Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

example: 2.5-1
Solve the circuit in Fig. 2.5-2 completely.
Solution
The 10 V independent voltage source across the node-pair fixes the 
terminal voltage of all elements at 10V. The current through 10 
W
Fig. 2.5-2 
Circuit for 
Example 2.5-1
5 A
10 V
10 
Ω
5 
Ω
+
–
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2.20
Basic Circuit Laws
will then be 10 V/10 
W
=
1 A and the current through 5 
W
will 
be 10 V/5 
W
=
2 A. These currents flow from top node to bottom 
node.
Now we apply KCL at the top node.
Current flowing into the positive polarity of 10V source 
=
5 A 
+ 
1 A 
+ 
2 A 
=
0.
\
Current flowing into the positive polarity of 10 V source 
=
2 A.
The circuit solution is marked in Fig. 2.5-3.
Power delivered by 5 A current source 
=
10 V
× 
5 A 
=
50 W
Power absorbed by 10 
W
resistor 
=
10 V
× 
1 A 
=
10 W
Power absorbed by 5 
W
resistor 
=
10 V
× 
2 A 
=
20 W
Power absorbed by 10 V current source 
=
10 V
× 
2 A 
=
20 W
Total power delivered 
=
Total power absorbed
example: 2.5-2
Find the circuit solution for the circuit in Fig. 2.5-4. 
Solution
The variable assignment and reference polarity assignment is 
shown in Fig. 2.5-5.
The terminal voltage of all elements will be v
x
with this polarity 
assignment. Now, i
1
=
v
x
/2.5 A and i
2
=
v
x 
/5 A. Applying KCL at 
the top node, we get
i
v
i
i e
v
v
v
v
v
x
x
x
x
x
x
1
2
2 0 2
0
2 5
2
0 2
5
0
0 4
2
5
− −
+ =
− +
+
= ⇒
= ⇒
=
(
.
)
. .,
.
( .
)
.
V
Now, i
1
=
2A and i
2
=
1 A. The circuit solution is 
marked in Fig. 2.5-6.
Power absorbed by 2.5 
W
resistor 
=
5 V 
× 
2 A 
=
10 W
Power absorbed by 5 
W
resistor 
=
5 V 
× 
1 A 
=
5 W
Power delivered by 2 A current source 
=
5 V 
× 
2 A 
=
10 W
Power delivered by the dependent current source 
=
5 V
× 
1 A 
=
5 W
Total power delivered 
=
Total power absorbed 
=
15 W
2.6 
analysIs of MultI-loop, MultI-node cIrcuIts
The method of analysis of multi-loop, multi-node circuits, using element relations along with KCL 
and KVL equations, is illustrated through worked examples in this section.
Fig. 2.5-3 
Circuit solution in 
Example 2.5-1
5 A
2 A
2 A
1 A
+ +
+
+
–
–
–
–
10 
Ω
5 
Ω
10 V
10 V
10 V
Fig. 2.5-6 
Circuit solution in 
Example 2.5-2
5 V
5 V
2 A
1 A
2.5 
Ω
5 
Ω
2 A
5 V
5 V
+
+
+
+
–
–
–
–
–1 A
Fig. 2.5-5 
Reference polarity 
assignment in the circuit 
in Example 2.5-2
–0.2 
v
x
2.5 
Ω
2 A
v
x
5 
Ω
+
–
Fig. 2.5-4 
Circuit for 
Example 2.5-2
–0.2 
v
x
i
1
i
2
2.5 
Ω
2 A
v
x
5 
Ω
+
+
+
+
–
–
–
–
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Analysis of Multi-Loop, Multi-Node Circuits 
2.21
example: 2.6-1
Find the voltage across the dependent current source (v
y
) in the circuit in Fig. 2.6-1.
Solution
This circuit has six elements, three nodes and three meshes. We 
need to find out only v
y
. Let us try to solve the circuit in terms of v
x
and v
y
without using any other new variables.
