G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Chapter 3 Instructor Notes
Chapter 3 presents the principal topics in the analysis of resistive (DC) circuits. The presentation
of node voltage and mesh current analysis is supported by several solved examples and drill exercises, with
emphasis placed on developing consistent solution methods, and on reinforcing the use of a systematic
approach. The aim of this style of presentation, which is perhaps more detailed than usual in a textbook
written for a non-majors audience, is to develop good habits early on, with the hope that the orderly
approach presented in Chapter 3 will facilitate the discussion of AC and transient analysis in Chapters 4
and 5. Make The Connection sidebars (pp. 75-77) introduce analogies between electrical and thermal
circuit elements. These analogies are encountered again in Chapter 5. A brief discussion of the principle
of superposition precedes the discussion of Thèvenin and Norton equivalent circuits. Again, the
presentation is rich in examples and drill exercises, because the concept of equivalent circuits will be
heavily exploited in the analysis of AC and transient circuits in later chapters. The Focus on Methodology
boxes (p.76 – Node Analysis; p. 86 – Mesh Analysis; pp. 103, 107, 111 – Equivalent Circuits) provide the
student with a systematic approach to the solution of all basic network analysis problems.
After a brief discussion of maximum power transfer, the chapter closes with a section on nonlinear
circuit elements and load-line analysis. This section can be easily skipped in a survey course, and may be
picked up later, in conjunction with Chapter 9, if the instructor wishes to devote some attention to load-line
analysis of diode circuits. Finally, those instructors who are used to introducing the op-amp as a circuit
element, will find that sections 8.1 and 8.2 can be covered together with Chapter 3, and that a good
complement of homework problems and exercises devoted to the analysis of the op-amp as a circuit
element is provided in Chapter 8.
The homework problems present a graded variety of circuit problems. Since the aim of this
chapter is to teach solution techniques, there are relatively few problems devoted to applications. We
should call the instructor's attention to the following end-of-chapter problems: 3.8 and 3.19 on the
Wheatstone bridge; 3.21, 3.22, 3.23, on three-wire residential distribution service; 3.24, 3.25, 3.26 on AC
three-phase electrical distribution systems; 3.28-3.31 on fuses; 3.62-66 on various nonlinear resistance
devices.
Learning Objectives
1. Compute the solution of circuits containing linear resistors and independent and
dependent sources using node analysis.
2. Compute the solution of circuits containing linear resistors and independent and
dependent sources using mesh analysis.
3. Apply the principle of superposition to linear circuits containing independent sources.
4. Compute Thévenin and Norton equivalent circuits for networks containing linear
resistors and independent and dependent sources.
5. Use equivalent circuits ideas to compute the maximum power transfer between a source
and a load.
6. Use the concept of equivalent circuit to determine voltage, current and power for
nonlinear loads using load-line analysis and analytical methods.
3.1
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Sections 3.1, 3.2, 3.3, 3.4: Nodal and Mesh Analysis
I
cus on Methodology: Node Voltage Analysis Method
1.
2.
3.
Select a reference node(usually ground). This node usually has most elements tied to it.
All other nodes will be referenced to this node.
Define the remaining n–1 node voltages as the independent or dependent variables.
Each of the m voltage sources in the circuit will be associated with a dependent variable.
If a node is not connected to a voltage source, then its voltage is treated as an
independent variable.
Apply KCL at each node labeled as an independent variable, expressing each current in
terms of the adjacent node voltages.
Focus on Methodology: Mesh Current Analysis Method
1.
2.
3.
4.
Define each mesh current consistently. Unknown mesh currents will be always defined
in the clockwise direction; known mesh currents (i.e., when a current source is present)
will always be defined in the direction of the current source.
In a circuit with n meshes and m current sources, n–m independent equations will result.
The unknown mesh currents are the n–m independent variables.
Apply KVL to each mesh containing an unknown mesh current, expressing each voltage
in terms of one or more mesh currents..
Solve the linear system of n–m unknowns.
Problem 3.1
Solution:
Known quantities:
Circuit shown in Figure P2.1 with mesh currents: I1
= 5 A, I2 = 3 A, I3 = 7 A.
Find:
The branch currents through:
a) R1,
b) R2,
c) R3.
Analysis:
a) Assume a direction for the current through
node A:
KCL:
R1 (e.g., from node A to node B). Then summing currents at
− I1 + I R1 + I 3 = 0
I R1 = I1 − I 3 = −2 A
This can also be done by inspection noting that the assumed direction of the current through
direction of I1 are the same.
R1 and the
b) Assume a direction for the current through R2 (e.g., from node B to node A). Then summing currents at
node B:
KCL:
I 2 + I R2 − I3 = 0
3.2
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
I R2 = I 3 − I 2 = 4 A
This can also be done by inspection noting that the assumed direction of the current through
direction of I3 are the same.
c) Only one mesh current flows through
direction, then:
R2 and the
R3. If the current through R3 is assumed to flow in the same
I R1 = I 3 = 7 A .
______________________________________________________________________________________
Problem 3.2
Solution:
Known quantities:
Circuit shown in Figure P3.1 with source and node voltages:
VS 1 = VS 2 = 110 V , V A = 103 V , VB = −107 V .
Find:
The voltage across each of the five resistors.
Analysis:
Assume a polarity for the voltages across R1 and R2 (e.g., from ground to node A, and from node B to
ground). R1 is connected between node A and ground; therefore, the voltage across R1 is equal to this node
voltage. R2 is connected between node B and ground; therefore, the voltage across R2 is equal to the
negative of this voltage.
VR1 = V A = 103 V, VR 2 = −VB = +107 V
The two node voltages are with respect to the ground which is given.
Assume a polarity for the voltage across R3 (e.g., from node B to node A). Then:
KVL:
V A + VR 3 + VB = 0
VR 3 = V A − VB = 210 V
Assume polarities for the voltages across R4 and R5 (e.g., from node A to ground , and from ground to
node B):
KVL:
− VS 1 + V R 4 + V A = 0
V R 4 = VS 1 − V A = 7 V
KVL:
− VS 2 − V B − V R 5 = 0
VR 5 = −VS 2 − VB = −3 V
______________________________________________________________________________________
Problem 3.3
Solution:
Known quantities:
Circuit shown in Figure P3.3 with known source currents and resistances, R1
3.3
= 3Ω, R2 = 1Ω, R3 = 6Ω .
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Find:
The currents I1, I2 using node voltage analysis.
Analysis:
At node 1:
§1 ·
v1 ⋅ ¨ + 1¸ + v2 ⋅ (− 1) = 1
©3 ¹
At node 2:
§ 1·
v1 ⋅ (− 1) + v 2 ⋅ ¨1 + ¸ = −2
© 6¹
Solving, we find that:
v1 = −1.5 V
v2 = −3 V
Then,
v1
= − 0 .5 A
3
v
i 2 = 2 = − 0 .5 A
6
i1 =
______________________________________________________________________________________
Problem 3.4
Solution:
Known quantities:
Circuit shown in Figure P3.3 with known source currents and resistances, R1
Find:
The currents I1, I2 using mesh analysis.
Analysis:
At mesh (a):
ia = 1 A
At mesh (b):
3 (ib − ia ) + ib + 6 (ib − ic ) = 0
At mesh (c):
ic = 2 A
Solving, we find that:
ib = 1.5 A
Then,
i1 = (ia − ib ) = −0.5 A
i2 = (ib − ic ) = −0.5 A
3.4
= 3Ω, R2 = 1Ω, R3 = 6Ω .
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
______________________________________________________________________________________
Problem 3.5
Solution:
Known quantities:
Circuit shown in Figure P3.5 with resistance values, current and voltage source values.
Find:
The current, i, through the voltage source using node voltage analysis.
Analysis:
At node 1:
v −v
v1
v −v
+ 1 2 + 1 3 =0
200
5
100
At node 2:
v2 − v1
+ i + 0 .2 = 0
5
At node 3:
−i +
v3 − v1 v3
+
=0
100
50
For the voltage source we have:
v3 − v2 = 50 V
Solving the system, we obtain:
v1 = −45.53 V , v2 = −48.69 V ,
v3 = 1.31 V and, finally, i = 491 mA .
______________________________________________________________________________________
Problem 3.6
Solution:
Known quantities:
The current source value, the voltage source value and the resistance values for the circuit shown in Figure
P3.6.
Find:
The three node voltages indicated in Figure P3.6 using node voltage analysis.
Analysis:
50 Ω
75 Ω
V1
V2
10V
+
V3
i
0.2 A
200 Ω
25 Ω
3.5
100 Ω
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
At node 1:
v1
v −v
+ 1 2 = 0 .2 A
200
75
At node 2:
v2 − v1 v2 v2 − v3
+
+
+i = 0
75
25
50
At node 3:
−i +
v3 − v2 v3
+
=0
50
100
For the voltage source we have:
v3 + 10 = v2
Solving the system, we obtain:
v1 = 14.24 V , v2 = 4.58V ,
v3 = −5.42 V and, finally, i = −254 mA .
______________________________________________________________________________________
Problem 3.7
Solution:
Known quantities:
The voltage source value, 3 V, and the five resistance values, indicated in Figure P3.7.
Find:
The current, i, drawn from the independent voltage source using node voltage analysis.
Analysis:
At node 1:
v1 − 3 v1 v1 − v2
+
+
=0
0.5 0.5 0.25
At node 2:
v2 − v1
v
+ 2 + 0 .5 = 0
0.25 0.75
Solving the system, we obtain:
v1 = 1.125 V , v2 = 0.75 V
Therefore, i =
3 − v1
= 3.75 A .
0 .5
______________________________________________________________________________________
Problem 3.8
Solution:
Known quantities:
The voltage source value, 15 V, and the four resistance values, indicated in Figure P3.8.
Find:
The voltage at nodes a and b,
Va and Vb , and their difference, Va − Vb using node voltage analysis.
3.6
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Analysis:
Using nodal analysis at the two nodes a and b, we write the equations
Vb − 15 Vb
+
=0
18
20
Va − 15 Va
+
=0
36
20
Rearranging the equations,
38Vb − 300 = 0
14Va − 75 = 0
Solving for the two unknowns,
Va = 5.36 V and Vb = 7.89 V
Therefore,
Va − Vb = −2.54 V
______________________________________________________________________________________
Problem 3.9
Solution:
Known quantities:
The voltage source value, 15 V, and the four resistance values, indicated in Figure P3.8.
Find:
The voltage at nodes a and b,
Va and Vb , and their difference, Va − Vb using mesh analysis.
Analysis:
Using mesh analysis at the two meshes a and b, we write the equations
36 (ia − ib ) + 20 (ia − ib ) = 15
18ib + 20ib + 20 (ib − ia ) + 36 (ib − ia ) = 0
Rearranging the equations,
15
+ ib
56
94ib − 56ia = 0
ia =
Solving for the two unknowns,
ia = 662 mA and ib
= 395 mA
Therefore,
Va = 20 (ib − ia ) = 5.36 V , Vb = 20ib = 7.89 V and Va − Vb = −2.54 V .
