𝟕𝟕
Problem 2.37
For the circuit shown in Figure P2.37, find the currents
i2 and i1 and the power supplied by the voltage source.
Solution:
Known quantities:
Circuit of Figure P2.37. R1=20Ω, R2=12Ω, R3=10Ω,
Is=3A, vs=7V.
Find:
i1, i2, and Pv.
Analysis:
Use KVL of the leftmost loop:
Use KCL where i1 meets is:
𝑣𝑣𝑠𝑠 − 𝑅𝑅1 ∗ 𝑖𝑖1 − 𝑅𝑅2 ∗ 𝑖𝑖2 − 𝑅𝑅3 ∗ 𝑖𝑖1 = 0
𝑣𝑣𝑠𝑠 = 𝑅𝑅1 ∗ 𝑖𝑖1 + 𝑅𝑅2 ∗ 𝑖𝑖2 + 𝑅𝑅3 ∗ 𝑖𝑖1
𝑖𝑖1 + 𝑖𝑖𝑠𝑠 − 𝑖𝑖2 = 0
Combine the two equations to solve for i1:
𝑖𝑖1 + 𝑖𝑖𝑠𝑠 = 𝑖𝑖2
𝑣𝑣𝑠𝑠 = 𝑅𝑅1 ∗ 𝑖𝑖1 + 𝑅𝑅2 ∗ (𝑖𝑖1 + 𝑖𝑖𝑠𝑠 ) + 𝑅𝑅3 ∗ 𝑖𝑖1
Solve for i2:
𝒊𝒊𝟏𝟏 = −𝟎𝟎. 𝟔𝟔𝟔𝟔𝟔𝟔
𝑖𝑖1 + 𝑖𝑖𝑠𝑠 = 𝑖𝑖2
Power delivered equals the vs times i1:
𝒊𝒊𝟐𝟐 = 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑
𝑷𝑷𝒗𝒗 = 𝒗𝒗𝒔𝒔 ∗ 𝒊𝒊𝟏𝟏 = −𝟒𝟒. 𝟖𝟖𝟖𝟖𝟖𝟖
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Problem 2.38
For the circuit shown in Figure P2.38, find the currents
i2 and i1 and the power supplied by the voltage sources.
Solution:
Known quantities:
Circuit of Figure P2.38. R1=18Ω, R2=10Ω, v1=15V, v2=6V.
Find:
i1, i2, Pv1,Pv2.
Analysis:
Use KVL of the rightmost loop:
Use KCL where i1 meets is:
𝑣𝑣2 + 𝑅𝑅1 ∗ (−𝑖𝑖1 ) − 𝑅𝑅2 ∗ 𝑖𝑖2 = 0
𝑣𝑣2 = 𝑅𝑅1 ∗ 𝑖𝑖1 + 𝑅𝑅2 ∗ 𝑖𝑖2
𝑖𝑖 − 𝑖𝑖1 − 𝑖𝑖2 = 0
𝑖𝑖1 + 𝑖𝑖2 = 𝑖𝑖
The assumption can be made that v1 is equal to the voltage across R2.
𝑣𝑣1 = 𝑅𝑅2 ∗ 𝑖𝑖2
𝒊𝒊𝟐𝟐 = 𝟏𝟏. 𝟓𝟓𝟓𝟓
The assumption can be made that the voltage drop across R1 is equal to the difference between v1 and v2.
𝑣𝑣1 − 𝑣𝑣2 = 𝑅𝑅1 ∗ 𝑖𝑖1
𝒊𝒊𝟏𝟏 = 𝟎𝟎. 𝟓𝟓𝟓𝟓
Solve for i:
𝑖𝑖1 + 𝑖𝑖2 = 𝑖𝑖
𝑖𝑖 = 2.0𝐴𝐴
Power delivered equals the voltage source times the current through it:
𝑷𝑷𝟏𝟏 = 𝒗𝒗𝟏𝟏 ∗ 𝒊𝒊 = 𝟑𝟑𝟑𝟑. 𝟎𝟎𝑾𝑾
𝑷𝑷𝟐𝟐 = 𝒗𝒗𝟐𝟐 ∗ (−𝒊𝒊𝟏𝟏 ) = −𝟑𝟑𝟑𝟑
Problem 2.39
Consider NiMH hobbyist batteries shown in the
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circuit of Figure P2.39.
a. If V1 = 12.0 V, R1 = 0.15 Ω and RL = 2.55Ωfind the load current IL and the power dissipated bythe load.
b. If we connect a second battery in parallel withbattery 1 that has voltage V2 = 12 V andR2 = 0.28 Ω, will
the load current IL increase ordecrease? Will the power dissipated by the loadincrease or decrease? By
how much?
Solution:
Known quantities:
Schematic of the circuit in Figure P2.39.
Find:
a) If V1 = 12.0V ,R1 = 0.15Ω,RL = 2.55Ω , the load current and the power dissipated by the load
b) If a second battery is connected in parallel with battery 1with V2 = 12.0V ,R2 = 0.28Ω , determine the
variations in the load current and in the power dissipated by the load due to the parallel connection with a
second battery.
Analysis:
IL =
a)
V1
12
12
= 4.444 A
=
=
R1 + RL
0.15 + 2.55
2.7
PLoad = I L2 RL = 50.4 W.
b) with another source in the circuit we must find the new power
dissipated by the load. To do so, we write KVL twice using mesh
currents to obtain 2 equations in 2 unknowns:
− v 2 + i2 R2 + (i2 − i L )R1 + v1 = 0
⇒

i L RL + i2 R2 = v 2
0.43 ⋅ i2 − 0.15 ⋅ i L = 0

0.28 ⋅ i2 + 2.55 ⋅ i L = 12
Solving the above equations gives us:
I 2 = 1.59
A
I L = 4.55 A
⇒
PLoad = I L2 RL = 52.79 W
This is an increase of 5%.
Problem 2.40
With no load attached, the voltage at the terminals
of a particular power supply is 50.8 V. When a 10-W
load is attached, the voltage drops to 49 V.
a. Determine vS and RS for this non ideal source.
b. What voltage would be measured at the terminals
in the presence of a 15-Ωload resistor?
c. How much current could be drawn from this power
supply under short-circuit conditions?
Solution:
Known quantities:
Open-circuit voltage at the terminals of the power source is 50.8 V; voltage drop with a 10-W load attached is 49 V.
Find:
a) The voltage and the internal resistance of the source
b) The voltage at its terminals with a 15-Ω load resistor attached
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c)
The current that can be derived from the source under short-circuit conditions.
Analysis:
(49V ) 2
= 10W ⇒ RL = 240Ω v s = 50.8V , the open-circuit voltage is
RL
(240)(50.8)
240
RL
− 240 = 8.82Ω
⇒ RS =
50.8 = 49
v S = 49 ⇒
49
RS + 240
RS + RL
(a)
RL
15
50.8 = 31.99V
vS =
8.82 + 15
RS + R L
(b)
v=
(c)
50.8
v
= 5.76 A
iCC (RL = 0) = S =
RS 8.82
Problem 2.41
A 220-V electric heater has two heating coils
which can be switched such that either coil can be used
independently or the two can be connected in series or
parallel, yielding a total of four possible configurations.
If the warmest setting corresponds to 2,000-W power
dissipation and the coolest corresponds to 300 W, find
a. The resistance of each of the two coils.
b. The power dissipation for each of the other two
possible arrangements.
Solution:
Known quantities:
Voltage of the heater, maximum and minimum power dissipation; number of coils, schematics of the configurations.
Find:
a) The resistance of each coil
b) The power dissipation of each of the other two possible arrangements.
Analysis:
(a) For the parallel connection, P = 2000 W. Therefore,
(220)2 + (220)2 = 220 2  1 + 1 
2000 =
( )

