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example: 5.1-3
The source function values for the three independent 
sources in Fig. 5.1-7 are V 
=
10 V, I
1
=
1 A and I
2
=
2 
A. The current in the resistor R is seen to be 1 A when 
the two current sources are switched off and 1.5 A 
when only I
2
is switched off and 2 A when all the three 
sources are active. Find what voltage must be applied 
by the voltage source if the current in R is to become 
zero with no change in current source values?
Solution
The current through R can be written as a linear combination of three source functions.
i.e., 
i aV bI
cI
=
+
+
1
2
Substituting the data stated in the problem we get three equations in three unknowns as follows:
1 10
1 5 10
2 10
2
=
=
+
=
+ +
a
a b
a b
c
.
Solving this system of equations, a 
=
0.1 , b 
=
0.5 and c 
=
0.25.
∴ =
+
+
i
V
I
I
0 1
0 5
0 25
1
2
.
.
.
The required voltage to make i zero with I
1 
=
1 A and I
2
=
2 A is obtained by
0 0 1
0 5 0 5
10
=
+
+
⇒ = −
.
.
.
V
V
V
example: 5.1-4
Find the value of V in the circuit in Fig. 5.1-8, such that the voltage source delivers zero power to the 
circuit by using Superposition Theorem.
Fig. 5.1-7 
Circuit for Example 5.1-3
I
1
I
2
R
i
V
A linear memoryless
network without any
independent sources
+
–

---

5.10
Circuit Theorems
Solution
The circuit is a linear one. Therefore, the current 
delivered by the voltage source can be expressed as 
a linear combination of the three source functions. 
Power delivered by the voltage source will be zero if 
the current delivered by it is zero. Thus, we want a 
value of V such that aV 
+ 
bI
1
+
cI
2
=
0, where a, b and c are the per unit contributions to the current 
delivered by the voltage source from the three source functions.
We determine the three contributions first by solving three single-source circuits shown in Fig. 5.1-9.
V
(a)
i
+
–
(b)
6 A
i
(c)
i
2 
Ω
2 
Ω
2 
Ω
2 
Ω
4 
Ω
4 
Ω
1 
Ω
1 
Ω
2 
Ω
2 
Ω
3 
Ω
4 
Ω
1 
Ω
3 
Ω
3 
Ω
3 A
Fig. 5.1-9 
Sub-circuits for applying Superposition Theorem in Example 5.1-4
The value of i in circuit of Fig 5.1-9 (a) is found to be 
−
×
+ +
+
×
+
+ +
= −
3
2
2
2 4
3 1
3 1
4
3 1
0 5
A
A
(
/ /(
))
(
)
.
The value of i in circuit of Fig 5.1-9 (b) is found to be 
−
×
+ +
+
×
+
+ +
= −
6
1
1
3 4
2 2
2 2
4
2 2
0 5
A
A
(
/ /(
))
(
)
.
The value of i in circuit of Fig 5.1-9 (c) is found to be 
V
V
4
2 2
3 1
6
+ +
+
=
(
) / /(
)
A
Therefore, current delivered by voltage source when all the three sources are active 
=
−
−
+ = −
0 5 0 5
6
6
6
.
.
V
V
A.
Therefore, the value of V such that the current (and power) delivered by the voltage source will be 
zero 
=
6 V.
We note that we did not have to resort to node analysis or mesh analysis in solving the three single-
source circuits shown in Fig. 5.1-9.