______________________________________________________________________________________
Problem 3.10
Solution:
Known quantities:
Circuit of Figure P3.10 with voltage source,
VS , current source, IS, and all resistances.
3.7
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Find:
a. The node equations required to determine the node voltages.
b. The matrix solution for each node voltage in terms of the known parameters.
Analysis:
a) Specify the nodes (e.g., A on the upper left corner of the circuit in Figure P3.10, and B on the right
corner). Choose one node as the reference or ground node. If possible, ground one of the sources in the
circuit. Note that this is possible here. When using KCL, assume all unknown current flow out of the node.
The direction of the current supplied by the current source is specified and must flow into node A.
KCL:
− IS +
Va − VS Va − Vb
+
=0
R2
R1
§ 1
§ 1 ·
V
1 ·
+ ¸¸ + Vb ¨¨ − ¸¸ = I S + S
Va ¨¨
R2
© R2 R1 ¹
© R1 ¹
KCL:
Vb − Va Vb − VS Vb − 0
+
+
=0
R1
R3
R4
§ 1
§ 1 ·
1
1 · VS
¸¸ =
+
Va ¨¨ − ¸¸ + Vb ¨¨ +
© R1 ¹
© R1 R3 R4 ¹ R3
b) Matrix solution:
IS +
Va =
VS
R2
VS
R3
1
1
+
R1 R2
1
−
R1
1
R1
§
V ·§ 1
1
1
1
1
1 · § V ·§ 1 ·
¨¨ I S + S ¸¸¨¨ +
+
+
+ ¸¸ − ¨¨ S ¸¸¨¨ − ¸¸
R2 ¹© R1 R3 R4 ¹ © R3 ¹© R1 ¹
R1 R3 R4
= ©
1
§1
1 ·§ 1
1
1 · § 1 ·§ 1 ·
−
¨¨ + ¸¸¨¨ +
+ ¸¸ − ¨¨ − ¸¸¨¨ − ¸¸
R1
© R1 R2 ¹© R1 R3 R4 ¹ © R1 ¹© R1 ¹
1
1
1
+
+
R1 R3 R4
−
1
1
V
+
IS + S
R1 R2
R2
§1
V ·
VS
1 ·§ V · § 1 ·§
1
¨¨ + ¸¸¨¨ S ¸¸ − ¨¨ − ¸¸¨¨ I S + S ¸¸
−
R2 ¹
R1
R3
© R1 R2 ¹© R3 ¹ © R1 ¹©
=
Vb =
1
1
1
§1
1 ·§ 1
1
1 · § 1 ·§ 1 ·
+
−
¨¨ + ¸¸¨¨ +
+ ¸¸ − ¨¨ − ¸¸¨¨ − ¸¸
R1 R2
R1
© R1 R2 ¹© R1 R3 R4 ¹ © R1 ¹© R1 ¹
1
1
1
1
−
+
+
R1
R1 R3 R4
Notes:
1. The denominators are the same for both solutions.
2. The main diagonal of a matrix is the one that goes to the right and down.
3. The denominator matrix is the "conductance" matrix and has certain properties:
a) The elements on the main diagonal [ i(row) = j(column) ] include all the conductance
connected to node i = j.
3.8
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
b) The off-diagonal elements are all negative.
c) The off-diagonal elements are all symmetric, i.e., the i j-th element = j i-th element. This is
true only because there are no controlled (dependent) sources in this circuit.
d) The off-diagonal elements include all the conductance connected between node i [row] and
node j [column].
______________________________________________________________________________________
Problem 3.11
Solution:
Known quantities:
Circuit shown in Figure P3.11
VS1 = VS 2 = 110 V
R1 = 500 mΩ R2 = 167 mΩ
R3 = 700 mΩ
R4 = 200 mΩ R5 = 333 mΩ
Find:
a. The most efficient way to solve for the voltage across R3. Prove your case.
b. The voltage across R3.
Analysis:
a) There are 3 meshes and 3 mesh currents requiring the solution of 3 simultaneous equations. Only one of
these mesh currents is required to determine, using Ohm's Law, the voltage across R3.
In the terminal (or node) between the two voltage sources is made the ground (or reference) node, then
three node voltages are known (the ground or reference voltage and the two source voltages). This leaves
only two unknown node voltages (the voltages across R1, VR1, and across R2, VR2). Both these voltages
are required to determine, using KVL, the voltage across R3, VR3.
A difficult choice. Choose node analysis due to the smaller number of unknowns. Specify the nodes.
Choose one node as the ground node. In KCL, assume unknown currents flow out.
b)
KCL:
KCL:
VR1 − VS1 VR1 − 0 VR1 − VR 2
+
+
=0
R4
R1
R3
VR 2 − (− VS 2 ) VR 2 − 0 VR 2 − VR1
+
+
=0
R5
R2
R3
§ 1
§ 1 · V
1
1 ·
¸¸ + VR 2 ¨¨ − ¸¸ = S1
+
VR1 ¨¨ +
© R1 R3 R4 ¹
© R3 ¹ R4
§ 1 ·
§ 1
V
1
1 ·
¸¸ = − S 2
+
+
VR1 ¨¨ − ¸¸ + VR 2 ¨¨
R5
© R3 ¹
© R5 R2 R3 ¹
3.9
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
1
1
1
1
1
1
+
+
=
+
+
= 8.43 Ω-1
−3
−3
−3
R1 R3 R4 500 ⋅10
700 ⋅10
200 ⋅10
1
1
1
1
1
1
+
+
=
+
+
= 10.42 Ω-1
−3
−3
−3
R5 R2 R3 333 ⋅10
167 ⋅10
700 ⋅10
1
1
=
= 1.43 Ω-1
R3 700 ⋅10 −3
VS 1
VS 2
110
110
=
= 550 A
=
= 330 A
−3
R4 200 ⋅10
R5 333 ⋅10 −3
− 1.43
550
VR1 =
− 330 10.42
(5731) − (472) = 61.30 V
=
8.43 − 1.43 (87.84 ) − (2.04 )
− 1.43 10.42
8.429
550
− 1.429 − 330 (− 2782 ) − (− 786 )
=
= −23.26 V
85.790
85.80
− VR1 + VR 3 + VR 2 = 0
VR 3 = VR1 − VR 2 = 84.59 V
VR1 =
KVL:
______________________________________________________________________________________
Problem 3.12
Solution:
Known quantities:
Circuit shown in Figure P3.12
VS 2 = kT
k = 10 V/°C
VS1 = 24 V
RS = R1 = 12 kΩ
R2 = 3 kΩ
R3 = 10 kΩ
.
R4 = 24 kΩ
VR 3 = −2.524 V
The voltage across R3, which is given, indicates the temperature.
Find:
The temperature, T.
Analysis:
Specify nodes (A between R1 and R3, C between R3 and R2) and polarities of voltages (VA from ground to
A, Vc from ground to C, and VR3 from C to A). When using KCL, assume unknown currents flow out.
KVL:
− V A + VR 3 + VC = 0
VC = V A − VR 3
Now write KCL at node C, substitute for VC, solve for VA:
KCL:
VC − VS1 VC − VA VC
+
+
=0
R2
R3
R4
3.10
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
§ 1
V A VS 1
1
1 ·
¸¸ = 0
−
+ (V A − VR 3 )¨¨
+
+
R3 R2
© R2 R3 R4 ¹
§ 1
VS 1
1
1 ·
24
1
1 ·
§ 1
¸¸
+ VR 3 ¨¨
+
+
+ (− 2.524)¨
+
+
¸
3
3
3
R
24 ⋅103 ¹
© R2 R3 R4 ¹ = 3 ⋅10
© 3 ⋅10 10 ⋅10
= 18.14 V
VA = 2
1
1
1
1
+
+
R2 R4
3 ⋅103 24 ⋅10 3
−
VC = V A − VR 3 = 18.14 − (− 2.524) = 20.66 V
Now write KCL at node A and solve for VS2 and T:
KCL:
VA − VS1 VA − VS 2 VA − VC
+
+
=0
R1
RS
R3
R
R
VS 2 = V A + S (V A − VS1 ) + S (V A − VC ) =
R1
R3
12 ⋅10 3
12 ⋅10 3
(
)
(18.14 − 20.66) = 9.26 V
18
.
14
−
24
+
12 ⋅10 3
10 ⋅10 3
V
9.26
T = S2 =
= 0.926 °C
10
k
= 18.14 +
______________________________________________________________________________________
Problem 3.13
Solution:
Known quantities:
Circuit shown in Figure P3.13
VS = 5 V
R2 = 1.8 kΩ
AV = 70
R1 = 2.2 kΩ
R3 = 6.8 kΩ
R4 = 220 Ω
Find:
The voltage across R4 using KCL and node voltage analysis.
Analysis:
A node analysis is not a method of choice because the dependent source is [1] a voltage source and [2] a
floating source. Both factors cause difficulties in a node analysis. A ground is specified. There are three
unknown node voltages, one of which is the voltage across R4. The dependent source will introduce two
additional unknowns, the current through the source and the controlling voltage (across R1) that is not a
node voltage. Therefore 5 equations are required:
3.11
G. Rizzoni, Principles and Applications of Electrical Engineering
[1] KCL
V1 − VS V1 − V3 V1 − V2
+
+
=0
R1
R3
R2
[2] KCL
V2 − V1
− I CS = 0
R2
[3] KCL
V3 − V1
V
+ I CS + 3 = 0
R3
R4
[4] KVL
[5] KVL
− VS + VR1 + V1 = 0
Problem solutions, Chapter 3
VR1 = VS − V1
V2 = V3 + AV VR1 = V3 + AV (VS − V1 )
Substitute using Equation [5] into Equations [1], [2] and [3] and eliminate V2 (because it only appears
− V3 − AV VR1 + V2 = 0
twice in these equations). Collect terms:
§1
§ 1
V V A
1
1 AV ·
1 ·
¸¸ + V3 ¨¨ − − ¸¸ + I CS (0) = S + S V
V1 ¨¨ +
+
+
R1
R2
© R1 R3 R2 R2 ¹
© R3 R2 ¹
§ 1 AV ·
§ 1 ·
V A
¸¸ + V3 ¨¨ ¸¸ + I CS (− 1) = − S V
V1 ¨¨ −
−
R2
© R2 R2 ¹
© R2 ¹
§ 1 ·
§ 1
1 ·
V1 ¨¨ − ¸¸ + V3 ¨¨ + ¸¸ + I CS (+ 1) = 0
© R3 ¹
© R3 R4 ¹
1
1
1
1
=
= 555.6 ⋅10 -6 Ω-1
=
= 147.1 ⋅10 -6 Ω-1
3
R2 1.8 ⋅10
R3 6.8 ⋅10 3
1
1
1
1
+
=
+
= 702.6 ⋅10 -6 Ω-1
3
R3 R2 6.8 ⋅10 1.8 ⋅10 3
1
1
1
1
1 AV
1 + 70
+
=
+
= 4.69 ⋅10 -3 Ω-1
+
=
= 39.44 ⋅10 -3 Ω-1
3
3
3
R3 R4 6.8 ⋅10
R2 R2 1.8 ⋅10
0.22 ⋅10
1
1
1 AV
1
1
1 + 70
+
+
+
=
+
+
= 40.05 ⋅10 -3 Ω-1
3
3
3
R1 R3 R2 R2 2.2 ⋅10
6.8 ⋅10 1.8 ⋅10
VS AV
(5)(70 )
=
= 194.4 mA
R2
1.8 ⋅ 103
VS VS AV
5
(5)(70)
+
=
+
= 196.7 mA
3
R1
R2
2.2 ⋅ 10 1.8 ⋅10 3
Solving, we have:
VR 4 = V3 = 5.1 mV
Note:
1. This solution was not difficult in terms of theory, but was terribly long and arithmetically cumbersome.
This was because the wrong method was used. There are only 2 mesh currents in the circuit; the
sources were voltage sources; therefore, a mesh analysis is the method of choice.