R1
R2
 R1 R2 
or,
1
1
5
.
+
=
R1 R2 121
For the series connection, P = 300 W. Therefore, 300 =
Solving, we find that
(220)2
R1 + R2
R1 = 131.7Ω and R2 = 29.6Ω .
or,
1
3
.
=
R1 + R2 484
(b) the power dissipated by R1 alone is:
PR1 =
(220)2 = 368W
R1
and the power dissipated by R2 alone is
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PR2 =
(220)2
R2
= 1635W .
Section 2.5, 2.6 Resistance and Ohm’s Law
Problem 2.42
For the circuits of Figure P2.42, determine the
resistor values (including the power rating) necessary
to achieve the indicated voltages. Resistors are
available in 1/8-, 1/4-, 1/2-, and 1-W ratings.
Solution:
Known quantities:
Circuits of Figure 2.42.
Find:
Values of resistance and power rating
Analysis:
(a)
20 =
Ra
(50)
Ra+15,000
Ra (50 - 20) = 20(15) × 103
Ra = 10 kΩ
2
 50 
Pa = I 2 R=
 (10 ,000 )=40 mW
 25000 
1
PRa = W
8
P1 = I 2 R = 60 mW
1
PR1 = W
8
(b)
 10,000 

2.25 = 5 × 
+
10
,
000
R
b 

Rb = 12.2kΩ
PRb =
1
W
8
PR2 =
c)

2.7 × 103
28.3 = 110 × 
3
3
 2.7 × 10 + 1 × 10 + RL
1
W
8



RL = 6.8kΩ
PR L = 1W
PR3 =
1
W
8
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PR4 =
1
W
2
Problem 2.43
A 1-hp motor is powered by a DC generator 150m away. Using the
resistance of copper wire chart, determine the minimum AWG that
can be used to power the motor.
Solution:
Known quantities:
Figure P2.43.
VM-min = 105V -> IM-FL = 7.10A
VM-max = 117V -> IM-FL = 6.37A
VG = 110V
Find:
The minimum AWG wire that will meet the specifications.
Assumptions:
Assume an ideal DC source.
Analysis:
The circuit that this setup is describing is essentially a voltage source with a resistor, an inductor, and another
resistor all in series.
150m is equal to 492.13 feet. This will help with the resistance later.
Assume the voltage of the motor is 105V. Use Ohm’s Law to calculate the Current through the first conductor:
110𝑉𝑉 − 105𝑉𝑉
𝐼𝐼𝑀𝑀 =
2𝑅𝑅
IM has to be 7.1 A or greater in order for the motor to draw full power:
2.5𝑉𝑉
7.1𝐴𝐴 =
𝑅𝑅
𝑅𝑅 = 0.352𝛺𝛺
Use the value of R along with the chart to find the correct AWG:
0.352𝛺𝛺
= 0.715𝛺𝛺/1000𝑓𝑓𝑓𝑓
0.49213 × 103 ft
This means that 8 AWG wire must be used.
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Problem 2. 44
Cheap resistors are fabricated by depositing a thin layer of carbon onto a non-conducting cylindrical substrate (see
Figure P2.50). If such a cylinder has radius a and length d, determine the thickness of the film required for a
resistance R if
a = 1 mm
R = 33 kΩ
σ = 1/ρ= 2.9 MS/m
d = 9 mm
Neglect the end surfaces of the cylinder and assume that the thickness is much smaller than the radius.
Solution:
Known quantities:
Figure P2.50. Diameter of the cylindrical substrate; length of the
substrate; conductivity of the carbon.
Find:
The thickness of the carbon film required for a resistance R of 33 kΩ.
Assumptions:
Assume the thickness of the film to be much smaller than the radius
Neglect the end surface of the cylinder.
Analysis:
d
d
≅
σ ⋅ A σ ⋅ 2πa ⋅ ∆t
9 ⋅ 10 −3 m
d
∆t =
=
= 14.97 ⋅ 10 −12 m = 14.97 pm
S
R ⋅ 2πa ⋅ σ
33 ⋅ 10 3 Ω ⋅ 2.9 ⋅ 10 6 ⋅ 2π ⋅ 1 ⋅ 10 −3 m
m
R=
Problem 2.45
The resistive elements of fuses, lightbulbs, heaters,
etc., are significantly nonlinear (i.e., the resistance is
dependent on the current through the element).
Assume the resistance of a fuse (Figure P2.51) is given
by the expression R = R0[1 + A(T −T0)] with
T −T0 = kP; T0 = 25◦C; A = 0.7[◦C]−1;
k = 0.35◦C/W; R0 = 0.11 Ω; and P is the power
dissipated in the resistive element of the fuse.
Determine the rated current at which the circuit will
melt and open, that is, “blow” (Hint: The fuse blows
when R becomes infinite.)
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Solution:
Known quantities:
Figure P2.51. The constants A and k; the open-circuit resistance.
Find:
The rated current at which the fuse blows, showing that this happens at:
1
.
I=
AkR0
Assumptions:
Here the resistance of the fuse is given by:
R = R0 [1 + A(T − T 0 )]
where T 0 , room temperature, is assumed to be 25°C.
We assume that:
T −T 0 = kP
where P is the power dissipated by the resistor (fuse).
Analysis:
R = R0 (1 + A ⋅ ∆T ) = R0 (1 + AkP) = R0 (1 + AkI 2 R)
R − R0 AkI 2 R = R0
R0
R=
→ ∞ when 1 − R0 AkI 2
2
1− R0 AkI
→0
1
I=
V −
°C
m
1
0.11 ) 2 =6.09 A.
0.35
= (0.7
a
Va
°C
AkR0
Problem 2.46
Use Kirchhoff’s current law and Ohm’s law to
determine the current in each of the resistors R4, R5,
andR6 in the circuit of Figure P2.52. VS = 10 V,
R1 = 20 Ω, R2 = 40 Ω, R3 = 10 Ω, R4 = R5 =
R6 = 15 Ω.
Solution:
Known quantities:
Circuit shown in Figure P2.52 with voltage source, Vs = 10V and resistors,
R1 = 20Ω,R2 = 40Ω,R3 = 10Ω,R4 = R5 = R6 = 15Ω .
Find:
The current in the 15-Ω resistors.
Analysis:
Since the 3 resistors must have equal currents,
I15Ω =
1
⋅I
3
and,
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I=
VS
10
10
=
=
= 303 mA
R1 + R2 || R3 + R4 || R5 || R6 20 + 8 + 5 33
Therefore, I 15Ω =
10
= 101 mA
99
Problem 2.47
Use Kirchhoff’s current law and Ohm’s law to
determine the unknown currents.
A
B
Solution:
Known quantities:
Circuit shown in Figure P2.13
R0=1Ω, R1=2Ω, R2=3Ω, R3=4Ω, vs=10V
Find:
a) i0 b) i1 c) i2 d) i3
Analysis:
Use KCL at node A:
Use KCL at node B:
Use Ohms Law for i0:
Use Ohms Law for i1:
Use Ohms Law for i2:
Use Ohms Law for i3:
𝑖𝑖0 + 𝑖𝑖2 − 𝑖𝑖1 = 0
𝑖𝑖3 + 6𝐴𝐴 − 𝑖𝑖2 = 0
𝑖𝑖0 =
10𝑉𝑉 − 𝑣𝑣𝐴𝐴
1Ω
𝑖𝑖1 =
𝑖𝑖2 =
𝑣𝑣𝐴𝐴
2Ω
𝑣𝑣𝐵𝐵 − 𝑣𝑣𝑎𝑎
3Ω
0 − 𝑣𝑣𝐵𝐵
4Ω
Use algebra to solve the six equations with six unknowns. All that the problem statement asks for are the currents.
The two voltages come along for free. (N.B. This solution is simply using the theory behind Nodal Analysis to
solve the problem):
𝒊𝒊𝟎𝟎 = 𝟏𝟏. 𝟖𝟖𝟖𝟖𝟖𝟖
𝒊𝒊𝟏𝟏 = 𝟒𝟒. 𝟎𝟎𝟎𝟎𝟎𝟎
𝒊𝒊𝟐𝟐 = 𝟐𝟐. 𝟐𝟐𝟐𝟐𝟐𝟐
𝒊𝒊𝟑𝟑 = −𝟑𝟑. 𝟕𝟕𝟕𝟕𝟕𝟕
Note that the negative value for i3 indicates that it is flowing in the opposite direction from that assumed.
𝑖𝑖3 =
Problem 2.48
Use Kirchhoff’s current law and Ohm’s law to
determine the power delivered by the voltage source.
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Solution:
Known quantities:
Circuit shown in Figure P2.48
k=0.25A/A2
Find:
Power supplied by the voltage source.
Analysis:
The 7 Ohm and 5 Ohm resistor are in series so i can be found by using Ohm’s law:
24𝑉𝑉 = 12 ∗ 𝑖𝑖
𝑖𝑖 = 2𝐴𝐴
The dependent current source current can therefore be found using i:
𝑖𝑖𝑑𝑑 = 0.25 ∗ (2𝐴𝐴)2
𝑖𝑖𝑑𝑑 = 1𝐴𝐴
Use KCL to determine the current through the voltage source:
𝑖𝑖𝑣𝑣 + 𝑖𝑖𝑑𝑑 − 𝑖𝑖 = 0
𝑖𝑖𝑣𝑣 = 1𝐴𝐴
Use P=v*i to determine the power supplied by the voltage source:
𝑃𝑃 = 24𝑉𝑉 ∗ 1𝐴𝐴
𝑷𝑷 = 𝟐𝟐𝟐𝟐𝟐𝟐
Problem 2.49
Assuming R0 = 2 Ω, R1 = 1Ω, R2 = 4/3 Ω R3 = 6 Ωand VS = 12 V in the circuit of Figure
P2.55, use Kirchhoff’s voltage law and Ohm’s law to find
a. ia, ib, and ic
b. the current through each resistor.
Solution:
Known quantities:
Schematic of the circuit shown in Figure P2.55 with resistors
R0 = 2Ω,R1 = 1Ω,R2 = 4 / 3Ω,R3 = 6Ω and voltage source VS = 12 V.
Find:
a) The mesh currents i a , i b , i c
b) The current through each resistor.
Analysis:
Applying KVL to mesh (a), mesh (b) and mesh (c):
 2i a + (i a − i b ) = 0
i a R 0 + (i a − i b )R 1 = 0



