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Chapter 3 Nodal and Mesh Equations - Circuit Theorems
3-52 Circuit Analysis I with MATLAB Applications
3.14 Exercises
Multiple Choice
1. The voltage across the resistor in the circuit of Figure 3.67 is
A.
B.
C.
D.
E.
Figure 3.67. Circuit for Question 1
2. The current in the circuit of Figure 3.68 is
A.
B.
C.
D.
E.
Figure 3.68. Circuit for Question 2
6 V
16 V
– 8 V
32 V
none of the above
8 A
6 V
8 A
i
- 2 A 5 A 3 A 4 A none of the above
+^ −
4 V
10 V i
Circuit Analysis I with MATLAB Applications 3-
Exercises
3. The node voltages shown in the partial network of Figure 3.69 are relative to some reference
node which is not shown. The current is
A.
B.
C.
D.
E.
Figure 3.69. Circuit for Question 3
4. The value of the current for the circuit of Figure 3.70 is
A.
B.
C.
D.
E.
Figure 3.70. Circuit for Question 4
i
- 4 A 8 ⁄ 3 A
- 5 A
- 6 A none of the above
+^ −
+^ −
4 V^ 8 V
i + − 8 V
8 V
6 V 13 V
6 V
12 V
i
- 3 A
- 8 A
- 9 A 6 A none of the above
12 V 6 Ω 8 A^3 Ω
3 Ω (^) i
Circuit Analysis I with MATLAB Applications 3-
Exercises
7. For the network of Figure 3.73, the Thevenin equivalent resistance to the right of terminals
a and b is
A.
B.
C.
D.
E.
Figure 3.73. Network for Question 7
8. For the network of Figure 3.74, the Thevenin equivalent voltage across terminals a and b is
A.
B.
C.
D.
E.
Figure 3.74. Network for Question 8
R TH
none of the above
a
b
R TH
2 Ω^2 Ω
V TH
– 3 V
– 2 V
1 V
5 V
none of the above
+^ −
2 Ω 2 A
2 V
a
b
Chapter 3 Nodal and Mesh Equations - Circuit Theorems
3-56 Circuit Analysis I with MATLAB Applications
9. For the network of Figure 3.75, the Norton equivalent current source and equivalent parallel
resistance across terminals a and b are
A.
B.
C.
D.
E.
Figure 3.75. Network for Question 9
10. In applying the superposition principle to the circuit of Figure 3.76, the current due to the
source acting alone is
A.
B.
C.
D.
E.
Figure 3.76. Network for Question 10
I N
R N
1 A 2, Ω
1.5 A 25, Ω
4 A 2.5, Ω
0 A 5, Ω
none of the above
5 Ω^ 2 A
a
b
2 A
i 4 V
8 A
- 1 A 4 A
- 2 A none of the above
8 A 2 Ω
+ − 4 V
i 2 Ω
Chapter 3 Nodal and Mesh Equations - Circuit Theorems
3-58 Circuit Analysis I with MATLAB Applications
Figure 3.79. Circuit for Problem 3
Figure 3.80. Circuit for Problem 4
Figure 3.81. Circuit for Problem 5
12 A 24 A
12 Ω^15 Ω
36 V
i (^) X
i (^6) Ω 5i^ X
18 A
12 A
240 V
36 A
120 V − + + −
24 A
4 Ω^3 Ω
v (^) 36A
12 A 24 A
18 A
12 Ω^15 Ω
36 V
i (^6) Ω −
i (^) X 5i (^) X
Circuit Analysis I with MATLAB Applications 3-
Exercises
Figure 3.82. Circuit for Problem 6
7. Compute the power absorbed by the resistor in the circuit of Figure 3.83 using any method.
Answer:
Figure 3.83. Circuit for Problem 7
8. Compute the power absorbed by the resistor in the circuit of Figure 3.84 using any
method. Answer:
Figure 3.84. Circuit for Problem 8
9. In the circuit of Figure 3.85:
a. To what value should the load resistor should be adjusted to so that it will absorb
maximum power? Answer:
12 V
12 Ω^15 Ω
− − + 24 V^10 Ω
8 Ω (^) i (^) X v (^10) Ω
10i (^) X
1.32 w
12 V
− + 24 V
− 36 V
73.73 w
12 V
+^ −
6 A
8 A
R LOAD
Circuit Analysis I with MATLAB Applications 3-
Exercises
12. Use the superposition principle to compute voltage in the circuit of Figure 3.88.
Answer:
Figure 3.88. Circuit for Problem 12
13.In the circuit of Figure 3.89, and are adjustable voltage sources in the range
V, and and represent their internal resistances. Table 3.4 shows the results
of several measurements. In Measurement 3 the load resistance is adjusted to the same value as
Measurement 1, and in Measurement 4 the load resistance is adjusted to the same value as Mea-
surement 2. For Measurements 5 and 6 the load resistance is adjusted to. Make the neces-
sary computations to fill-in the blank cells of this table.
Answers: , , ,
TABLE 3.4 Table for Problem 13
Measurement Switch Switch (V) (V) (A)
1 Closed Open 48 0 16
2 Open Closed 0 36 6
3 Closed Open 0 − 5
4 Open Closed 0 − 42
5 Closed Closed 15 18
6 Closed Closed 24 0
v (^6) Ω 21.6 V
12 A 18 A 24 A
4 Ω^6 Ω
12 Ω^15 Ω
+^ −^ 36 V
v (^6) Ω
v (^) S1 v (^) S
- 50 ≤ V ≤ 50 R (^) S1 R (^) S
1 Ω
S 1 S (^2) v (^) S1 v (^) S2 i (^) LOAD
– 15 V – 7 A 11 A – 24 V
Chapter 3 Nodal and Mesh Equations - Circuit Theorems
3-62 Circuit Analysis I with MATLAB Applications
Figure 3.89. Network for Problem 13
14. Compute the efficiency of the electrical system of Figure 3.90. Answer:
Figure 3.90. Electrical system for Problem 14
15. Compute the regulation for the 2st floor load of the electrical system of Figure 3.91.
Answer:
Figure 3.91. Circuit for Problem 15
Resistive Load
Adjustable
S 1 S 2
v (^) S
v (^) S
R S
R S
i (^) LOAD v (^) LOAD
480 V
1st FloorLoad
100 A
2nd FloorLoad
v (^) S i^1 i^2^ 80 A
480 V
1st FloorLoad
100 A
2nd FloorLoad
V S i^1 i^2^ 80 A
Chapter 3 Nodal and Mesh Equations - Circuit Theorems
3-64 Circuit Analysis I with MATLAB Applications
3.15 Answers to Exercises
Multiple Choice
1. E The current entering Node A is equal to the current leaving that node. Therefore, there is no
current through the resistor and the voltage across it is zero.
2. C From the figure below,. Also, and. Then,
a n d. T h e r e f o r e ,
3. A From the figure below we observe that the node voltage at A is relative to the reference
node which is not shown. Therefore, the node voltage at B is relative to the
same reference node. The voltage across the resistor is and the direc-
tion of current through the resistor is opposite to that shown since Node B is at a higher
potential than Node C. Thus
8 A
6 V
8 A
A
8 A 8 A
V AC = 4 V V AB = V BC = 2 V V AD = 10 V
V BD = V AD – V AB= 10 – 2 = 8 V V CD = V BD – V BC = 8 – 2 = 6 V
i = 6 ⁄ 2 = 3 A
+^ −
4 V
10 V i
A B^ C
D
6 V
6 + 12 = 18 V
V BC = 18 – 6 = 12 V
i = – 12 ⁄ 3 = – 4 A
+^ −
+^ −
4 V^ 8 V
i + − 8 V
8 V
6 V 13 V
6 V
12 V
A
C B
Circuit Analysis I with MATLAB Applications 3-
Answers to Exercises
4. E We assign node voltages at Nodes A and B as shown below.
At Node A
and at Node B
These simplify to
and
Multiplication of the last equation by 2 and addition with the first yields and thus
5. E Application of KCL at Node A of the circuit below yields
or
Also by KVL
12 V 6 Ω 8 A^3 Ω
A (^) 3 Ω B i
V A – 12
V A
V A – V B
V B – V A
V B
3 ---V^ A
– 3 ---V B= 2
– 3 -- - V A
+ 3 ---V B = 8
V B = 18
i = – 18 ⁄ 3 = – 6 A
2 A 2 Ω
v
v (^) X
2v (^) X
A
v 2 ---^
v – 2v (^) X
v – v (^) X= 2
Circuit Analysis I with MATLAB Applications 3-
Answers to Exercises
8. A Replacing the current source and its parallel resistance with an equivalent voltage source
in series with a resistance we get the network shown below.