2. In general, a node analysis will have fewer unknowns (because one node is the ground or reference
node) and will, in such cases, be preferable.
______________________________________________________________________________________
3.12
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Problem 3.14
Solution:
Known quantities:
The values of the resistors and of the voltage sources
(see Figure P3.14).
Find:
The voltage across the 10 Ω resistor in the circuit of
Figure P3.14 using mesh current analysis.
Analysis:
For mesh (a):
ia (50 + 20 + 20) − ib (20) − ic (20) = 12
For mesh (b):
− ia (20 ) + ib (20 + 10) − ic (10) + 5 = 0
For mesh (c):
− ia (20) − ib (10) + ic (20 + 10 + 15) = 0
Solving,
ia = 127.5 mA
ib = −67.8 mA
and
ic = 41.6 mA
vR4 = 10 (ib − ic ) = 10 (− 0.109 ) = −1.09 V .
______________________________________________________________________________________
Problem 3.15
Solution:
Known quantities:
The values of the resistors, of the
voltage source and of the current
source in the circuit of Figure P3.15.
Find:
The voltage across the current source
using mesh current analysis.
Analysis:
For mesh (a):
ia (20 + 30) + ib (− 30) = 3
For meshes (b) and (c):
ia (− 30) + ib (10 + 30) + ic (30 + 20) = 0
For the current source:
ic − ib = 0.5
Solving,
3.13
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
ia = −133 mA , ib = −322 mA and ic = 178 mA .
Therefore,
v = ic (30 + 20) = 8.89 V .
______________________________________________________________________________________
Problem 3.16
Solution:
Known quantities:
The values of the resistors and of
the voltage source in the circuit of
Figure P3.16.
Find:
The current i through the
resistance R4 mesh current
analysis.
Analysis:
For mesh (a):
ia (50 + 1200) + ib (− 1200) = 5.6
For meshes (b) and (c):
ia (− 1200) + ib (1200 + 330) + ic (440) = 0
For the current source:
ic − ib = 0.2v x = 0.2 (1200 (ia − ib )) = 240 (ia − ib )
Solving,
ia = 136 mA , ib = 137 mA and ic = −106 mA .
Therefore,
i = ic = −106 mA .
______________________________________________________________________________________
Problem 3.17
Solution:
Known quantities:
The values of the resistors, of the voltage source and of the current source in the circuit of Figure P3.5.
Find:
The current through the voltage source using mesh
current analysis.
Analysis:
For mesh (a):
ia (100 + 5) + ib (− 5) + 50 = 0
For the current source:
ib − ic = 0.2
3.14
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
For meshes (b) and (c):
− ia (5) + ib (200 + 5) + ic (50) = 50
Solving,
ia = −465 mA , ib = 226 mA and ic = 26 mA .
Therefore,
i = ic − ia = 491 mA .
______________________________________________________________________________________
Problem 3.18
Solution:
Known quantities:
The values of the resistors and of the current source in the circuit of Figure P3.6.
Find:
The current through the voltage source in the circuit of Figure P3.6 using mesh current analysis.
Analysis:
50 Ω
75 Ω
10V
ia
+
i
0.2 A
200 Ω
ib
25 Ω
100 Ω
ic
For mesh (a):
ia (100) + 10 = 0
For mesh (b):
ib (200 + 75 + 25) + ic (− 25) + 0.2 (− 200) = 0
For mesh (c):
ib (− 25) + ic (50 + 25) = 10
Solving,
ia = −100 mA , ib = 148 mA and ic = 183 mA .
Therefore,
i = ic − ia = 283 mA
______________________________________________________________________________________
Problem 3.19
Solution:
Known quantities:
The values of the resistors in the circuit of Figure P3.19.
3.15
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Find:
The current in the circuit of Figure P3.19 using mesh current analysis.
Analysis:
1/2 Ω
1Ω
i2
I
i
i1
1/5 Ω
i3
1/4 Ω
1/3 Ω
Since I is unknown, the problem will be solved in terms of this current.
For mesh #1, it is obvious that:
i1 = I
For mesh #2:
§ 1 1·
§ 1·
i1 (− 1) + i2 ¨1 + + ¸ + i3 ¨ − ¸ = 0
© 2 5¹
© 5¹
For mesh #3:
§ 1·
§ 1·
§1 1 1·
i1 ¨ − ¸ + i2 ¨ − ¸ + i3 ¨ + + ¸ = 0
© 4¹
© 5¹
© 4 3 5¹
Solving,
i2 = 0.645 I
i3 = 0.483I
i = i3 − i2
Then,
and
i = 0.483I − 0.645 I = −0.163I
______________________________________________________________________________________
Problem 3.20
Solution:
Known quantities:
The values of the resistors of the circuit in Figure P3.20.
Find:
The voltage gain,
AV =
v2
, in the circuit of Figure P3.20
v1
using mesh current analysis.
Analysis:
Note that
v=
i1 − i2
2
For mesh #1:
3.16
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
§ 1·
§ 1·
i1 ¨1 + ¸ + i2 ¨ − ¸ + i3 (0 ) = v1
© 2¹
© 2¹
For mesh #2:
§ 1·
§1 1 1·
§ 1·
i1 ¨ − ¸ + i2 ¨ + + ¸ + i3 ¨ − ¸ = 2v
© 4¹
©2 4 4¹
© 2¹
For mesh #3:
or
i1 (− 1.5) + i2 (2) + i3 (− 0.25) = 0
§1 1·
§ 1·
i1 (0) + i2 ¨ − ¸ + i3 ¨ + ¸ = −2v
©4 4¹
© 4¹
or
i1 (1) + i2 (− 1.25) + i3 (0.5) = 0
i3 = −0.16v1
Solving,
1
from which
v2 = i3 = −0.04v1
4
v
and
AV = 2 = −0.04
v1
______________________________________________________________________________________
Problem 3.21
Solution:
Known quantities:
Circuit in Figure P3.21 and the values of the voltage sources,
VS1 = VS 2 = 450 V , and the values of the 5
resistors:
R1 = 8 Ω
R2 = 5 Ω
R4 = R5 = 0.25 Ω
R3 = 32 Ω
Find:
The voltages across R1, R2 and R3 using KCL and node analysis.
Analysis:
Choose a ground/reference node. The node common to the two voltage sources is the best choice. Specify
polarity of voltages and direction of the currents.
KCL:
KCL:
VR1 − VS 1 VR1 − 0 VR1 − VR 2
+
+
=0
R4
R1
R3
VR 2 − (− VS 2 ) VR 2 − 0 VR 2 − VR1
+
+
=0
R5
R2
R3
Collect terms in terms of the unknown node voltages:
§ 1
§ 1 · V
1
1 ·
VR1 ¨¨ + + ¸¸ + VR 2 ¨¨ − ¸¸ = S1
© R4 R1 R3 ¹
© R3 ¹ R4
§ 1 ·
§ 1
V
1
1 ·
+ ¸¸ = − S 2
VR1 ¨¨ − ¸¸ + VR 2 ¨¨ +
R5
© R3 ¹
© R5 R2 R3 ¹
3.17
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Evaluate the coefficients of the unknown node voltages:
VS1 VS 2 450
=
=
= 1.8 kA
R4
R5 0.25
1
1
=
= 0.03125 Ω−1
R3 32
1
1
1
1
1 1
+
+
=
+ +
= 4.14 Ω−1
R1 R4 R3 0.25 8 32
1
1
1
1
1 1
+
+
=
+ +
= 4.23 Ω−1
R5 R2 R3 0.25 5 32
VR1 =
1800
− 31.25 ⋅ 10 −3
− 1800
4.23
4.16
− 31.25 ⋅10 −3
− 31.25 ⋅ 10 −3
4.23
4.156
1800
−3
− 31.25 ⋅10
− 1800
= −422.2 V
17.59
− VR1 + VR 3 + VR 2 = 0
VR 2 =
KVL:
= 429.5 V
VR 3 = VR1 − VR 2 = 852.0 V
______________________________________________________________________________________
Problem 3.22
Solution:
Known quantities:
Circuit in Figure P3.22 with the values of the voltage sources,
VS1 = VS 2 = 115 V , and the values of the
5 resistors:
R1 = R2 = 5 Ω
R3 = 10 Ω
R4 = R5 = 200 mΩ
Find:
The new voltages across R1, R2 and R3, in case F1 "blows" or opens using KCL and node analysis.
Analysis:
Specify polarity of voltages. The ground is already specified. The current through the fuse F1 is zero.
KCL:
KCL:
VR1 − 0 VR1 − VR 2
+
=0
R1
R3
VR 2 − (− VS 2 ) VR 2 − 0 VR 2 − VR1
+
+
=0
R5
R2
R3
0+
Collect terms in unknown node voltages:
3.18
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
§ 1
§ 1 ·
1 ·
VR1 ¨¨ + ¸¸ + VR 2 ¨¨ − ¸¸ = 0
© R1 R3 ¹
© R3 ¹
§ 1 ·
§ 1
V
1
1 ·
+ ¸¸ = − S 2
VR1 ¨¨ − ¸¸ + VR 2 ¨¨ +
R5
© R3 ¹
© R5 R2 R3 ¹
1
1
=
= 0.1 Ω−1
R3 10
1
1
+
= 0.3 Ω−1
R1 R3
VS 2
115
=
= 575 A
R5 200 ⋅ 10 -3
0
− 0 .1
VR1 =
− 575 5.3
(0) − (57.5) = −36.39 V
=
0.3 − 0.1 (1.59 ) − (0.01)
− 0.1 5.3
0 .3
VR 2 =
1
1
1
+
+
= 5.3 Ω−1
R5 R2 R3
0
− −0.1 − 575 (− 172.5) − (0 )
=
= −109.2 V
1.58
1.58
− VR1 + VR 3 + VR 2 = 0
KVL:
VR 3 = VR1 − VR 2 = 72.81 V
− VS1 + VR 4 + VF + VR1 = 0
KVL:
VR 4 = I1 R4 = 0
VF = 115 − 0 − (− 36.39) = 151.4 V
Note the voltages are strongly dependent on the loads (R1, R2 and R3) connected at the time the fuse
blows. With other loads, the result will be quite different.