4
(i a − i b )R 1 − i b R 2 + (i c − i b )R 3 = 0 ⇒ (i a − i b ) − i b + 6(i c − i b ) = 0
3


 VS = (i c − i b )R 3
 6(i c − i b ) = 12
Solving the system we obtain:
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 IR
0
i a = 2 A 
I

R
i b = 6 A ⇒  1
i = 8 A
 IR 2
c
I
 R3
= ia = 2 A
(positive in the direction of i a )
= i b − i a = 4 A (positive in the direction of i b )
= ib = 6 A
(positive in the direction of i b )
= i c − i b = 2 A (positive in the direction of i c )
Problem 2.50
Assuming R0 = 2Ω, R1 = 2 Ω, R2 = 5 Ω, R3 = 4 A, and VS = 24 V in the circuit of FigureP2.55, use Kirchhoff’s voltage law and
Ohm’s law to find
a. ia, ib, and ic.
b. The voltage across each resistance.
Solution:
Known quantities:
Schematic of the circuit shown in Figure P2.55 with resistors
R0 = 2Ω,R1 = 2Ω,R2 = 5Ω,R3 = 4Ω and voltage source VS = 24 V.
Find:
a) The mesh currents i a , i b , i c
b) The current through each resistor.
Analysis:
Applying KVL to mesh (a), mesh (b) and mesh (c):
 2i a + 2(i a − i b ) = 0
i a R 0 + (i a − i b )R 1 = 0


(i a − i b )R 1 − i b R 2 + (i c − i b )R 3 = 0 ⇒  2(i a − i b ) − 5i b + 4 (i c − i b ) = 0


 VS = (i c − i b )R 3
 4(i c − i b ) = 24
Solving the system we obtain:
VR 0 = R 0 ia = 4 V
i a = 2 A 

VR = R 1 (ib − ia ) = 4 V
i b = 4 A ⇒  1
i = 10 A VR 2 = R 2 ib = 20 V
c
(⊕ down)
(⊕ down)
(⊕ up)
V = R (i − i ) = 24 V (⊕ up)
b
3 c
 R3
Problem 2.51
Assume that the voltage source in the circuit of
Figure P2.55 is now replaced by a current source, and
R0 = 1 Ω, R1 = 3 ΩR2= 2Ω, R3 = 4 Ω, and
IS = 12 A. Use Kirchhoff’s voltage law and Ohm’s
law to determine the voltage across each resistance.
Solution:
Known quantities:
Schematic of the circuit shown in Figure P2.55 with resistors
R0 = 1Ω, R1 = 3Ω, R2 = 2Ω, R3 = 4Ω
and
current source IS = 12 A.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
Find:
The voltage across each resistance.
Analysis:
Applying KVL to mesh (a), mesh (b) and mesh (c):
i a R 0 + (i a − i b )R 1 = 0
ia + 3(ia − ib ) = 0


