Problems
4.47
6. Find
V
1
,
V
2
and
V
3
such that
v
1
=
10 V,
v
2
=
10 V and
v
3
=
20 V in the circuit in Fig. 4.11-6. With
these values of
V
1
,
V
2
and
V
3
, find the power delivered by all voltage sources and power dissipated
by all resistors. Use nodal analysis.
V
2
V
3
V
1
v
1
v
2
v
3
20
Ω
10
Ω
10
Ω
10
Ω
5
Ω
5
Ω
R
+
+
+
–
–
–
Fig. 4.11-6
7. (i) The nodal conductance matrix of the circuit in Fig. 4.11-7 is given as
2
Ω
+
–
2
Ω
10
Ω
R
k
1
i
x
i
x
v
x
i
y
k
3
v
x
k
2
i
y
I
5
Ω
5
Ω
Fig. 4.11-7
Y
=
−
−
−
0 1
0 3 0 13
0 2
0 8
0 47
0 1
0 6
0 7
.
.
.
.
.
.
.
.
.
S. Find
k
1
,
k
2
and
k
3
and solve the circuit completely by nodal analysis if
I
=
1 A.
8. Find
k
1
,
k
2
and
k
3
such that the nodal conductance matrix for the circuit in Fig. 4.11-8 is lower triangular.
Find the power delivered by independent sources and dependent sources. Use nodal analysis.
+
+
+
–
–
–
5 V
10 V
10
Ω
R
k
1
i
x
i
x
i
y
v
x
k
2
v
x
k
3
i
y
4
Ω
20
Ω
5
Ω
2
Ω
5
Ω
Fig. 4.11-8
9. Find all dependent source
coefficients such that the
Y
-matrix of the circuit in Fig. 4.11-9 is
diagonal.
+
+
+
+
+
+
–
–
–
–
–
–
1
Ω
1
Ω
R
k
1
i
y
v
x
v
z
v
y
k
2
v
z
k
6
v
z
k
5
v
y
k
3
v
x
k
4
v
z
4
Ω
5
Ω
5
Ω
5
Ω
Fig. 4.11-9
4.48
Nodal Analysis and Mesh Analysis
of Memoryless Circuits
10. Find
k such that
v
is zero in the circuit in Fig. 4.11-10. Solve the circuit completely with this value
of
k. Use nodal analysis.
1
Ω
4 V
10 V
2
Ω
2
Ω
2
Ω
1
Ω
v
x
v
kv
x
+
+
+
++
–
–
–
––
Fig. 4.11-10
11. Find
k such that
v
is zero in the circuit in Fig. 4.11-11. Solve the circuit completely for this value
of
k. Use nodal analysis.
5 V
2 A
1 A
10 V
2
Ω
1
Ω
2.5
Ω
v
x
v
kv
x
+
+
+
+
+
–
–
–
–
–
Fig. 4.11-11
12. Find the node voltages and resistor currents in the circuit in Fig. 4.11-12 by nodal analysis.
2
Ω
6
Ω
2
Ω
5
Ω
10
Ω
7 A
R
i
x
3
i
x
i
y
2
i
y
10 V
+
+
–
–
Fig. 4.11-12
13. The nodal conductance matrix of the circuit in Fig. 4.11-13 is given below. Find the values of all
resistances in the circuit.
i
3
i
1
i
2
Z
=
Ω
14
–4
–6
12
–4
–3
–6
–3
13
Fig. 4.11-13
14. Express all the resistor currents in the directions as marked in the form linear combinations of
V
1
,
V
2
and
V
3
for the circuit shown in Fig. 4.11-14. Use Mesh Analysis.
4
Ω
4
Ω
5
Ω
6
Ω
5
Ω
+
+
– ––
3
Ω
V
2
V
1
V
3
Fig. 4.11-14
C i r c u i t T h e o r e m s
CHAPTER OBJECTIVES
• To derive Superposition Theorem from the property of linearity of elements.
• To explain the two key theorems – Superposition Theorem and Substitution Theorem in
detail.
• To derive other theorems like Compensation Theorem, Thevenin’s Theorem, Norton’s
Theorem, Reciprocity Theorem and Maximum Power Transfer Theorem from these two
key principles.
• To provide illustrations of applications of circuit theorems in circuit analysis through solved
examples.
• To emphasise the use of Compensation Theorem, Thevenin’s Theorem and Norton’s Theorem
in circuits containing dependent sources as a pointer to their applications in the study of
electronic circuits.
IntroductIon
Circuit Analysis involves determination of element voltages and currents for all elements of the
circuit using element equations and interconnection equations. Kirchhoff’s
Current Law equations
at all nodes and Kirchhoff’s Voltage Law equations in all loops along with element
v
-
i relationship
equations will yield the necessary set of equations.
However, we need systematic procedures for exploiting these equations. Node analysis and mesh
analysis were two such systematic procedures we took up for detailed study in the last chapter. In this
chapter, we discuss some circuit theorems and circuit transformations that increase our efficiency in
solving circuits and that render further insight into certain features of a
linear circuit. These theorems
constitute a basic set of tools that enhance the analyst’s efficiency in solving circuits.
The previous chapter showed the following:
(1) All the element voltages and element currents in a circuit can be obtained from its node
voltages that are governed by a matrix equation
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