______________________________________________________________________________________
Problem 3.23
Solution:
Known quantities:
Circuit in Figure P3.22 and the values of the voltage sources,
VS1 = VS 2 = 120 V , and the values of the 5
resistors:
R1 = R2 = 2 Ω
R3 = 8 Ω
R4 = R5 = 250 mΩ
Find:
The voltages across R1, R2,
R3, and F1 in case F1 "blows" or opens using KCL and node analysis.
Analysis:
Specify polarity of voltages. The ground is already specified. The current through the fuse F1 is zero.
KCL:
0+
VR1 − 0 VR1 − VR 2
+
=0
R1
R3
3.19
G. Rizzoni, Principles and Applications of Electrical Engineering
KCL:
Problem solutions, Chapter 3
VR 2 − (− VS 2 ) VR 2 − 0 VR 2 − VR1
+
+
=0
R5
R2
R3
§ 1
§ 1 ·
1 ·
VR1 ¨¨ + ¸¸ + VR 2 ¨¨ − ¸¸ = 0
© R1 R3 ¹
© R3 ¹
§ 1 ·
§ 1
V
1
1 ·
+ ¸¸ = − S 2
VR1 ¨¨ − ¸¸ + VR 2 ¨¨ +
R5
© R3 ¹
© R5 R2 R3 ¹
1 1
= = 0.125 Ω−1
R3 8
1
1
+
= 0.625 Ω−1
R1 R3
VS 2
120
=
= 480 A
R5 250 ⋅ 10 -3
0
− 0.125
VR1 =
4.625
(0 ) − (60) = −20.87 V
=
− 0.125 (2.89 ) − (0.016 )
− 0.125 4.625
− 480
0.625
0.625
VR 2 =
KVL:
KVL:
1
1
1
+
+
= 4.625 Ω−1
R5 R2 R3
0
− 0.125 − 480 (− 300 ) − (0 )
=
= −104.35 V
2.87
2.87
− VR1 + VR 3 + VR 2 = 0
VR 3 = VR1 − VR 2 = 83.48 V
− VS1 + VR 4 + VF + VR1 = 0
VR 4 = I1 R4 = 0
VF = 120 − 0 − (− 20.87 ) = 140.9 V
______________________________________________________________________________________
Problem 3.24
Solution:
Known quantities:
The values of the voltage sources,
VS1 = VS 2 = VS 3 = 170 V , and the values of the 6 resistors in the
circuit of Figure P3.24:
RW 1 = RW 2 = RW 3 = 0.7 Ω
R1 = 1.9 Ω
R2 = 2.3 Ω
R3 = 11 Ω
Find:
a. The number of unknown node voltages and mesh currents.
b. Unknown node voltages.
Analysis:
If the node common to the three sources is chosen as the ground/reference node, and the series resistances
are combined into single equivalent resistances, there is only one unknown node voltage. On the other
3.20
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
hand, there are two unknown mesh currents. A node analysis is the method of choice! Specify polarity of
voltages and direction of currents.
Req1 = RW 1 + R1 = 2.6 Ω
Req 2 = RW 2 + R2 = 3.0 Ω
Req3 = RW 3 + R3 = 11.7 Ω
KCL:
VN − VS1 VN − (− VS 2 ) VN − VS 3
+
+
=0
Req1
Req 2
Req 3
VS 1 VS 2 VS 3
170 170 170
−
+
−
+
Req1 Req 2 Req 3
VN =
= 2.6 3.0 11.7 = 28.94 V
1
1
1
1
1
1
+
+
+
+
Req1 Req 2 Req 3
2.6 3.0 11.7
KVL:
− VS1 + I1 RW 1 + I1 R1 + V N = 0
I1 =
VS1 − VN 170 − 28.94
=
= 54.26 A
RW 1 + R1
2.6
______________________________________________________________________________________
Problem 3.25
Solution:
Known quantities:
The values of the voltage sources,
VS 1 = VS 2 = VS 3 = 170 V , the common node voltage,
VN = 28.94 V ,and the values of the 6 resistors in the circuit of Figure P3.24:
RW 1 = RW 2 = RW 3 = 0.7 Ω
R1 = 1.9 Ω
R2 = 2.3 Ω
R3 = 11 Ω
Find:
The current through and voltage across R1.
Analysis:
KVL:
− VS1 + I1 RW 1 + I1 R1 + V N = 0
I1 =
VS1 − VN 170 − 28.94
=
= 54.26 A
RW 1 + R1
2.6
= I1 R1 = (54.26)(1.9) = 103.1 V
OL:
VR1
______________________________________________________________________________________
Problem 3.26
Solution:
Known quantities:
The values of the voltage sources,
VS1 = VS 2 = VS 3 = 170 V , and the values of the 6 resistors in the
circuit of Figure P3.24:
3.21
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
RW 1 = RW 2 = RW 3 = 0.7 Ω
R1 = 1.9 Ω
R2 = 2.3 Ω
R3 = 11 Ω
Find:
The mesh (or loop) equations and any additional equation required to determine the current through R1 in
the circuit shown in Figure P3.24.
Analysis:
KVL:
− VS1 + I1RW 1 + I1R1 + (I1 − I 2 )R2 + (I1 − I 2 )RW 2 − VS 2 = 0
KVL:
VS 2 + (I 2 − I1 )RW 2 + (I 2 − I1 )R2 + I 2 R3 + I 2 RW 3 + VS 3 = 0
I1 (R1 + RW 1 + R2 + RW 2 ) + I 2 (− R2 − RW 2 ) = VS1 + VS 2
I1 (− R2 − RW 2 ) + I 2 (R2 + RW 2 + R3 + RW 3 ) = −VS 2 − VS 3
I R1 = I 1 =
(VS1 + VS 2 )
− (R2 + RW 2 )
− (VS 2 + VS 3 ) (R2 + RW 2 + R3 + RW 3 )
(R1 + RW 1 + R2 + RW 2 )
− (R2 + RW 2 )
(R2 + RW 2 + R3 + RW 3 )
− (R2 + RW 2 )
______________________________________________________________________________________
Problem 3.27
Solution:
Known quantities:
The values of the voltage sources,
VS1 = 90 V, VS 2 = VS 3 = 110 V , and the values of the 6 resistors
in the circuit of Figure P3.24:
RW 1 = RW 2 = RW 3 = 1.3 Ω
R1 = 7.9 Ω
R2 = R3 = 3.7 Ω
Find:
The branch currents, using KVL and loop analysis.
Analysis:
Three equations are required. Voltages will be summed around the 2 loops that are meshes, and KCL at the
common node between the resistances. Assume directions of the branch currents and the associated
polarities of the voltages. After like terms are collected:
KVL:
KVL:
KCL:
− VS1 + I1RW 1 + I1R1 + (I1 − I 2 )R2 + (I1 − I 2 )RW 2 − VS 2 = 0
VS 2 + (I 2 − I1 )RW 2 + (I 2 − I1 )R2 + I 2 R3 + I 2 RW 3 − VS 3 = 0
I 3 = I1 − I 2
Plugging in the given parameters results in the following system of equations:
14.2 I1 − 5.0 I 2 = 200
5.0 I1 − 10.0 I 2 = 0
I 3 = I1 − I 2
Solving the system of equations gives:
I1 = 17.09 A
I 2 = 8.55 A
3.22
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
I 3 = 8.55 A
Hence, the assumed polarity of the second and third branch currents is actually reversed.
______________________________________________________________________________________
Problem 3.28
Solution:
Known quantities:
The values of the voltage sources,
Figure P3.22:
R1
VS1 = VS 2 = 115 V , and the values of the 5 resistors in the circuit of
5
R2
R3 10
R4
R5
200 m
Find:
The voltages across R1, R2 and R3, under normal conditions, i.e., no blown fuses using KVL and a mesh
analysis.
Analysis:
KVL:
I1 R1 R4
R1I3
I2 R2
R2 I3 Vs2
R5
-I1R1 I2 R2
Vs1
R1 R2
R3 I3
0
Rearranging the above equations:
5.2 I1 − 5I 3 = 115
5.2 I 2 − 5I 3 = 115
5I1 + 5I 2 − 20 I 3 = 0
I1 = I 2
I1 = 2 I 3
−5
115 0
115 5.2 − 5
0
5 − 20 − 11960
=
= 42.6 A
I1 = I 2 =
−5
5.2 0
− 280.8
0 5.2 − 5
5
5 − 20
I 3 = 21.3 A
VR1 = R1 (I1 − I 3 ) = 106.5 V
VR 2 = R2 (I 3 − I 2 ) = −106.5 V
VR 3 = R3 I 3 = 213 V
______________________________________________________________________________________
3.23
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Problem 3.29
Solution:
Known quantities:
The values of the voltage sources,
VS1 = VS 2 = 110 V , and the values of the 5 resistors in the circuit of
Figure P3.22:
R1 = 100 Ω
R2 = 22 Ω
R3 = 70 Ω
R4 = R5 = 13 Ω
Find:
The voltage across R1 using KVL and mesh analysis.
Analysis:
KVL:
I1 (R1 + R4 ) − R1 I 3 = VS1
I 2 (R2 + R5 ) − R2 I 3 = VS 2
-I1 R1 − I 2 R2 + (R1 + R2 + R3 )I 3 = 0
Rearranging the above equations:
113I1 − 100 I 3 = 110
35 I 2 − 22 I 3 = 110
100 I1 + 22 I 2 − 192 I 3 = 0
110
110
0
I1 =
113
0
100
0
35
22
0
35
22
− 100
− 22
− 192 − 935660
=
= 2.64 A
− 100 − 354668
− 22
− 192
I 3 = 1.88 A
VR1 = R1 (I 1 − I 3 ) = 75.89 V
______________________________________________________________________________________
Problem 3.30
Solution:
Known quantities:
The values of the voltage sources,
VS1 = VS 2 = 115 V , and the values of the 5 resistors in the circuit of
Figure P3.22:
R1 = 5 Ω
R2 = 5 Ω
R3 = 10 Ω
R4 = R5 = 0.2 Ω
Find:
The voltage across R1 using KVL and mesh analysis.
3.24
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Analysis:
Specify polarity of voltages. The ground is already specified. The current through the fuse F1 is zero.