(i a − i b )R 1 − i b R 2 + (i c − i b )R 3 = 0 ⇒ 3(i a − ib ) − 2ib + 4(ib

i = 12 A
i c = IS
c
− ic ) = 0
Solving the system we obtain:
16

ia = 3 A VR 0


64

VR
A⇒ 1
ib =
9

VR 2
=
1
2
A
i
c
V
 R3


= R 0 ia = 5.33 V
(⊕ down)
= R 1 (ib − ia ) = 5.33 V
(⊕ down)
= R 2 ib = 14.22 V
(⊕ up)
= R 3 (ic − ib ) = 19.55 V (⊕ up)
Problem 2.52
The voltage divider network of Figure P2.58 is
expected to provide 5 V at the output. The resistors,
however, may not be exactly the same; that is, their
tolerances are such that the resistances may not be
exactly 5 kΩ
a. If the resistors have +/-10 percent tolerance, find the worst-case output voltages.
b. Find these voltages for tolerances of +/-5 percent.
Given: v = 10 V, R1 = 5 kΩ, R2 = 5 kΩ.
Solution:
Known quantities:
Schematic of voltage divider network shown of Figure P2.58.
Find:
The worst-case output voltages for ±10 percent
tolerance
b) The worst-case output voltages for ±5percent tolerance
a)
Analysis:
a) 10% worst case: low voltage
R2 = 4500 Ω, R1 = 5500 Ω
4500
5 = 2.25V
vOUT,MIN =
4500 + 5500
10% worst case: high voltage
R2 = 5500 Ω, R1 = 4500 Ω
5500
5 = 2.75V
vOUT,MAX =
4500 + 5500
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b) 5% worst case: low voltage
R2 = 4750 Ω, R1 = 5250 Ω
4750
vOUT,MIN =
5 = 2.375V
4750 + 5250
5% worst case: high voltage
R2 = 5250 Ω, R1 = 4750 Ω
5250
5 = 2.625V
vOUT,MAX =
5250 + 4750
Problem 2.53
Find the equivalent resistance seen by the source and
the currents i and i1and v2in the circuit of Figure P2.53.
Solution:
Known quantities:
Circuits of Figure 2.53.
Find:
Equivalent resistance and current i and i1and v2
Analysis:
Simplify resistors:
(9Ω||72Ω) + 2Ω = 𝑅𝑅𝑒𝑒𝑒𝑒
Find i:
𝑹𝑹𝒆𝒆𝒆𝒆 = 𝟏𝟏𝟏𝟏Ω
𝑖𝑖 =
9𝑉𝑉
9𝑉𝑉
=
𝑅𝑅𝑒𝑒𝑒𝑒 10Ω
𝒊𝒊 = 𝟎𝟎. 𝟗𝟗𝟗𝟗
Find i1 using current division:
𝑖𝑖1 =
(0.9𝐴𝐴) ∗ 72Ω
(72Ω + 9Ω)
𝒊𝒊𝟏𝟏 = 𝟎𝟎. 𝟖𝟖𝟖𝟖
The voltage drop across the 72Ω resistor (v2) is the same as the voltage drop across the 9Ω resistor.
𝒗𝒗𝟐𝟐 = 𝟎𝟎. 𝟖𝟖𝟖𝟖 × 𝟗𝟗𝛀𝛀 = 𝟕𝟕. 𝟐𝟐𝟐𝟐
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Problem 2.54
Find the equivalent resistance seen by the source
and the current i in the circuit of Figure P2.45.
Solution:
Known quantities:
Circuits of Figure 2.45.
Find:
Equivalent resistance and current i
Analysis:
Step1:
(4||4) + 22 = 24 Ω
Step 2:
24||8 = 6 Ω
Therefore, the equivalent circuit is as shown in the figure:
Further,
(4 + 6)||90 = 9 W
The new equivalent circuit is shown below.
Thus,
Rtotal = 10Ω .
This means the circuit is delivering 5A (50V/10Ω).
We can now find the current i by the current divider rule as
follows:
 10 
i=
 (5) = 0.5 A
 10 + 90 
Problem 2.55
In the circuit of Figure P2.55, the power absorbed
by the 15-Ω resistor is 15 W. Find R.
Solution:
Known quantities:
Circuits of Figure P2.55.
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Find:
Resistance R
Analysis:
Combining the elements to the right of the 15 Ω resistor, we compute
Req = ((4||4) + 6)||24 + 4 = 10 Ω .
The power dissipated by the 15-Ω resistor is
P15 Ω =
v152 Ω
= 15 W ,
15
therefore,
Using the Ohms Law:
v15Ω= 15 V and i1 = 1 A.
i2 =
15V
= 1.5 A .
10Ω
Applying KCL, we can find iR:
iR = i1 + i2 = 2.5 A .
Using KVL:
− 25 + 2.5 R + 15 = 0
Therefore, R = 4Ω.
Problem 2.56
Find the equivalent resistance between terminals a
and b in the circuit of Figure P2.47.
Solution:
Known quantities:
Circuits of Figure 2.47.
Find:
Equivalent resistance.
Analysis:
2Ω + 2Ω = 4Ω
6Ω 12Ω = 4Ω
4Ω 4Ω = 2Ω
4Ω 4Ω = 2Ω
2Ω + 2Ω = 4Ω
Req = 3Ω + 4Ω 4Ω = 5Ω
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Problem 2.57
For the circuit shown in Figure P2.48, find the
equivalent resistance seen by the source. How much
power is delivered by the source?
Solution:
Known quantities:
Circuits of Figure 2.48.
Find:
a) Equivalent resistance
b) Power delivered.
Analysis:
(a)
2Ω + 1Ω = 3Ω
3Ω 3Ω = 1.5Ω
4Ω + 1.5Ω + 5Ω = 10.5Ω
10.5Ω 6Ω = 3.818Ω
Req = 3.818Ω + 7Ω = 10.818Ω
(b)
P=
14 2
= 18.12 W
10.818
Problem 2.58
For the circuit shown in Figure P2.58, find the
equivalent resistance seen by the current source and the number of nodes.
A
Solution:
B
C
Known quantities:
Circuits of Figure P2.58.
R1 = 2Ω, R2 = 3Ω, R3 = 85Ω,
R4 = 2Ω, R5 = 4Ω
Find:
The equivalent resistance seen by the current source.
The number of nodes in the circuit.
D
Analysis:
Simplify left side:
𝑅𝑅1 ||𝑅𝑅2 + 𝑅𝑅3 = 𝑅𝑅𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
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𝑅𝑅𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 86.2Ω
Simplify right side:
𝑅𝑅4 + 𝑅𝑅5 = 𝑅𝑅𝑟𝑟𝑟𝑟𝑟𝑟ℎ𝑡𝑡
𝑅𝑅𝑟𝑟𝑟𝑟𝑟𝑟ℎ𝑡𝑡 = 6Ω
Combine:
𝑅𝑅𝑒𝑒𝑒𝑒 = 𝑅𝑅𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 ||𝑅𝑅𝑟𝑟𝑟𝑟𝑟𝑟ℎ𝑡𝑡
𝑹𝑹𝒆𝒆𝒆𝒆 = 𝟓𝟓. 𝟔𝟔𝟔𝟔𝛀𝛀
Count the nodes:
𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵 = 𝟒𝟒
Nodes A, B, C, & D as shown above
Problem 2.59
For the circuit shown in Figure P2.59, find the number of
nodes, the equivalent resistance seen by the voltage source,
and the current through R7.
Solution:
A
Known quantities:
B
Circuit of Figure P2.59. R1=10Ω, R2=5 Ω, R3=8 Ω,
R4=2 Ω, R5=4 Ω, R6=2 Ω, R7=1 Ω,R8=10 Ω vs=20V.
Find:
C
The equivalent resistance seen by the voltage source.
The current through R7.
D
Analysis:
Count the nodes:
Combine R1 and R2:
𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵 = 𝟒𝟒
Nodes A, B, C, & D as shown above.
𝑅𝑅1 ||𝑅𝑅2 = 3.33Ω
R4-6are in parallel with each other and in series with R1||2:
𝑅𝑅4 �|𝑅𝑅5 |�𝑅𝑅6 + 𝑅𝑅1||2 = 4.13Ω
The previous resistance is in parallel with R3:
R7 is in parallel with R8:
𝑅𝑅3 ||4.13Ω = 2.72Ω
𝑅𝑅8 ||R 7 = 0.91Ω
These last two resistances are in series and are equal to the equivalent resistance:
𝑹𝑹𝒆𝒆𝒆𝒆 = 𝟑𝟑. 𝟔𝟔𝟔𝟔𝛀𝛀
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The current through R7||8 will be the same as the total current through the circuit. Find the total current then use
current division to find the current through R7:
𝑰𝑰𝑹𝑹𝟕𝟕
𝑣𝑣𝑠𝑠
= 5.51𝐴𝐴
𝑅𝑅𝑒𝑒𝑒𝑒
𝑹𝑹𝟕𝟕𝟕𝟕
= 𝑰𝑰 ∗
= 𝟓𝟓. 𝟎𝟎𝟎𝟎𝟎𝟎
𝑹𝑹𝟕𝟕
𝐼𝐼 =
Problem 2.60
For the circuit shown in Figure P2.60, find the
equivalent resistance seen by the current source
and count the number of nodes in the circuit.
A
B
C
Solution:
Known quantities:
Circuit of Figure P2.60.
R1 = 12Ω, R2 = 5 Ω, R3 = 8 Ω, R4 = 2 Ω,
R5 = 4 Ω, R6 = 2 Ω, R7 = 1 Ω, R8 = 10 Ω,
R9 = 10 Ω
F
E
D
Find:
The equivalent resistance seen by the current source.
The number of nodes in the circuit.
Analysis:
Simplify starting on the right side:
𝑅𝑅6 ||(𝑅𝑅2 + 𝑅𝑅7 ) = 𝑅𝑅𝑟𝑟𝑟𝑟𝑟𝑟ℎ𝑡𝑡
Simplify middle:
𝑅𝑅𝑟𝑟𝑟𝑟𝑟𝑟ℎ𝑡𝑡 = 1.5Ω
�𝑅𝑅𝑟𝑟𝑟𝑟𝑟𝑟ℎ𝑡𝑡 + 𝑅𝑅8 �||𝑅𝑅5 = 𝑅𝑅𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
Finish with the left:
𝑅𝑅𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 2.968Ω
((𝑅𝑅𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 + 𝑅𝑅9 )||𝑅𝑅4 + 𝑅𝑅1 )||𝑅𝑅3 = 𝑅𝑅𝑒𝑒𝑒𝑒
Count the nodes:
𝑹𝑹𝒆𝒆𝒆𝒆 = 𝟓𝟓. 𝟎𝟎𝟎𝟎𝛀𝛀
𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵 = 𝟔𝟔
See nodes A, B, C, D, E, & F above
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Problem 2.61
A
B
C
For the circuit shown in Figure P2.61, find the
number of nodes in the circuit, the power delivered
by the voltage source and the equivalent resistance
seen by the voltage source.
Solution:
Known quantities:
D
Circuit of Figure P2.61.
R1 = 9Ω, R2 = 4 Ω, R3 = 4 Ω,
R4 = 5 Ω, R5 = 4 Ω, vs = 10V
Find:
The number of nodes in the circuit.
The power delivered by the voltage source.
The equivalent resistance seen by the voltage source.
Analysis:
Count the nodes:
𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵 = 𝟒𝟒
See nodes A, B, C, & D above
Simplify starting on the right side:
𝑅𝑅5 ||𝑅𝑅4 = 𝑅𝑅𝑟𝑟𝑟𝑟𝑟𝑟ℎ𝑡𝑡
Simplify middle:
𝑅𝑅𝑟𝑟𝑟𝑟𝑟𝑟ℎ𝑡𝑡 = 2.22Ω
(𝑅𝑅𝑟𝑟𝑟𝑟𝑟𝑟ℎ𝑡𝑡 + 𝑅𝑅3 )||𝑅𝑅2 = 𝑅𝑅𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
Finish with the left side:
𝑅𝑅𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 2.43Ω
𝑅𝑅𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 + 𝑅𝑅1 = 𝑅𝑅𝑒𝑒𝑒𝑒
Find power delivered using watts law:
𝑹𝑹𝒆𝒆𝒆𝒆 = 𝟏𝟏𝟏𝟏. 𝟒𝟒𝟒𝟒𝛀𝛀
P=
v s2
Req
10 2
=
= 8.75
11.43
𝑷𝑷 = 𝟖𝟖. 𝟕𝟕𝟕𝟕𝑾𝑾
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
Problem 2.62
Determine the equivalent resistance of the infinite
network of resistors in the circuit of Figure P2.62.
Solution:
Known quantities:
Schematic of the circuit shown in Figure P2.62.
Find:
The equivalent resistance Req of the infinite network of resistors.
Analysis:
We can see the infinite network of resistors as the equivalent to the circuit
in the picture:
Therefore,
(
)
Req = R + R || Req + R = 2R +
(
RReq
R + Req
)
Req = 1 + 3 R
= 2.73R
Problem 2.63
For the circuit shown in Figure P2.63, find the
number of nodes in the circuit, the power delivered
by the current source and the equivalent resistance
seen by the current source.
A
B
C
Solution:
Known quantities:
Circuit of Figure P2.63.
is = 5A, R1 = 10Ω, R2 = 7 Ω.
R3 = 8 Ω, R4 = 4 Ω, R5 = 2 Ω
D
Find:
The number of nodes in the circuit.
The power delivered by the current source.
The equivalent resistance seen by the current source.
Analysis:
Count the nodes:
𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵 = 𝟒𝟒
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
See nodes A, B, C, & D above
Simplify the right side:
𝑅𝑅5 + 𝑅𝑅4 = 𝑅𝑅𝑟𝑟𝑟𝑟𝑟𝑟ℎ𝑡𝑡
Simplify the left side:
𝑅𝑅𝑟𝑟𝑟𝑟𝑟𝑟ℎ𝑡𝑡 = 6Ω
𝑅𝑅1 ||𝑅𝑅2 + 𝑅𝑅3 = 𝑅𝑅𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
Combine both sides:
𝑅𝑅𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 12.12Ω
𝑅𝑅𝑟𝑟𝑟𝑟𝑟𝑟ℎ𝑡𝑡 ||𝑅𝑅𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 𝑅𝑅𝑒𝑒𝑒𝑒
𝑹𝑹𝒆𝒆𝒆𝒆 = 𝟒𝟒. 𝟎𝟎𝟎𝟎𝛀𝛀
Find power delivered using watts law:
P = i s2 Req
= 5 2 ⋅ 4.01 =
𝑷𝑷 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑
Problem 2.64
In the circuit of Figure P2.64, find the equivalent
resistance looking in at terminals a and b if terminals c
and d are open and again if terminals c and d are
shorted together. Also, find the equivalent resistance
looking in at terminals c and d if terminals a and b are
open and if terminals a and b are shorted together.
Solution:
Known quantities:
Schematic of the circuit shown in Figure P2.64.
Find:
The equivalent resistance at terminals a,b in the case that terminals c,d are a) open b) shorted; the same for terminals
c,d with respect to terminals a,b.
Analysis:
With terminals c-d open, Req= (360 + 540) (180 + 540)Ω = 400Ω ,
with terminals c-d shorted, Req= (360 180) + (540 540)Ω = 390Ω ,
with terminals a-b open, Req= (540 + 540) (360 + 180)Ω = 360Ω ,
with terminals a-b shorted, Req= (360 540) + (180 540)Ω = 351Ω .
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
Problem 2.65
In the circuit of Figure P2.64, find the equivalent
resistance looking in at terminals a and b if terminals c
is shorted to terminal a and terminal d is shorted to terminal
b.
Solution:
Known quantities:
Schematic of the circuit shown in Figure P2.64.
Find:
The equivalent resistance seen by terminals a and b.
Analysis:
Shorting a to c eliminates the 360Ω resistor. Shorting b to d eliminates
the rightmost 540Ω resistor. This leaves the remaining two resistors in
parallel with each other.
540Ω||180Ω = 𝑅𝑅𝑒𝑒𝑒𝑒
𝑹𝑹𝒆𝒆𝒆𝒆 = 𝟏𝟏𝟏𝟏𝟏𝟏𝛀𝛀
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Problem 2.66
A
For the circuit shown in Figure P2.66, find the
number of nodes in the circuit and the equivalent
resistance seen by the voltage source.
Solution:
Known quantities:
C
Circuit of Figure P2.66.
R1 = 12Ω, R2 = 5 Ω, R3 = 8 Ω,
R4 = 2 Ω, R5 = 4 Ω
D
Find:
The equivalent resistance seen by the voltage source.
The number of nodes in the circuit.
B
Analysis:
Count the nodes:
𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵 = 𝟒𝟒
See nodes A, B, C, & D above
Simplify R2, R3, and R5 using a wye-delta transformation (N.B. The nodes are labeled A, B, & C to match the nodes
in figure 2.69a. Node D is the common node of the wye:
R x = wye = R2 = 5Ω
R y − wye = R5 = 4Ω
R z − wye = R3 = 8Ω
Now using equations 2.37 through 2.39
R1− delta =
Rx R y + Rx Rz + R y Rz
Ry
=
92
= 23Ω (this goes from A to C in parallel with the R1 of the circuit
4
R2− delta =
92
= 11.5Ω (this goes directly from A to B)
Rz
R3− delta =
92
= 18.4Ω (this goes from C to B in parallel with the R4 of the circuit
Rx
R AC = R1 R1− delta = 7.89Ω
RCB = R4 R3− delta = 1.80Ω
Req = R AB = (R AC + RCB ) R2− delta = 5.25Ω
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
Problem 2.67
Determine the voltage between nodes A and B in
the circuit shown in Figure P2.67.
VS = 12 V
R1 = 11 kΩ
R3 = 6.8 kΩ
R2 = 220 kΩ
R4 = 0.22 MΩ
Solution:
Known quantities:
Schematic of the circuit shown in Figure P2.67 with source voltage, Vs = 12V ;
and resistances,
R1 = 11kΩ, R2 = 220kΩ, R3 = 6.8kΩ, R4 = 0.22MΩ
Find:
The voltage between nodes A and B.
Analysis:
The same current flows through R1 and R3. Therefore, they are connected in series. Similarly, R2 and R4 are
connected in series.
SPECIFY THE ASSUMED POLARITY OF THE VOLTAGE BETWEEN NODES A AND B. THIS WILL HAVE
TO BE A GUESS AT THIS POINT.
