Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

Kirchhoff’s current law is applicable to supernodes too
Therefore, the KCL equation for the supernode containing node-1 and node-2 is obtained as 
−
+
+
+
=
i t
i t
i t
i t
1
6
2
5
0
( )
( )
( )
( )
. Obviously, this must be the sum of KCL equations written for node-1 
and node-2. Similarly, the KCL equation for the supernode in the circuit in Fig. 2.2-2(b) must be the 
sum of KCL equations written for node-1, node-2 and node-5. This may be verified.
The total number of KCL equations that can be written for an n-node circuit is equal to 
n
n
n
n
n
C
C
C
C
1
2
3
1
+
+
+ +
−
. This series has a sum equal to 2
n
-
2. 
Only (n 
- 
1) independent equations from these 2
n
-
2 KCL equations can be used for solving the 
circuit.
example: 2.2-1
Find the power delivered by all the sources in the circuit in
Fig. 2.2-3. 
Solution
Currents through the voltage sources and voltage across the 
current sources have to be obtained first. The circuit with all 
nodes and reference directions for variables identified is shown in
Fig. 2.2-4. 
Applying KCL at node-A, we get –5 – (–2) – i
1
=
0 
�
i
1
=
–3 A
Applying KCL at node-B, we get (–2) 
+
(–2) – i
2
=
0 
�
i
2
=
– 4 A
Applying KCL at node-C, we get 5 – (–2) 
+
i
3
=
0 
�
i
3
=
–7 A
Fig. 2.2-3 
Circuit for 
Example 2.2-1
V
2
V
1
I
1
I
3
5 A
–2 A
–10 V
–2 A
10 V
5 V
I
2
V
3
+
+
+
–
–
–
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2.12
Basic Circuit Laws
Applying KVL in the loop I
1
-
V
1
-
V
3
, we get v
1
–10–5 
=
0 
�
v
1
=
15 V
Applying KVL in the loop I
3
-
V
1
-
V
2
, we get v
3
–10 
+
(–10) 
=
0 
�
v
3
=
20 V
Applying KVL in the loop I
2
-
V
2
-
V
3
, we get –v
2
– (–10) 
–5 
=
0 
�
v
2
=
5 V
Power delivered by an element is given by –vi, where v 
and i are its voltage and current variables, respectively, as 
per passive sign convention.
\
Power delivered by I
1
source 
=
–v
1
× 
5 A 
=
–75 W
Power delivered by I
2
source 
=
–v
2
× 
(–2) A 
=
10 W
Power delivered by I
3
source 
=
–v
3
× 
(–2) A 
=
40 W
Power delivered by V
1
source 
=
–10 V 
× 
i
1
=
30 W
Power delivered by V
2
source 
=
–(–10 V) 
× 
i
2
=
– 40 W
Power delivered by V
3
source 
=
–5 V 
× 
i
3
=
35 W
example: 2.2-2
Express i
2
(t), i
4
(t) and i
5
(t) in terms of i
1
(t), i
3
(t) and i
6
(t) in the circuit shown in Fig. 2.2-5. 
Solution
Applying KCL at node-1, we get 
−
+
+
= ⇒
=
i t
i t
i t
i t
1
4
6
4
0
( )
( )
( )
( )
−
i t
i t
1
6
( )
( )
Applying KCL at node-3, we get 
−
+
−
= ⇒
=
i t
i t
i t
i t
5
3
6
5
0
( )
( )
( )
( )
−
i t
i t
3
6
( )
( )
Applying KCL at node-2, we get 
−
+
+
= ⇒
=
i t
i t
i t
i t
4
2
5
2
0
( )
( )
( )
( )
−
+
+
= ⇒
=
−
=
−
i t
i t
i t
i t
4
5
1
3
( )
( )
( )
( )
example: 2.2-3
Fig. 2.2-6 shows the connecting wires in a part of a circuit. Some of the currents are specified for t 
≥
0 
in Fig. 2.2-6. The current i
2
is seen to be a constant in time. The current i
3
is seen to approach zero as 
t 
→
∞
. Find i
1
, i
2
and i
3
for t 
≥
0. 
Solution
Applying KVL at first node, we get 3
5 1
0
1
1
+ −
−
= ⇒ =
−
−
i
e
i
e
t
t
(
)
2 5
−
−
i
e
i
e
t
A i
2
is stated to be a constant in time. Let i
2
=
A. i
3
is 
stated to approach zero as t 
→
∞
. This implies that there is no DC 
component in i
3
. It may contain both e
-
t
and e
-
3t
components. Let 
i
Ce
De
A
t
t
3
3
=
+
−
−
Then, applying KCL at the second node, we get 
− +
−
− + =
− +
+ −
−
−
+ =
−
−
−
−
−
i
e
i
i
i e
e
e
Ce
De
A
i
t
t
t
t
t
1
3
3
2
3
3
3 1
0
2 5
3 3
0
(
)
. .,
.