KVL:
KVL:
I1 = 0
− VS 2 + (I 2 − I 3 )R2 + I 2 R5 = 0
I 3 R1 + I 3 R3 + (I 3 − I 2 )R2 = 0
I 2 (R2 + R5 ) + I 3 (− R2 ) = VS 2
I 2 (− R2 ) + I 3 (R1 + R2 + R3 ) = 0
R2 + R5 = 5.2 Ω
R1 + R2 + R3 = 20 Ω
115 − 5
0 20 (2300) − (0)
I2 =
=
= 29.11 A
5.2 − 5 (104) − (25)
− 5 20
5.2 115
−5 0
(0) − (− 575) = 7.28 A
I3 =
=
5.2 − 5
79
− 5 20
VR1 = I R1 R1 = − I 3 R1 = −36.39 V
VR 2 = I R 2 R2 = (I 3 − I 2 )R2 = −109.16 V
VR 3 = I R 3 R3 = I 3 R3 = 72.78 V
KVL:
− VS1 + I1 R4 + VF + VR1 = 0
VF = 151.39 V
______________________________________________________________________________________
Problem 3.31
Solution:
Known quantities:
The values of the voltage sources,
VS1 = VS 2 = 115 V , and the values of the 5 resistors in the circuit of
Figure P3.22:
R1 = 4 Ω
R2 = 7.5 Ω
R3 = 12.5 Ω
R4 = R5 = 1 Ω
Find:
The voltages across R1, R2,
R3, and across the open fuse using KVL and mesh analysis.
Analysis:
Specify polarity of voltages. The ground is already specified. The current through the fuse F1 is zero.
I1 = 0
3.25
G. Rizzoni, Principles and Applications of Electrical Engineering
KVL:
Problem solutions, Chapter 3
I 2 (R2 + R5 ) − R2 I 3 = VS 2
− I 2 R2 + (R1 + R2 + R3 )I 3 = 0
Rearranging the above equations:
8.5 I 2 − 7.5 I 3 = 115
7.5 I 2 − 24 I 3 = 0
115 − 7.5
0 − 24
− 2760
I2 =
=
= 18.68 A
8.5 − 7.5 − 147.75
7.5 − 24
I 3 = 5.84 A
VR1 = − I 3 R1 = −23.35 V
VR 2 = R2 (I 3 − I 2 ) = −96.3 V
VR 3 = R3 I 3 = 73 V
VF = VS1 − VR1 = 138.35 V
______________________________________________________________________________________
3.26
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Section 3.5: Superposition
Problem 3.32
Solution:
Known quantities:
The values of the voltage sources,
VS1 = 110 V , VS 2 = 90 V and the values of the 3 resistors in the
circuit of Figure P3.32:
R1 = 560 Ω
R2 = 3.5 kΩ
R3 = 810 Ω
Find:
The current through R1 due only to the source VS2.
Analysis:
Suppress VS1. Redraw the circuit. Specify polarity of VR1. Choose ground.
KCL:
OL:
− VR1−2 − 0 − V R1−2 − 0 − VR1− 2 − (− VS 2 )
+
+
=0
R1
R2
R3
VS 2
90
R3
810
VR1−2 =
=
= 33.61 V
1
1
1
1
1
1
+
+
+
+
R1 R2 R3 560 3500 810
V
33.61
= 60.02 mA
I R1−2 = R1−2 =
560
R1
______________________________________________________________________________________
Problem 3.33
Solution:
Known quantities:
The values of the current source, of the voltage source and of the resistors in the circuit of Figure P3.33:
I B = 12 A
RB = 1 Ω
VG = 12 V
RG = 0.3 Ω
R = 0.23 Ω
Find:
The voltage across
R1 using superposition.
Analysis:
Specify a ground node and the polarity of the voltage across
R. Suppress the
voltage source by replacing it with a short circuit. Redraw the circuit.
KCL:
− IB +
VR − I VR − I VR − I
+
+
=0
RB
RG
R
3.27
G. Rizzoni, Principles and Applications of Electrical Engineering
VR − I =
Problem solutions, Chapter 3
IB
12
=
= 1.38 V
1
1
1 1 1
1
+
+
+
+
RB RG R 1 0.3 0.23
Suppress the current source by replacing it with an open circuit.
KCL:
VR −V VR −V − VG VR −V
+
+
=0
RB
RG
R
VG
12
RG
0 .3
=
= 4.61 V
VR −V =
1
1
1 1 1
1
+
+
+
+
RB RG R 1 0.3 0.23
VR = VR − I + VR −V = 5.99 V
Note: Superposition essentially doubles the work required to solve this problem. The voltage across
R can
easily be determined using a single KCL.
______________________________________________________________________________________
Problem 3.34
Solution:
Known quantities:
The values of the voltage sources and of the resistors in the circuit of Figure P3.34:
VS1 = VS 2 = 12 V
R1 = R2 = R3 = 1 kΩ
Find:
The voltage across
R2 using superposition.
Analysis:
Specify the polarity of the voltage across
R2 . Suppress the voltage source VS1 by replacing it with a short
circuit. Redraw the circuit.
1
1 kΩ = 0.5 kΩ
2
R2
(12)(1000) = 8 V
=
R2 + Req 1000 + 500
Req = R1 R3 =
V R 2 − 2 = VS 2
Suppress the voltage source
VS 2 by replacing it with a short circuit. Redraw the
circuit.
1
1 k Ω = 0 .5 k Ω
2
Req
(12 V )(0.5 kΩ) = −4 V
= −VS1
=
R1 + Req 1 kΩ + 0.5 kΩ
Req = R2 R3 =
VR 2−1
VR 2 = VR 2−1 + V R 2−2 = −4 V + 8 V = 4 V
3.28
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Note: Although superposition is necessary to solve some circuits, it is a very inefficient and very
cumbersome way to solve a circuit. This method should, if at all possible, be avoided. It must be used when
the sources in a circuit are AC sources with different frequencies, or where some sources are DC and others
are AC.
______________________________________________________________________________________
Problem 3.35
Solution:
Known quantities:
The values of the voltage sources and of the resistors in the circuit of Figure P3.35:
VS1 = VS 2 = 450 V
R1 = 7 Ω
R2 = 5 Ω
Find:
The component of the current through
R3 = 10 Ω
R4 = R5 = 1 Ω
R3 that is due to VS2, using superposition.
Analysis:
Suppress VS1 by replacing it with a short circuit. Redraw the circuit. A solution using equivalent resistances
R1 and R4 are in parallel:
RR
(7 )(1) = 0.875 Ω
R14 = 1 4 =
R1 + R4 7 + 1
R14 is in series with R3 :
R143 = R14 + R3 = 0.875 + 10 = 10.875 Ω
looks reasonable.
Req = R5 + (R2 R143 ) = R5 +
R2 R143
(5)(10.875) = 4.425 Ω
= 1+
R2 + R143
5 + 10.875
VS 2
450
=
= 101.695 A
Req 4.425
OL:
IS =
CD:
I R 3−2 =
I S R2
(101.695)(5) = 32.03 A
=
5 + 10.875
R2 + R143
______________________________________________________________________________________
Problem 3.36
Solution:
Known quantities:
The values of the voltage sources and of the resistors in the circuit of Figure P3.24:
VS1 = VS 2 = VS 3 = 170 V
RW 1 = RW 2 = RW 3 = 0.7 Ω
R1 = 1.9 Ω
Find:
The current through
R2 = 2.3 Ω
R3 = 11 Ω
R1 , using superposition.
3.29
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Analysis:
Req1 = RW 1 + R1 = 2.6 Ω
Req 2 = RW 2 + R2 = 3 Ω
Req 3 = RW 3 + R3 = 11.7 Ω
Specify the direction of I1. Suppress VS2 and VS3. Redraw circuit.
Req = Req1 +
I I −1 =
Req 2 Req 3
Req 2 + Req 3
= 4.99 Ω
VS 1
= 34.08 A
Req
Suppress VS1 and VS3. Redraw circuit.
KCL:
VA − (− VS 2 ) VA
V
+
+ A =0
Req 2
Req1 Req3
VS 2
= −70.54 V
VA = −
Req 2 Req 2
1+
+
Req1 Req 3
I I −2 = −
VA
= 27.13 A
Req1
Suppress VS1 and VS2. Redraw circuit.
KCL:
V A − (− VS 3 ) V A − 0 V A − 0
+
+
=0
Req 3
Req1
Req 2
VS 3
VA = −
1+
I I −3 = −
Req 3
Req1
+
Req3
= 18.09 V
Req 2
VA
= −6.96 A
Req1
I = I I −1 + I I − 2 + I I −3 = 54.25 A
Note: Superposition should be used only for special conditions, as stated in the solution to Problem 3.34. In
the problem above a better method is:
a.
mesh analysis using KVL (2 unknowns)
b.
node analysis using KCL (1 unknown but current must be obtained using OL).
________________________________________________________________________
Problem 3.37
Solution:
Known quantities:
The values of the resistors, of the voltage source and of the current source in the circuit of Figure P3.5.
Find:
The current through the voltage source using superpoistion.
Analysis:
(1) Suppress voltage source V. Redraw the circuit.
3.30
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
For mesh (a):
ia (100 + 5) + ib (− 5) = 0
For the current source:
ib − ic = 0.2
For meshes (b) and (c):
− ia (5) + ib (200 + 5) + ic (50) = 0
Solving,
ia = 2 mA , ib = 39 mA and ic = −161 mA .
Therefore,
i1 = ic − ia = −163 mA .
(2) Suppress current source I. Redraw the circuit.
For mesh (a):
ia (100 + 5) + ib (− 5) + 50 = 0
For mesh (b):
− ia (5) + ib (200 + 5 + 50) = 50
Solving,
ia = −467mA and ib = 187 mA .
Therefore,
i2 = ib − ia = 654 mA .
Using the principle of superposition,
i = i1 + i2 = 491 mA
________________________________________________________________________
Problem 3.38
Solution:
Known quantities:
The values of the resistors, of the voltage source and of the current source in the circuit of Figure P3.6.
Find:
The current through the voltage source using superposition.
Analysis:
(1) Suppress voltage source V. Redraw the circuit.
50 Ω
75 Ω
ia
i1
0.2 A
200 Ω
ib
25 Ω
3.31
ic
100 Ω
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
For mesh (a):
ia = 0
For mesh (b):
ib (200 + 75 + 25) + ic (− 25) − 40 = 0
For mesh (c):
ib (− 25) + ic (25 + 100) = 0
Solving,
ib = 136 mA and ic = 27 mA .
Therefore,
i1 = ic = 27 mA .
(2) Suppress current source I. Redraw the circuit.
50 Ω
75 Ω
10V
ia
+
i2
200 Ω
ib
25 Ω
ic
100 Ω
For mesh (a):
ia (50) − 10 = 0
For mesh (b):
ib (200 + 75 + 25) + ic (− 25) = 0
For mesh (c):
ib (− 25) + ic (25 + 100) = −10
Solving,
ia = 200 mA , ib = −6.8 mA and ic = −81 mA .
Therefore,
i2 = ic − ia = −281 mA .
Using the principle of superposition,
i = i1 + i2 = −254 mA
________________________________________________________________________
3.32
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Problem 3.39
Solution:
Known quantities:
The voltage source value, 3 V, and the five resistance values, indicated in Figure P3.7.
Find:
The current, i, drawn from the independent voltage source using superposition.
Analysis:
(1) Suppress voltage source V. Redraw the circuit.
At node 1:
v1
v
v −v
+ 1 + 1 2 =0
0.5 0.5 0.25
At node 2:
v2 − v1
v
+ 2 + 0 .5 = 0
0.25 0.75
Solving the system, we obtain:
v1 = −0.075 V , v2
Therefore,
i1 = −
= −0.15 V
v1
= 150 mA .