By Ohm’s law,
and thus
9. D The Norton equivalent current source is found by placing a short across the terminals a
and b. This short shorts out the resistor and thus the circuit reduces to the one shown
below.
By KCL at Node A,
and thus
The Norton equivalent resistance is found by opening the current sources and looking to
the right of terminals a and b. When this is done, the circuit reduces to the one shown below.
+^ −
2 V
a
b
− + 4 V
i
i = 42 ----------- +–^22 - = 0.5 A
v (^) TH = v (^) ab = 2 × 0.5 +( – 4 )= – 3 V I (^) N 5 Ω
2 A
a
b
2 A
I SC = I N A
I N + 2 = 2
I N = 0
R N
Chapter 3 Nodal and Mesh Equations - Circuit Theorems
3-68 Circuit Analysis I with MATLAB Applications
Therefore, and the Norton equivalent circuit consists of just a resistor.
10. B With the source acting alone, the circuit is as shown below.
We observe that and thus the voltage drop across each of the resistors to the
left of the source is with the indicated polarities. Therefore,
Problems
1. We first replace the parallel conductances with their equivalents and the circuit simplifies to that
shown below.
Applying nodal analysis at Nodes 1, 2, and 3 we get:
Node 1:
Node 2:
a
b
R N = 5 Ω 5 Ω
4 V
+ − 4 V
i 2 Ω
A
B
v (^) AB = 4 V 2 Ω 4 V 2 V i = – 2 ⁄ 2 = – 1 A
12 A 18 A 24 A
4 Ω –^1 6 Ω^ –^1
12 Ω –^1
v (^18) A
v 1 v 2 v (^3) 1 2 3
15 Ω –^1
16v 1 – 12v 2 = 12
- 12 v 1 + 27v 2 – 15v 3 =– 18
Chapter 3 Nodal and Mesh Equations - Circuit Theorems
3-70 Circuit Analysis I with MATLAB Applications
Also,
Simplifying the above equations, we get:
Addition of the first two equations above and multiplication of the third by yields
and by adding the last two equations we get
or
Check with MATLAB:
format rat R=[1/3 −3/20 7/30; −1/12 3/20 −1/15; 1 0 −1]; I=[36 −18 36]'; V=R\I; fprintf('\n'); disp('v1='); disp(V(1)); disp('v2='); disp(V(2)); disp('v3='); disp(V(3))
v1=
v2=
v3=
3. We assign node voltages , , , and current as shown in the circuit below. Then,
v 2 – v (^1) --------------- 12 - v 2 – v (^3)
- --------------- 15 - = – 18
v 1 – v 3 = 36
3 ---v^1
20 ------v^2
- – 15 ------v 3 = – 18 v 1 – v 3 = 36
12 ------v^3 =^9
v 3 = v (^6) Ω= 108 -------- 5 - = 21.6V
v 1 v 2 v 3 v 4 i (^) Y
Circuit Analysis I with MATLAB Applications 3-
Answers to Exercises
and
Simplifying the last two equations above, we get
and
Next, we observe that , and. Then and by
substitution into the last equation above, we get
or
Thus, we have two equations with two unknowns, that is,
v (^1) ---- 4 - v 1 – v (^2)
- --------------- 12 - + 18 – 12 = 0
v 2 – v (^1) --------------- 12 - v 2 – v (^3) --------------- 12 - v 2 – v (^4)
12 A 24 A
12 Ω^15 Ω
36 V
i (^) X
i (^6) Ω 5i^ X
18 A
v 1 v^2 v^3
v (^4) i (^) Y
3 ---v^1
1
60 ------v^2
15 ------v^3
i (^) X= v ---------------^1 12 – v^2 - v 3 = 5i (^) X v 4 = 36 V v 3 = 12 -----^5 -^ ( v 1 – v 2 )
1
60 ------v^2
15 ------^
× 12 ------ ( v 1 – v 2 )
3 ---v^1
- 12 ------v 2 = – 6 1
- 9 ---v (^1)