Specify the polarities of the voltage across R3 and R4 which will be determined using voltage division. The actual
polarities are not difficult to determine. Do so.
VD :VR 3 =
VS R3
[12V ][6.8kΩ] = 4.584V
=
11 + 6.8kΩ
R1 + R3
[
]
V S R4
[
12V ] 0.22 × 10 3 kΩ
=
= 6V
VD :VR 4 =
R2 + R4
220 + 0.22 × 10 3 kΩ
KVL :− VR 3 + V AB + VR 4 = 0 ∴V AB = VR 3 − VR 4 = −1.416V
(
)
The voltage is negative indicating that the polarity of V AB is opposite of that specified.
A solution is not complete unless the assumed positive direction of a current or assumed positive polarity of a
voltage IS SPECIFIED ON THE CIRCUIT.
Problem 2.68
For the circuit shown in Figure P2.68, find the
number of nodes in the circuit, the power delivered
by the voltage source and the equivalent resistance
seen by the voltage source.
A
B
D
C
Solution:
Known quantities:
E
Circuit of Figure P2.68.
R1=12Ω, R2=5 Ω, R3=8 Ω,
R4=2 Ω, R5=4 Ω, R6=2 Ω,
R7=1 Ω, R8=R9=10 Ω, vs=15V
F
Find:
The number of nodes in the circuit.
The power delivered by the voltage source.
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The equivalent resistance seen by the voltage source.
Analysis:
Count the nodes:
𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵 = 𝟔𝟔
See nodes A, B, C, D, E, & F above
R9 is eliminated due to the short.
Start at the right and work left:
𝑅𝑅8 ||(𝑅𝑅6 + 𝑅𝑅7 ) = 𝑅𝑅𝑟𝑟𝑟𝑟𝑟𝑟ℎ𝑡𝑡
𝑅𝑅𝑟𝑟𝑟𝑟𝑟𝑟ℎ𝑡𝑡 = 2.3Ω
(𝑅𝑅𝑟𝑟𝑟𝑟𝑟𝑟ℎ𝑡𝑡 + 𝑅𝑅4 )||𝑅𝑅5 = 𝑅𝑅𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
𝑅𝑅𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 2.1Ω
𝑅𝑅𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 + (𝑅𝑅3 ||(𝑅𝑅1 + 𝑅𝑅2 )) = 𝑅𝑅𝑒𝑒𝑒𝑒
𝑹𝑹𝒆𝒆𝒆𝒆 = 𝟕𝟕. 𝟓𝟓𝟓𝟓𝛀𝛀
Find power delivered using watts law:
P=
v s2
15 2
=
= 29.94W
Req 7.51
𝑷𝑷 = 𝟐𝟐𝟐𝟐. 𝟗𝟗𝟗𝟗
Problem 2.69
For the circuit shown in Figure P2.69, find the
voltages vac and vbd. Also find the power delivered
by the voltage source.
Solution:
Known quantities:
Circuit of Figure 2.69. R1=12Ω, R2=5Ω, R3=8Ω, R4=2Ω,
R5=4Ω, R6=2Ω, R7=1Ω,vs=15V.
Find:
vac
vbd
The power delivered by the voltage source.
Analysis:
Find Req by adding all the resistors in series:
𝑅𝑅𝑒𝑒𝑒𝑒 = 𝑅𝑅1 + 𝑅𝑅2 + 𝑅𝑅3 + 𝑅𝑅4 + 𝑅𝑅5 + 𝑅𝑅6 + 𝑅𝑅7
𝑅𝑅𝑒𝑒𝑒𝑒 = 34Ω
Use the equivalent resistance to find the current through the loop:
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15𝑉𝑉 = 𝑅𝑅𝑒𝑒𝑒𝑒 ∗ 𝐼𝐼
Find voltage across R1, R2, and R3 using the current:
𝐼𝐼 = 0.44𝐴𝐴
𝑣𝑣𝑎𝑎𝑎𝑎 = (𝑅𝑅1 + 𝑅𝑅2 + 𝑅𝑅3 ) ∗ 𝐼𝐼
Do the same with R2-6:
𝒗𝒗𝒂𝒂𝒂𝒂 = 𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎𝟎𝟎
𝑣𝑣𝑏𝑏𝑏𝑏 = (𝑅𝑅2 + 𝑅𝑅3 + 𝑅𝑅4 + 𝑅𝑅5 + 𝑅𝑅6 ) ∗ 𝐼𝐼
𝒗𝒗𝒃𝒃𝒃𝒃 = 𝟗𝟗. 𝟐𝟐𝟐𝟐𝟐𝟐
Find power delivered:
v s2
15 2
=
= 6.618
P=
34
Req
𝑷𝑷 = 𝟔𝟔. 𝟔𝟔𝟔𝟔𝟔𝟔
Sections 2.7, 2.8: Practical Sources and Measuring Devices
Problem 2.70
A thermistor is a nonlinear device which changes its terminal resistance value as its surrounding
temperature changes. The resistance and temperature generally have a relation in the form of
Rth(T ) = R0e−β(T−T0)
where Rth = resistance at temperature T, Ω
R0 = resistance at temperature T0 = 298 K, Ω
β = material constant, K−1
T, T0 = absolute temperature, K
a. If R0 = 300Ω and β = −0.01 K−1, plot Rth(T ) asa function of the surrounding temperature T for350 ≤ T ≤ 750.
b. If the thermistor is in parallel with a 250-Ω resistor, find the expression for the equivalent resistance and
plot Rth(T ) on the same graph for part a.
Solution:
Known quantities:
Parameters R0 = 300 Ω (resistance at temperature T0 = 298 K),
250Ω.
and β = −0.01 K -1, value of the second resistor =
Find:
a) Plot Rth (T ) versus T in the range 350 ≤ T ≤ 750 [°K]
b) The equivalent resistance of the parallel connection with the 250-Ω resistor; plot Req (T ) versus T in the
range 350 ≤ T ≤ 750 [°K] for this case on the same plot as part a.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
Assumptions:
− β T−T
Rth (T ) = R0e ( 0 ).
Analysis:
a) Rth (T ) = 300e
−0.01⋅(T−298)
b) Req (T ) = Rth (T ) || 250Ω =
1500 e−0.01 (T−298)
T−298)
5+ 6 e−0.01 (
The two plots are shown below.
In the above plot, the solid line is for the thermistor alone; the
dashed line is for the thermistor-resistor combination.
Problem 2.71
A moving-coil meter movement has a meter
resistance rM= 200 Ω, and full-scale deflection is
caused by a meter current Im= 10 μA. The movement
must be used to indicate pressure measured by the
sensor up to a maximum of 100 kPa. See Figure P2.71.
a. Draw a circuit required to do this, showing all
appropriate connections between the terminals of
the sensor and meter movement.
b. Determine the value of each component in the
circuit.
c. What is the linear range, that is, the minimum and
maximum pressure that can accurately be
measured?
Solution:
Known quantities:
Meter resistance of the coil; meter current for full scale deflection; max
measurable pressure.
Find:
a) The circuit required to indicate the pressure measured by a sensor
b) The value of each component of the circuit; the linear range
c) The maximum pressure that can accurately be measured.
Assumptions:
Sensor characteristics follow what is shown in Figure P2.71
Analysis:
a) The upper and lower terminals of the sensor is attached to the upper and lower terminals of the meter, respectively.
A series resistor to drop excess voltage is required.
b) At full scale, meter:
I mˆ FS = 10µA
rmˆ = 200Ω
0.L. :Vmˆ FS = I mˆ FS rmˆ = 2mV .
at full scale, sensor (from characteristics):
PFS = 100 kPa
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VTFS = 10mV
KVL :− VTFS + VRFS + Vmˆ FS = 0
VRFS = VTFS − Vmˆ FS = 10mV − 2mV = 8mV
I RFS = I TFS = I m?FS = 10µA
Ohm law: R =
V RFS
8mV
=
= 800Ω .
I RFS 10µA
c) from sensor characteristic: 30 kPa –100 kPa.
Problem 2.72
The circuit of Figure P2.72 is used to measure the
internal impedance of a battery. The battery being
tested is a NiMH battery cell.
a. A fresh battery is being tested, and it is found that
the voltage Vout , is 2.28 V with the switch open and
2.27 V with the switch closed. Find the internal
resistance of the battery.
b. The same battery is tested one year later, and Vout is
found to be 2.2 V with the switch open but 0.31 V
with the switch closed. Find the internal resistance
of the battery.
Solution:
Known quantities:
Schematic of the circuit shown in Figure P2.72; voltage at terminals
switch open and closed for fresh battery; same voltages for the same
battery after 1 year.
with
Find:
The internal resistance of the battery in each case.
Analysis:
a)
b)
 10 
Vout = 
Voc
 10 + rB 
V