0 .5
(2) Suppress current source I. Redraw the
circuit.
At node 1:
v1 − 3 v1
v1
+
+
=0
0.5 0.5 (0.25 + 0.5 + 0.25)
Solving,
v1 = 1.2 V
3 − v1
Therefore, i1 =
= 3 .6 A .
0 .5
Using the principle of superposition,
i = i1 + i2 = 3.75 A
________________________________________________________________________
Problem 3.40
Solution:
Known quantities:
Circuit in Figure P3.12
VS 2 = kT
k = 10 V/°C
VS1 = 24 V
RS = R1 = 12 kΩ
R2 = 3 kΩ
R3 = 10 kΩ
R4 = 24 kΩ
.
VR 3 = −2.524 V
3.33
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
The voltage across R3, which is given, indicates the temperature.
Find:
The temperature, T using superposition.
Analysis:
(1) Suppress voltage source
VS 2 . Redraw the circuit.
For mesh (a):
ia (24k ) + ib (− 12k ) + ic (− 12k ) = 24
For mesh (b):
ia (− 12k ) + ib (46k ) + ic (− 10k ) = 0
For mesh (c):
ia (− 12k ) + ib (− 10k ) + ic (25k ) = 0
Solving,
ia = 2.08 mA , ib = 0.83 mA and ic = 1.33 mA .
Therefore,
VR 3, S 2 = 10000(ib − ic ) = −5 V .
(2) Suppress voltage source
VS1 . Redraw the circuit.
For mesh (a):
ia (24k ) + ib (− 12k ) + ic (− 12k ) + 10T = 0
For mesh (b):
ia (− 12k ) + ib (46k ) + ic (− 10k ) = 10T
For mesh (c):
ia (− 12k ) + ib (− 10k ) + ic (25k ) = 0
Solving,
ia = −0.52T mA , ib = 0.029T mA and ic = −0.2381T mA .
Therefore,
VR 3, S1 = 10000(ib − ic ) = 2.671T V .
Using the principle of superposition,
VR 3 = VR 3,S 2 + VR 3,S 1 = −5 + 2.671T = −2.524 V
Therefore,
T = 0.926 °C.
________________________________________________________________________
Problem 3.41
Solution:
Known quantities:
The values of the resistors and of the voltage sources (see Figure P3.14).
Find:
The voltage across the 10 Ω resistor in the circuit of Figure P3.14 using superposition.
3.34
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Analysis:
(1) Suppress voltage source
VS1 . Redraw the circuit.
For mesh (a):
ia (50 + 20 + 20) − ib (20) − ic (20 ) = 0
For mesh (b):
− ia (20 ) + ib (20 + 10) − ic (10) + 5 = 0
For mesh (c):
− ia (20) − ib (10) + ic (20 + 10 + 15) = 0
Solving,
ia = −73.8 mA , ib = −245 mA and ic = −87.2 mA .
Therefore,
V10Ω ,S 1 = 10(ib − ic ) = −1.578 V .
(2) Suppress voltage source
VS 2 . Redraw the circuit.
For mesh (a):
ia (50 + 20 + 20) − ib (20) − ic (20) = 12
For mesh (b):
− ia (20) + ib (20 + 10) − ic (10) = 0
For mesh (c):
− ia (20) − ib (10) + ic (20 + 10 + 15) = 0
Solving,
ia = 201 mA , ib = 177 mA and
ic = 129 mA .
Therefore,
V10Ω , S1 = 10(ib − ic ) = 0.48 V .
Using the principle of superposition,
V10Ω = V10 Ω ,S 2 + V10Ω , S1 = −1.09 V .
________________________________________________________________________
Problem 3.42
Solution:
Known quantities:
The values of the resistors, of the voltage
source and of the current source in the
circuit of Figure P3.15.
Find:
The voltage across the current source using
superposition.
Analysis:
(1) Suppress voltage source. Redraw the circuit.
3.35
G. Rizzoni, Principles and Applications of Electrical Engineering
For mesh (a):
ia (20 + 30) + ib (− 30 ) = 0
For meshes (b) and (c):
ia (− 30) + ib (10 + 30) + ic (30 + 20) = 0
For the current source:
ic − ib = 0.5
Solving,
ia = −208 mA , ib = −347 mA and ic = 153 mA .
Therefore,
vV = ic (30 + 20) = 7.65 V .
(2) Suppress current source. Redraw the circuit.
For mesh (a):
ia (20 + 30) + ib (− 30) = 3
For mesh (b):
ia (− 30) + ib (90) = 0
Solving,
ia = 75 mA and ib = 25 mA .
Therefore,
v I = ib (30 + 20) = 1.25 V .
Using the principle of superposition,
v = vV + v I = 8.9 V .
3.36
Problem solutions, Chapter 3
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Section 3.6: Equivalent circuits
Focus on Methodology: Computation of equivalent resistance
of a one-port network that does not contain dependent sources
1.
2.
3.
Remove the load, leaving the load terminals open circuited.
Zero all independent voltage and current sources
Compute the total resistance between load terminals, with the load removed. This
resistance is equivalent to that which would be encountered by a current source
connected to the circuit in place of the load.
Focus on Methodology: Computing the Thevenin voltage
1.
2.
3.
4.
Remove the load, leaving the load terminals open circuited.
Define the open-circuit voltage vOC across the open load terminals.
Apply any preferred method (e.g.: nodal analysis) to solve for vOC.
The Thevenin voltage is vT = vOC.
Focus on Methodology: Computing the Norton current
1.
2.
3.
4.
Replace the load with a short circuit.
Define the short-circuit current iSC to be the Norton equivalent current.
Apply any preferred method (e.g.: nodal analysis) to solve for iSC.
The Norton current is iN = iSC.
Problem 3.43
Solution:
Known quantities:
The schematic of the circuit (see Figure P3.1).
Find:
The Thévenin equivalent resistance seen by resistor
Norton (short-circuit) current when
R3 , the Thévenin (open-circuit) voltage and the
R3 is the load.
Analysis:
(1) Remove the load, leaving the load terminals open circuited, and the
voltage sources. Redraw the circuit.
RT = R1 || R4 + R2 || R5 =
RR
R1 R4
+ 2 5
R1 + R4 R2 + R5
(2) Remove the load, leaving the load terminals open circuited. Redraw
the circuit.
For node #1:
v1 v1 − VS1
+
=0
R1
R4
For node #2:
3.37
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
v 2 v 2 + VS 2
+
=0
R2
R5
Solving the system,
v1 =
R1
VS 1
R1 + R4
v2 = −
R2
VS 2
R2 + R5
Therefore,
vOC = v1 − v2 =
R2
R1
VS 2
VS 1 +
R2 + R5
R1 + R4
(3) Replace the load with a short circuit. Redraw the circuit.
For mesh (a):
ia (R1 + R4 ) − R1ic = VS1
For mesh (b):
ib (R2 + R5 ) − R2 ic = VS 2
For mesh (c):
− R1ia − R2ib + ic (R1 + R2 ) = 0
Solving the system,
(R1 R2 + R1 R5 + R2 R5 )VS1 + R1 R2VS 2
R1 R4 (R2 + R5 ) + R2 R5 (R1 + R4 )
R R V + (R1 R2 + R1 R4 + R2 R4 )VS 2
ib = 1 2 S1
R1 R4 (R2 + R5 ) + R2 R5 (R1 + R4 )
R (R + R5 )VS1 + R2 (R1 + R4 )VS 2
ic = 1 2
R1 R4 (R2 + R5 ) + R2 R5 (R1 + R4 )
ia =
Therefore,
R1 (R2 + R5 )VS 1 + R2 (R1 + R4 )VS 2
R1 R4 (R2 + R5 ) + R2 R5 (R1 + R4 )
________________________________________________________________________
iSC = ic =
Problem 3.44
Solution:
Known quantities:
The schematic of the circuit (see Figure P3.6).
Find:
The Thévenin equivalent resistance seen by resistor
Norton (short-circuit) current when
R5 , the Thévenin (open-circuit) voltage and the
R5 is the load.
Analysis:
(1) Remove the load, leaving the load terminals open circuited, and the voltage sources. Redraw the circuit.
3.38
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
50 Ω
75 Ω
200 Ω
25 Ω
RT
RT = 25 Ω || (75 Ω + 200 Ω ) = 22.92 Ω
(2) Remove the load, leaving the load terminals open circuited. Redraw the circuit.
50 Ω
75 Ω
V1
0.2 A
200 Ω
V2
25 Ω
For node #1:
v1 v1 − v2
+
= 0 .2
200
75
For node #2:
v2 − v1 v2 v2 − v3
+ +
+ i10V = 0
75
25
50
For node #3:
v3 − v2
= i10V
50
For the voltage source:
v3 + 10 = v2
Solving the system,
v1 = 13.33 V , v2 = 3.33 V and v3 = −6.67 V .
Therefore,
vOC = v3 = −6.67 V .
(3) Replace the load with a short circuit. Redraw the circuit.
3.39
10V
+
V3
+
voc
_
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
50 Ω
75 Ω
10V
ia
+
0.2 A
200 Ω
ib
25 Ω
ic
isc
For mesh (a):
ia (50) = 10
For mesh (b):
ib (300) − ic (25) = 40
For mesh (c):
ib (25) − ic (25) = 10
Solving the system,
ia = 200 mA , ib = 109 mA and ic = −291 mA .
Therefore,
iSC = ic = −291 mA .
________________________________________________________________________
Problem 3.45
Solution:
Known quantities:
The schematic of the circuit (see Figure P3.7).
Find:
The Thévenin equivalent resistance seen by resistor
R5 , the Thévenin (open-circuit) voltage and the
Norton (short-circuit) current when
R5 is the load.
Analysis:
(1) Remove the load, leaving the load terminals open circuited, and the voltage sources. Redraw the circuit.
RT = 0.5 Ω + 0.25 Ω + (0.5 Ω || 0.5 Ω ) = 1 Ω
(2) Remove the load, leaving the load terminals open circuited. Redraw the circuit.
For node #1:
v1 − 3 v1 v1 − v2
+
+
=0
0.5 0.5 0.25
For node #2:
v2 − v1
+ 0 .5 = 0
0.25
Solving the system,
3.40
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
v1 = 1.375 V and v2 = 1.25 V .
Therefore,
vOC = v2 = 1.25 V .
(3) Replace the load with a short circuit. Redraw the
circuit.
For mesh (a):
ia (0.5 + 0.5) − ib (0.5) = 3
For meshes (b) and (c):
− ia (0.5) + ib (0.5 + 0.25) + ic (0.5) = 0
For the current source:
ib − ic = 0.5
Solving the system,
ia = 3.875 A , ib = 1.75 A and ic = 1.25 A .
Therefore,
iSC = ic = 1.25 A .
________________________________________________________________________
Problem 3.46
Solution:
Known quantities:
The schematic of the circuit (see Figure P3.12).