ee
A
e
C
e
D
t
t
., (
)
(
)
(
)
+ +
−
+
− −
=
−
−
1
5
3
0
3
Fig. 2.2-4 
Circuit with nodes 
and reference 
directions identified
v
2
v
1
I
1
I
3
C
D
B
A
5 A
–2 A
–10 V
–2 A
10 V
5 V
I
2
i
2
i
1
i
3
v
3
+
+
+
+
+
+
–
–
–
–
–
–
V
2
V
1
V
3
Fig. 2.2-5 
Circuit for 
Example 2.2-2
i
1
(
t
)
i
4
(
t
)
i
2
(
t
)
i
5
(
t
)
i
3
(
t
)
i
6
(
t
)
1
2
3
4
6
4
5
2
3
1
Fig. 2.2-6 
Part of a circuit 
referred to in 
Example: 2.2-3 
i
1
i
3
i
2
3 A
3(1 – e
–3
t
) A
5(1 – e
–
t
) A
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Interconnections of Ideal Sources 
2.13
KCL is true at all t. Hence, the last equation must be valid for all t 
≥
0. No time-varying function 
can be equal to a constant unless that function itself is a constant. Thus, (A 
+ 
1) term in the last 
equation cannot be equalled by e
-
t
and e
-
3t
terms for all t 
≥
0. Therefore, (A 
+ 
1) has to be zero.
\
A 
=
-
1 
�
i
2
=
-
1 A.
A term involving e
-
t 
cannot get cancelled by another term that 
involves e
-3
t 
for all t 
≥
0. Therefore, coefficient of e
-
t
must be zero 
and coefficient of e
-3
t
also must be zero. Therefore, (5 
- 
C) 
=
0 and 
(3 
+ 
D) 
=
0.
\
C 
=
5 and D 
=
-
3 
�
i
3
=
(5 e
-
t
-
3 e
-3
t
) A. 
The solution is marked in Fig. 2.2-7.
2.3 
InterconnectIons of Ideal sources
Interconnecting two or more ideal voltage sources in a loop may lead to a degenerate circuit. Similarly, 
interconnecting two or more ideal current sources in series may 
lead to a degenerate circuit.
Consider the interconnection of two ideal independent 
voltage sources as shown in Fig. 2.3-1. KVL requires that 
−
+
=
v t
v t
s
s
1
2
0
( )
( )
. Therefore, v t
s1
( ) has to be equal to v t
s1
( ) at 
all time instants. If they are not equal to each other, then either 
KVL has to yield or the ideal sources have to yield. KVL cannot 
yield since it is only a disguised form of law of conservation 
of energy. Therefore, KVL has to be obeyed by a circuit at all
instants.
There are two ways out of this impasse for a case where v t
s1
( ) 
≠
v t
s2
( ). The first is to declare that two ideal independent voltage sources cannot be connected in 
parallel unless their terminal voltages are equal to each other at all instants of time. That is, we call 
such connections illegal if v t
s1
( ) 
≠
v t
s2
( ). But, that is a kind of escaping the issue!
The correct way to resolve the issue is to recognise that ideal independent voltage source is a 
model for a practical electrical device, and, as in the case of any model, this model too has its range 
of applicability. Connecting a practical voltage source in parallel with another practical voltage 
source is a context in which they cannot be modelled by ideal independent voltage source model 
satisfactorily. In fact, the model is not satisfactory even when v t
s1
( ) 
=
v t
s2
( ). This is because there is 
no way to find out the current that will flow in the circuit. Any current can flow at any instant in such a
circuit.
Thus, the only correct way to model a circuit that involves parallel connections of voltage sources 
(more generally, loops comprising only voltage sources) is to take into account the parasitic elements 
that are invariably associated with any practical voltage source. A somewhat detailed model for the 
two-source system is shown in Fig. 2.3-2.
L
i1 
and L
i2 
represent the internal inductance of the sources, C
i1
and C
i2
represent the terminal 
capacitance of the sources and R
i1
and R
i2
represent the internal resistance of the sources. L
c
and R
c
represent the inductance and resistance of connecting wires. Obviously, two practical voltage sources 
can be connected in parallel even if their open-circuit electromotive forces are not equal at all t; only 
that they cannot be modelled by ideal independent voltage source model.
Fig. 2.2-7 
Solution for 
Example 2.2-3
–1 A
3 A
3(1 – e
–3
t
) A
5(1 – e
–
t
) A
(5e
–
t
– 3e
–3
t
) A
(2 – 5e
–
t
) A
i
1
i
2
i
3
Fig. 2.3-1 
Two ideal 
independent 
voltage sources 
in parallel
v
s1
(
t
)
v
s2
(
t
)
+
+
–
–
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2.14
Basic Circuit Laws
V
s1
(
t
)
L
i1
L
c
L
c
C
i1
R
c
R
i2
L
i2
v
s2
(
f
)
R
c
C
i2
+
–
+
–
R
i1
Fig. 2.3-2 
A detailed model for a circuit with two voltage sources in parallel
Two ideal independent current sources in series raise a similar issue (see Fig. 2.3-3). KCL requires 
that i t
i t
t
s
s
1
2
( )
( )
.
=
for all Even if this condition is satisfied, there is no way to obtain the voltages 
appearing across the current sources. Therefore, the correct model to be employed for practical 
current sources that appear in series in a circuit is a detailed model that takes into account the 
parasitic elements associated with any practical device. 
More generally, if there is a node in a circuit where only 
current sources are connected, then those current sources 
cannot be modelled by ideal independent current source 
model. 
Similar situations may arise in modelling practical 
dependent sources by ideal dependent source models. In 
all such cases we have to make the model more detailed in 
order to resolve the conflict that arises between Kirchhoff’s 
laws and ideal nature of the model.