 2.28 
rB = 10 oc −1 = 10
−1 = 0.044Ω

V
2.27 
 out 
V

 2.2

rB = 10 oc −1 = 10
−1 = 60.97Ω
 0.31 
 Vout 
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Problem 2.73
Consider the practical ammeter, described in
Figure P2.73, consisting of an ideal ammeter in series
with a 1-kΩ resistor. The meter sees a full-scale
deflection when the current through it is 30 μA. If we
desire to construct a multirange ammeter reading
full-scale values of 10 mA, 100 mA, and 1 A,
depending on the setting of a rotary switch, determine
appropriate values of R1, R2, and R3.
Solution:
Known quantities:
Ammeter shown in Figure P2.73; Current for full-scale deflection;
desired full scale values.
Find:
Value of the resistors required for the given full scale ranges.
Analysis:
We desire R1,R2, R3 such that Ia= 30 µA for I = 10 mA, 100 mA, and 1
A, respectively. We use conductances to simplify the arithmetic:
1
1
Ga =
=
S
1000
Ra
1
G1,2,3 =
R 1,2,3
By the current divider rule:
Ga
I
Ia =
Ga + Gx
or:
I 
1
1  Ia 
G x = G a   − G a or
=


Gx Ga  I − I a 
 Ia 
 I 
Rx = Ra  a  .
 I − Ia 
We can construct the following table:
x
I
Rx (Approx.)
1
10-2 A
3Ω
2
10-1 A
0.3 Ω
3
100 A
0.03 Ω
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Problem 2.74
A circuit that measures the internal resistance of a
practical ammeter is shown in Figure P2.74, where
RS = 50,000 Ω, VS = 12 V, and Rp is a variable
resistor that can be adjusted at will.
a. Assume that ra<<50,000 Ω. Estimate the current i.
b. If the meter displays a current of 150 μA when
Rp= 15 Ω, find the internal resistance of the meter ra.
Solution:
Known quantities:
Schematic of the circuit shown in Figure P2.74; for part b: value of
R p and current displayed on the ammeter.
Find:
The current i; the internal resistance of the meter.
Assumptions:
ra << 50 kΩ
Analysis:
a) Assuming that ra << 50 kΩ
12
V
i≈ s =
= 240 µA
50000
Rs
b) With the same assumption as in part a)
Rp
imeter = 150⋅(10)-6 =
i
ra + R p
or:
150⋅(10)-6=
15
240 ⋅10 -6 .
ra +15
Therefore, ra = 9 Ω.
Problem 2.75
A practical voltmeter has an internal resistance rm.What is the value of rmif the meter reads 11.81 V when connected as shown in
Figure P2.75.
Solution:
Known quantities:
Voltage read at the meter; schematic of the circuit shown in Figure P2.75 with
source voltage, Vs = 12V and source resistance, Rs = 25kΩ .
Find:
The internal resistance of the voltmeter.
Analysis:
Using the voltage divider rule:
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rm
(12)
rm + R s
V = 11.81 =
Therefore,
rm
= 1.55 MΩ.
Problem 2.76
Using the circuit of Figure P2.75, find the voltage
that the meter reads if VS = 24 V and RS has the
following values:
RS = 0.2rm, 0.4rm, 0.6rm, 1.2rm, 4rm, 6rm, and 10rm.
How large (or small) should the internal resistance of
the meter be relative to RS?
Solution:
Known quantities:
Circuit shown in Figure P2.75 with source voltage, Vs = 24V ; and ratios
between R s and
rm .
Find:
The meter reads in the various cases.
Analysis:
By voltage division:
rm
V=
(24)
rm + R s
Rs
0.2 rm
V
20 V
0.4 rm
17.14 V
0.6 rm
15 V
1.2 rm
10.91 V
4 rm
4.8 V
6 rm
10
3.43 V
2.18 V
rm
For a voltmeter, we always desire
rm
>> R s .
Problem 2.77
A voltmeter is used to determine the voltage across
a resistive element in the circuit of Figure P2.77. The
instrument is modeled by an ideal voltmeter in parallel
with a 120-kΩresistor, as shown. The meter is placed
to measure the voltage across R4. Assume R1 = 8 kΩ
R2 = 22 kΩR3 = 50 kΩ, RS = 125 kΩ, and
IS = 120 mA. Find the voltage across R4 with and
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
without the voltmeter in the circuit for the following
values:
a. R4 = 100 Ω
b. R4 = 1 kΩ
c. R4 = 10 kΩ
d. R4 = 100 kΩ
Solution:
Known quantities:
Schematic of the circuit shown in Figure P2.77, values of the
components.
Find:
The voltage across R4 with and without the voltmeter for the
following values:
a) R4 = 100Ω
b) R4 = 1kΩ
c) R4 = 10kΩ
d) R4 = 100kΩ .
Assumptions:
The voltmeter behavior is modeled as that of an ideal voltmeter in parallel with a 120- kΩ resistor.
Analysis:
We develop first an expression for VR 4 in terms of R 4 . Next, using current division:



RS

 I R1 = I S 

 RS + R1 + R2 || (R3 + R4 ) 


R2
 I = I 

R1 
 R4
 R2 + R3 + R4 

Therefore,

 

RS
R2
 ⋅ 

I R4 = I S 
 RS + R1 + R2 || (R3 + R4 )   R2 + R3 + R4 
V R4 = I R4 R4

 
R S R4
R2
⋅
= I S 


 RS + R1 + R2 (R3 + R4 )   R2 + R3 + R4
2.129 × 10 3 ⋅ R4
=
R4 + 68.877 × 10 3



Without the voltmeter:
a) VR 4 = 3.08 V
b) VR 4 = 30.47 V
c) VR 4 = 269.91 V
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
d) VR 4 = 1260.7 V.
Now we must find the voltage drop across R4 with a 120-kΩ resistor across R4 . This is the voltage that the
voltmeter will read.