Find:
The Thévenin equivalent resistance seen by resistor
R3 , the
Thévenin (open-circuit) voltage and the Norton (short-circuit)
current when R3 is the load.
Assumption:
As in P3.12, we assume
T = 0.926 °C, so that VS 2 = 9.26 V .
Analysis:
(1) Remove the load, leaving the load terminals open circuited, and the voltage sources. Redraw the circuit.
RT = 12 kΩ || 12 kΩ + 3 kΩ || 24 kΩ = 8.67 kΩ
(2) Remove the load, leaving the load terminals open circuited. Redraw the circuit.
For node #1:
v1 − 24 v1 − 9.26
+
=0
12000
12000
For node #2:
v2 − 24
v2
+
=0
3000 24000
Solving the system,
v1 = 16.63 V and v2 = 21.33 V .
Therefore,
vOC = v1 − v2 = −4.7 V .
3.41
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
(3) Replace the load with a short circuit. Redraw the circuit.
For mesh (a):
ia (24k ) − ib (12k ) − ic (12k ) = 24 − 9.26
For mesh (b):
− ia (12k ) + ib (36k ) = 9.26
For mesh (c):
− ia (12k ) + ic (15k ) = 0
Solving the system,
ia = 1.71 mA , ib = 0.83 mA and
ic = 1.37 mA .
Therefore,
iSC = ib − ic = −0.54 mA .
________________________________________________________________________
Problem 3.47
Solution:
Known quantities:
The schematic of the circuit (see Figure P3.14).
Find:
The Thévenin equivalent resistance seen by resistor R4 ,
the Thévenin (open-circuit) voltage and the Norton (shortcircuit) current when R4 is the load.
Analysis:
(1) Remove the load, leaving the load terminals open
circuited, and the voltage sources. Redraw the circuit.
RT = R2 || (R3 + (R1 || R5 )) = 20 Ω || (20 Ω + (50 Ω || 15 Ω )) = 12.24 Ω
(2) Remove the load, leaving the load terminals open circuited. Redraw the circuit.
For node #1:
v1 − 12 v1 − v2
+
+ i5V = 0
50
20
For node #2:
v2 − v1 v2
+
=0
20
20
For node #3:
v3
− i5V = 0
15
For the 5-V voltage source:
v1 − v3 = 5
Solving the system,
v1 = 5.14 V , v2 = 2.57 V , v1 = 0.13 V and i5V = 8.95 mA .
3.42
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Therefore,
vOC = v2 − v3 = 2.44 V .
(3) Replace the load with a short circuit. Redraw the circuit.
For mesh (a):
ia (90) − ib (20) − ic (20) = 12
For mesh (b):
− ia (20 ) + ib (20 ) + 5 = 0
For mesh (c):
− ia (20) + ic (35) = 0
Solving the system,
ia = 119.5 mA , ib = −130.5 mA and
ic = 68.3 mA .
Therefore,
iSC = ic − ib = 198.8 mA .
________________________________________________________________________
Problem 3.48
Solution:
Known quantities:
The schematic of the circuit (see Figure P3.15).
Find:
The Thévenin equivalent resistance seen by resistor
R5 ,
the Thévenin (open-circuit) voltage and the Norton (shortcircuit) current when R5 is the load.
Analysis:
(1) Remove the load, leaving the load terminals open circuited, and the voltage sources. Redraw the circuit.
RT = 30 Ω + 10 Ω + (20 Ω || 30 Ω ) = 52 Ω
(2) Remove the load, leaving the load terminals open circuited. Redraw the circuit.
For node #1:
v1 − 3 v1 v1 − v2
+ +
=0
20
30
10
For node #2:
v2 − v1
= 0 .5
10
Solving the system,
v1 = 7.8 V and v2 = 12.8 V .
Therefore,
vOC = v2 = 12.8 V .
(3) Replace the load with a short circuit.
Redraw the circuit.
3.43
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
For mesh (a):
ia (20 + 30) − ib (30) = 3
For meshes (b) and (c):
− ia (30 ) + ib (30 + 10) + ic (30 ) = 0
For the current source:
ic − ib = 0.5
Solving the system,
ia = −92 mA , ib = −254 mA and ic = 246 mA .
Therefore,
iSC = ic = 246 mA .
________________________________________________________________________
Problem 3.49
Solution:
Known quantities:
The schematic of the circuit (see Figure P3.33).
Find:
The Thévenin equivalent resistance seen by resistor R , the Thévenin (opencircuit) voltage and the Norton (short-circuit) current when R is the load.
Analysis:
(1) Remove the load, leaving the load terminals open circuited, and the voltage
sources. Redraw the circuit.
RT = 1 Ω || 0.3 Ω = 0.23 Ω
(2) Remove the load, leaving the load terminals
open circuited. Redraw the circuit.
For node #1:
v1 v1 − 12
+
= 12
1
0 .3
Solving,
v1 = 12 V .
Therefore,
vOC = v1 = 12 V .
(3) Replace the load with a short circuit. Redraw the circuit.
For mesh (a):
ia (1 + 0.3) − ib (0.3) − 12(1) + 12 = 0
For mesh (b):
− ia (0.3) + ib (0.3) = 12
Solving the system,
ia = 12 A and ib = 52 A .
Therefore,
3.44
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
iSC = ib = 52 A .
________________________________________________________________________
Problem 3.50
Solution:
Known quantities:
The schematic of the circuit (see Figure P3.35).
Find:
The Thévenin equivalent resistance seen by resistor
R3 , the Thévenin
(open-circuit) voltage and the Norton (short-circuit) current when R3
is the load.
Analysis:
(1) Remove the load, leaving the load terminals open circuited, and the voltage sources. Redraw the circuit.
RT = 1 Ω || 7 Ω + 1 Ω || 5 Ω = 1.71 Ω
(2) Remove the load, leaving the load terminals open circuited.
Redraw the circuit.
For node #1:
v1 − 450 v1
+ =0
1
7
For node #2:
v2 + 450 v2
+ =0
1
5
Solving the system,
v1 = 393.75 V and v2 = −375 V .
Therefore,
vOC = v1 − v2 = 768.75 V .
(3) Replace the load with a short circuit. Redraw the circuit.
For mesh (a):
ia (1 + 7 ) − ic (7 ) = 450
For mesh (b):
ib (5 + 1) − ic (5) = 450
For mesh (c):
− ia (7 ) − ib (5) + ic (7 + 5) = 0
Solving the system,
ia = 450 A , ib = 450 A and ic = 450 A .
Therefore,
iSC = ic = 450 A .
________________________________________________________________________
3.45
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Problem 3.51
Solution:
Known quantities:
The values of the voltage source,
VS = 110 V , and the values of the 4 resistors in the circuit of Figure
P3.51:
R1 = R2 = 930 mΩ
R3 = 100 mΩ
RS = 19 mΩ
Find:
The change in the voltage across the total load, when the customer connects the third load R3 in parallel
with the other two loads.
Analysis:
Choose a ground. If the node at the bottom is chosen as ground (which grounds one terminal of the ideal
source), the only unknown node voltage is the required voltage. Specify directions of the currents and
polarities of voltages.
Without R3:
I S + I1 + I 2
KCL:
=0
OL:
VRS VR1 VR 2
+
+
=0
RS
R1 R2
VOi − VS VOi − 0 VOi − 0
+
+
=0
RS
R1
R2
VS
RS
RS
110
=
= 105.7 V
VOi =
1
1
1 RS 1.04086
+
+
RS R1 R2
With R3:
IS
KCL:
+ I1 + I 2 + I 3 = 0
OL:
VRS VR1 VR 2 VR 3
+
+
+
=0
RS
R1 R2
R3
VOf − VS
VOf − 0 VOf − 0 VOf − 0
+
+
=0
RS
R1
R2
R3
VS
RS
RS
110
=
= 89.37 V
VOf =
1
1
1
1 RS 1.04086 + 0.19
+
+
+
RS R1 R2 R3
Therefore, the voltage decreased by:
∆ Vo = VOf − VOi = −16.33 V
+
Notes:
3.46
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
1.
"Load" to an EE usually means current rather than resistance.
2.
Additional load reduces the voltage supplied to the customer because of the additional voltage dropped
across the losses in the distribution system.
________________________________________________________________________
Problem 3.52
Solution:
Known quantities:
The values of the voltage source,
VS = 450 V , and the values of the 4 resistors in the circuit of Figure
P3.52
R1 = R2 = 1.3 Ω
R3 = 500 mΩ
RS = 19 mΩ
Find:
The change in the voltage across the total load, when the customer connects the third load
with the other two loads.
Analysis:
See Solution to Problem 3.40 for a detailed mathematical analysis.
R3 in parallel
∆ Vo = VOf − VOi = −15.6 V
________________________________________________________________________
Problem 3.53
Solution:
Known quantities:
The circuit shown in Figure P3.53, the values of the terminal voltage,
VT , before and after the application
of the load, respectively VT = 20 V and VT = 18 V , and the value of the load resistor RL = 2.7 kΩ .
Find:
The internal resistance and the voltage of the ideal source.
Analysis:
KVL:
− VS + I T RS + VT = 0
VS = VT = 20 V
If IT = 0:
If VT = 18
OL:
V:
IT =
VT
= 6.67 mA
RL
and
RS =
VS − VT
= 300 Ω
IT
Note that RS is an equivalent resistance, representing the various internal losses of the source and is not
physically a separate component. VS is the voltage generated by some internal process. The source voltage
can be measured directly by reducing the current supplied by the source to zero, i.e., no-load or open-circuit
conditions. The source resistance cannot be directly measured; however, it can be determined, as was done
above, using the interaction of the source with an external load.
________________________________________________________________________
3.47
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Problem 3.54
Solution:
Known quantities:
The values of the voltage source,
VCC = 20 V , and the values of the 2 resistors in the circuit of Figure
P3.54:
R1 = 1.3 MΩ
R2 = 220 kΩ
Find:
The Thévenin equivalent circuit with respect to the port shown in Figure P3.54.
Analysis:
The Thévenin equivalent voltage is the open circuit voltage [with I = 0]
between the terminals of the port. Specify the polarity of the voltage.
KCL:
VTH − VCC VTH
+
+I =0
R1
R2
VTH =
VCC
R1
1
1
+
R1 R2
= 2.895 V
Note that, since I = 0, R1 and R2 are effectively in series,
using a VD relation would be easier.
Suppress the ideal, independent voltage source, by shorting it. Determine the
equivalent resistance with respect to the terminals of the port.
REQ = R1 R2 =
(1.3 ⋅10 )(220 ⋅10 ) = 188.2 kΩ
6
3
1.3 ⋅10 6 + 220 ⋅10 3
______________________________________________________________________________________
Problem 3.55
Solution:
Known quantities:
The values of the battery voltage,
VB = 11 V , the value of the generator voltage, VG = 12 V , and the
values of the 3 resistors in the circuit of Figure P3.55:
RB = 0.7 Ω
RG = 0.3 Ω
RL = 7.2 Ω
Find:
a. The Thévenin equivalent of the circuit to the right of the terminal pair of port x-x'.
b. The terminal voltage of the battery, i.e., the voltage between x and x'.
3.48
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Analysis:
a. Specify the polarity of the voltage:
VD:
VTH =
VG RL
(12)(7.2) = 11.52 V
=
RG + RL 0.3 + 7.2
Suppress generator source:
R EQ = RL RG =
b.
(7.2)(0.3) = 288 mΩ
0 .3 + 7 .2
Specify the polarity of the terminal voltage. Choose a ground.
KCL:
VT − VB VT − VTH
+
=0
RB
REQ
VB VTH
+
RB REQ
= 11.37 V
VT =
1
1
+
RB REQ
______________________________________________________________________________________
Problem 3.56
Solution:
The values of the battery voltage,
VB = 11 V , the value of the generator voltage, VG = 12 V , and the
values of the 3 resistors in the circuit of Figure P3.56:
RB = 0.7 Ω
RG = 0.3 Ω
RL = 7.2 Ω
Find:
a. The Thévenin equivalent of the circuit to the left of the terminal pair of port y-y'.
b. The terminal voltage of the generator, i.e., the voltage between y and y'.
Analysis:
a. Specify the polarity of the Thévenin equivalent voltage:
VD:
VTH =
VB R L
(11)(7.2) = 10.03 V
=
R B + R L 0 .7 + 7 .2
Suppress generator source:
RT = R L RB =
b.
(7.2)(0.7 ) = 638 mΩ
0 .7 + 7 .2
Specify the polarity of the terminal voltage. Choose a ground.
KCL:
VT − VG VT − vT
+
=0
RG
RT
VG vT
+
RG RT
VT =
= 11.37 V
1
1
+
RG RT
3.49
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Section 3.7: Maximum power transfer
Problem 3.57
Solution:
Known quantities:
The values of the voltage and of the resistor in the equivalent circuit of Figure P3.57:
VTH = 12 V
Req = 8 Ω
Assumptions:
Assume the conditions for maximum power transfer exist.
Find:
a. The value of RL .
b.
c.
The power developed in RL .
The efficiency of the circuit, that is the ratio of power absorbed by the load to power supplied by the
source.
Analysis:
a. For maximum power transfer:
R L = Req = 8 Ω
b.
VRL =
VD:
PRL
VTH RL
(12)(8) = 6 V
=
8+8
Req + RL
V 2 RL (6 )2
=
=
= 4.5 W
RL
8
P0
PRL
I S2 RL
RL
c. η =
=
= 2
=
= 0.5 = 50%
2
PS PRe q + PRL I S Req + I S RL Req + RL
________________________________________________________________________
Problem 3.58
Solution:
Known quantities:
The values of the voltage and of the resistor in the equivalent circuit of Figure P3.57:
VTH = 300 V
Req = 600 Ω
Assumptions:
Assume the conditions for maximum power transfer exist.
Find:
a. The value of RL .
b.
c.
The power developed in RL .
The efficiency of the circuit, that is the ratio of power absorbed by the load to power supplied by the
source.
Analysis:
a. For maximum power transfer:
3.50
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
R L = R eq = 600 Ω
b.
VRL =
VD:
PRL
VTH RL
(35)(600) = 17.5 V
=
Req + RL 600 + 600
V 2 RL (17.5)2
=
=
= 510.4 mW
RL
600
P0
PRL
I 2R
RL
=
= 2 S L2 =
= 0.5 = 50%
PS PRe q + PRL I S Req + I S RL Req + RL
________________________________________________________________________
c.
η=
Problem 3.59
Solution:
Known quantities:
The values of the voltage source, VS
source,
= 12 V , and of the resistance representing the internal losses of the
RS = 0.3 Ω , in the circuit of Figure P3.59.
Find:
a. Plot the power dissipated in the load as a function of the load resistance. What can you conclude from
your plot?
b. Prove, analytically, that your conclusion is valid in
Power dissipated in the load
all cases.
120
Analysis:
− VS = IRS + IR = 0
b.
VS
RS + R
R [Ω] I [A]
0
40
0.1
30
0.3
20
0.9
10
2.1
5
KVL:
I=
PR = I R =
2
VS2 R
(R + RS )
2
PR = I 2 R
80
Power [W]
a.
100
PR [W]
0.0
90.0
120.0
90.0
52.5
60
40
20
0
= VS2 R (R + RS )
−2
dPR
−2
−3
= VS2 (1)(R + RS ) + VS2 (R )(− 2 )(R + RS ) (1) = 0
dR
(R + RS )1 − 2 R = 0 Ÿ R = RS
3.51
0
0.5
1
1.5
Load Resistance [Ohm]
2
2.5
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Section 3.8: Nonlinear circuit elements
Problem 3.60
Solution:
Known quantities:
The two nonlinear resistors, in the circuit of Figure P3.60, are characterized by:
ia = 2va3
ib = vb3 + 10vb
Find:
The node voltage equations in terms of v1 and v2.
Analysis:
At node 1,
At node 2,
But
v1
+ ia = 1Ÿ v1 + 2va3 = 1
1
ib − ia = 26 Ÿ vb3 + 10vb − 2va3 = 26
va = v1 − v2 and vb = v2 . Therefore, the node equations are
v1 + 2(v1 − v 2 ) = 1
3
and
v23 + 10v2 − 2(v1 − v2 ) = 26
________________________________________________________________________
3
Problem 3.61
Solution:
Known quantities:
The characteristic curve I-V shown in Figure P3.61, and the values of the voltage, VT
resistance, RT
= 15 V , and of the
= 200 Ω , in the circuit of Figure P3.61.
Find:
a. The operating point of the element that has the characteristic curve shown in Figure P3.61.
b. The incremental resistance of the nonlinear element at the operating point of part a.
c. If VT were increased to 20 V, find the new operating point and the new incremental resistance.
Analysis:
a.
KVL:
− 15 + 200 I + V = 0
− 15 + 200(0.0025V 2 ) + V = 0
Solving for V and I,
I = 52.2 mA
V = 4.57 V or − 6.57 V
The second voltage value is physically impossible.
Rinc = 10(0.0522)−0.5 = 43.8 Ω
c. I = 73 mA
V = 5.40 V
Rinc = 37 Ω
________________________________________________________________________
b.
3.52
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Problem 3.62
Solution:
Known quantities:
The characteristic curve
the resistance, R
I-V shown in Figure P3.62, and the values of the voltage, VS = 450 V , and of
= 9 Ω , in the circuit of Figure P3.62.
Find:
The current through and the voltage across the nonlinear device.
Analysis:
The I-V characteristic for the nonlinear device is given. Plot the circuit I-V
characteristic, i.e., the DC load line.
KVL:
− VS + I D R + V D = 0
VS − VD 450 − VD
=
=
R
9
if VD = 450 V
=0A
ID =
= 50 A
VD = 0
if
The DC load line [circuit characteristic] is linear. Plotting the two intercepts above and connecting them
with a straight line gives the DC load line. The solution for V and I is at the intersection of the device and
circuit characteristics:
I DQ ≈ 26 A V DQ ≈ 210 V .
________________________________________________________________________
Problem 3.63
Solution:
Known quantities:
The I-V characteristic shown in Figure P3.63, and the values of the voltage, VS
resistance, R
= VT = 1.5 V , and of the
= Req = 60 Ω , in the circuit of Figure P3.63.
Find:
The current through and the voltage across the nonlinear device.
Analysis:
The solution is at the intersection of the device and circuit characteristics.
The device I-V characteristic is given. Determine and plot the circuit I-V
characteristic.
KVL:
− VS + I D R + VD = 0
VS − VD 1.5 V − VD
=
=
R
60 Ω
=0A
if VD = 1.5 V
ID =
= 25 mA
if
VD = 0
The DC load line [circuit characteristic] is linear. Plotting the two intercepts above and connecting them
with a straight line gives the DC load line. The solution is at the intersection of the device and circuit
characteristics, or "Quiescent", or "Q", or "DC operating" point:
I DQ ≈ 12 mA VDQ ≈ 0.77 V .
________________________________________________________________________
3.53
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Problem 3.64
Solution:
Known quantities:
The I-V characteristic shown in Figure P3.64 as a function of pressure.
VS = VT = 2.5 V
R = Req = 125 Ω
Find:
The DC load line, the voltage across the device as a function of pressure, and the current through the
nonlinear device when p = 30 psig.
Analysis:
Circuit characteristic [DC load line]:
KVL:
− VS + I D R + VD = 0
VS − VD 2.5 V − VD
=
=
R
125 Ω
=0A
if VD = 2.5 V
ID =
= 20 mA
if
VD = 0
The circuit characteristic is a linear relation. Plot the two intercepts and connect with a straight line to plot
the DC load line. Solutions are at the intersections of the circuit with the device characteristics, i.e.:
p [psig]
VD [V]
10
20
25
30
40
2.14
1.43
1.18
0.91
0.60
The plot is nonlinear.
At p = 30 psig:
VD = 1.08 V
I D = 12.5 mA .
________________________________________________________________________
Problem 3.65
Solution:
Known quantities:
The I-V characteristic of the nonlinear device in the circuit shown in Figure P3.65:
I D = I 0e
vD
VT
I 0 = 10 −15 A
VT = 26 mV
VS = VT = 1.5 V
R = Rea = 60 Ω
Find:
An expression for the DC load line. The voltage across and current through the nonlinear device.
3.54
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 3
Analysis:
Circuit characteristic [DC load line]:
KVL:
− VS + I D R + VD = 0
VS − VD 1.5 − VD
=
R
60
§I ·
§ I ·
VD = VT ln¨¨ D ¸¸ = 0.026 ⋅ ln¨ D−15 ¸
© 10 ¹
© I0 ¹
[1]
ID =
[2]
Iterative procedure:
Initially guess V D = 750 mV . Note this voltage must be between zero and the value of the source
voltage.
Then:
a. Use Equation [1] to compute a new ID.
b. Use Equation [2] to compute a new VD.
c. Iterate, i.e., go step a. and repeat.
VD [mV] ID [mA]
750
12.5
784.1
11.93
782.9
11.95
782.9
11.95
…
…
I DQ ≈ 11.95 mA VDQ ≈ 782.9 mV .
________________________________________________________________________
Problem 3.66
Solution:
Known quantities:
The I-V characteristic shown in Figure P3.66 as a function of pressure.
VS = VT = 2.5 V
R = Req = 125 Ω
Find:
The DC load line, and the current through the nonlinear device when p
= 40 psig.
Analysis:
Circuit characteristic [DC load line]:
KVL:
− VS + I D R + VD = 0
VS − VD 2.5 V − VD
=
=
125 Ω
R
=0 A
if VD = 2.5 V
ID =
= 20 mA
if
VD = 0
The circuit characteristic is a linear relation that can be plotted by plotting the
two intercepts and connecting them with a straight line. Solutions are at the
intersections of the circuit and device characteristics.
At p = 40 psig:
VD = 0.60 V
I D = 15.2 mA
________________________________________________________________________
3.55