 

R2
RS
I R 4 = I S 
 ⋅ 

 RS + R1 + R2 || (R3 + (R4 || 120kΩ)   R2 + R3 + (R4 || 120kΩ) 
VR4 = I R4 R4

 

R S R4
R2
 ⋅ 

= I S 
(
)
+
+
+
Ω
+
+
Ω
R
R
R
||
R
(
R
||
120
k
R
R
(
R
||
120
k
)
1
2
3
4
3
4
 S
  2

(120000 + R4 )⋅ R4
= 11.272 × 10 −3 ⋅
R4 + 43.760 × 10 3
With the voltmeter:
a) VR4 = 3.09 V
b) VR4 = 30.47 V
c) VR4 = 272.58 V
d) VR4 = 1724.99 V.
Problem 2.78
An ammeter is used as shown in Figure P2.78. The
ammeter model consists of an ideal ammeter in series
with a resistance. The ammeter model is placed in the
branch as shown in the figure. Find the current through R5
both with and without the ammeter in the circuit for the
following values, assuming that RS = 20 Ω, R1 =800 Ω, R2 = 600 ΩR3 = 1.2 kΩR4 = 150 Ω, and
VS = 24 V.
a. R5 = 1 kΩ
b. R5 = 100 Ω
c. R5 = 10 Ω
d. R5 = 1 Ω
Solution:
Known quantities:
Schematic of the circuit shown in Figure P2.78, value of the components.
Find:
The current through R5 both with and without the ammeter, for the
following values of the resistor R5 :
a) R5 = 1kΩ
b) R5 = 100Ω
c) R5 = 10Ω
d) R5 = 1Ω .
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
Analysis:
First we should find an expression for the current through R5 in terms of R5 and the meter resistance, Rm . By
the voltage divider rule we have:
VR3 =
V R1 =
And
I R5 =
Therefore,
I R5 =
R3 (R4 + R5 + Rm ) ⋅ VR1
R3 (R4 + R5 + Rm ) + R2
R1 (R3 (R4 + R5 + Rm ) + R2 ) ⋅ Vs
R1 (R3 (R4 + R5 + Rm ) + R2 ) + Rs
VR 3
R4 + R5 + Rm
R1 (R3 (R4 + R5 + Rm ) + R2 ) ⋅ Vs
⋅
R3 (R4 + R5 + Rm )
R1 (R3 (R4 + R5 + Rm ) + R2 ) + RS R3 (R4 + R5 + Rm ) + R2
⋅
1
R4 + R5 + Rm
=
553.846(R5 + Rm + 550 ) ⋅ 24 1200(R5 + Rm + 150 )
1
⋅
⋅
573.846(R5 + Rm + 558.579 ) 1800(R5 + Rm + 550 ) (R5 + Rm + 150 )
=
15.4424
R5 + Rm + 558.579
Using the above equation will give us the following table:
with meter in without meter
circuit
in circuit
a
9.75 mA
9.91 mA
b
22.59 mA
23.45 mA
c
26.02 mA
27.16 mA
d
26.42 mA
27.60 mA
Problem 2.79
Shown in Figure P2.79 is an aluminum
cantilevered beam loaded by the force F. Strain gauges
R1, R2, R3, and R4 are attached to the beam as shown
in Figure P2.79 and connected into the circuit shown.
The force causes a tension stress on the top of the
beam that causes the length (and therefore the
resistance) of R1 and R4 to increase and a compression
stress on the bottom of the beam that causes the length
(and therefore the resistance) of R2 and R3 to decrease.
This causes a voltage of 50 mV at node B with respect
to node A. Determine the force if
Ro = 1 kΩ VS = 12 V L = 0.3 m
w = 25 mm h = 100 mm Y = 69 GN/m2
Solution:
Known quantities:
Schematic of the circuit and geometry of the beam shown in Figure P2.79, characteristics of the material, reads on
the bridge.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
Find:
The force applied on the beam.
Assumptions:
Gage Factor for Strain gauge is 2
Analysis:
R1and R2 are in series; R3 and R4 are in series.
Voltage Division: VR 2 =
Voltage Division: V R4 =
VS R2
VS (R0 − ∆R)
V (R − ∆R)
=
= S 0
R1 + R2 R0 + ∆R + R0 − ∆R
2R0
V S R4
VS ( R0 + ∆R )
V ( R + ∆R )
=
= S 0
2 R0
R3 + R4 R0 − ∆R + R0 + ∆R
KVL: −VR 2 − VBA + VR 4 = 0
VBA = VR4 − VR2 =
VS ( R0 + ∆R) VS ( R0 − ∆R) VS 2∆R
V 2 * 6 LF
−
=
= VS GFε = S 2
2 R0
2 R0
2 R0
wh Y
assuming GF=2 for aluminum.
V wh 2Y
F = BA
=
Vs 12 L
0.050V (0.025m)(0.100m) 2 69 × 10 9
12V (12 )0.3m
N
m 2 = 19.97kN .
Problem 2.80
Shown in Figure P2.79 is a structural steel
cantilevered beam loaded by a force F. Strain gauges
R1, R2, R3, and R4 are attached to the beam as shown
and connected into the circuit shown. The force causes
a tension stress on the top of the beam that causes the
length (and therefore the resistance) of R1 and R4 to
increase and a compression stress on the bottom of the
beam that causes the length (and therefore the
resistance) of R2 and R3 to decrease. This generates a
voltage between nodes B and A. Determine this
voltage if F = 1.3 MN and
Ro = 1 kΩVS = 24 V L = 1.7 m
w = 3 cm h = 7 cm Y = 200 GN/m2
Solution:
Known quantities:
Schematic of the circuit and geometry of the beam shown in Figure P2.79,
characteristics of the material, reads on the bridge.
Find:
The force applied on the beam.
Assumptions:
Gage Factor for Strain gauge is 2
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 2
Analysis:
R1and R2 are in series; R3 and R4 are in series.
VS R2
VS (R0 − ∆R)
V (R − ∆R)
=
= S 0
R1 + R2 R0 + ∆R + R0 − ∆R
2R0
V (R + ∆R)
VS R4
VS (R0 − ∆R)
= S 0
=
VD: VR 4 =
2R0
R3 + R4 R0 − ∆R + R0 + ∆R
−V
−
V
+
V
=
0
KVL:
R2
BA
R4
VD: VR 2 =
VBA = VR4 − VR2 =
VS ( R0 + ∆R) VS ( R0 − ∆R) VS 2∆R
V ⋅ GF ⋅ 6 LF
−
=
= VS GFε = S
2 R0
2 R0
2 R0
wh 2Y
Assuming GF=2 for aluminum.
VBA
24V ⋅ 2 ⋅ 6 ⋅1.7m ⋅ 1.3 × 10 6 N
= 21.6 mV
=
9 N
2
(0.03 m)(0.07 m) 200 × 10 2
m
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