We note that v
AD
=
v
CD
+
v
AC
=
(10
-
v
x
) V.
\
v
AB
=
v
AD
– v
BD
=
(10
-
v
x
) – v
y
V.
v
BC
=
v
BD
– v
CD
=
(v
y
–10) V.
v
CA
=
v
x
V.
Note that we have essentially employed KVL in order to arrive at these relations. Now, we 
can express all the currents at node-A and node-B in terms of v
x
and v
y
. The currents going 
away from node-A are (v
AB
/3) A, (v
AD
/8) A and (
-
v
CA
/1) A. The sum of these three terms must be 
zero.
∴
− −
+
−
−
=
10
3
10
8
1
0
v
v
v
v
x
y
x
x
The currents going away from node-B are (
-
v
AB
/3) A, (v
BC
/5) A and v
x
. Sum of these terms must 
be zero.
∴
− + +
+
+
=
10
3
5
0
v
v
v
v
x
y
y
x
Simplifying these two equations, we get
35v
x
+
8v
y
=
110
20v
x
+
8v
y
=
80
Solving these two equations, we get v
x
=
2 V; v
y
=
5 V
Therefore, the voltage across the dependent current source 
=
5 V.
example: 2.6-2
Find the power delivered by the voltage and current sources in the 
circuit shown in Fig. 2.6-2.
Solution
We need to find out the current through the 20 V voltage 
source and the voltage across the 5 A current source. Refer to
Fig. 2.6-3.
Fig. 2.6-1 
Circuit for 
Example 2.6-1
8 
Ω
3 
Ω
1 
Ω
5 
Ω
v
x 
A
v
x
v
y
A
D
B
C
+
+
+
10 V
–
–
–
Fig. 2.6-2 
Circuit for 
Example 2.6-2
+
10 
Ω
10 
Ω
5 
Ω
5 
Ω
5 A
20 V
–
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2.22
Basic Circuit Laws
+
+
+
+
10 
Ω
10 
Ω
5 
Ω
5 
Ω
5 A
20 V
i
i
1
i
2
i
3
A
B
C
–
–
–
–
Fig. 2.6-3 
Circuit in Example 2.6-2 with variables assigned
KVL in the first mesh gives
− +
+
=
⇒
=
−
20 10
0
20 10
i V
V
i
AC
AC
(
) V.
∴ =
−
= −
i
i
i
1
20 10
5
4 2 A.
KCL has to be satisfied at node-A. 
∴ = − = − −
= −
i
i i
i
i
i
2
1
4 2
3
4
(
)
.
A
Applying KCL at node-B, we get, 
− + − = ⇒ = + =
+
i
i
i
i
i
2
3
3
2
5 0
5
3 1
(
) A
Therefore, V
BC
=
5
× 
(3i 
+ 
1) 
=
15i 
+
5 V.
Now we apply KVL on the outer loop of the circuit to get,
− +
+
+
=
−
+
+
−
+
+ =
=
20 10
10
0
20 10
30
40
15
5
0
55
2
i
i
V
i e
i
i
i
i e
i
BC
. .,
(
) (
)
. .,
555
1
∴ =
i
A
And, V
BC
=
15i 
+
5 
=
20 V.
Therefore, the current delivered by the voltage source is 1 A and the voltage appearing the current 
source is 20 V.
Therefore, the power delivered by the voltage source 
=
20 V
× 
1 A 
=
20 W
The power delivered by the current source 
=
20 V
× 
5 A 
=
100 W
example: 2.6-3
Find i
x 
in the circuit
shown in Fig. 2.6-4. 
Solution
Voltage across 2 
W
resistance is 2i
x
V with positive polarity at the 
top terminal. Therefore, the current through 4 
W
is 2i
x
/4 
=
0.5i
x
A 
from the top terminal to the bottom terminal.
Therefore, the current through 3 
W
must be 1.5i
x
A from left to 
right by KCL applied to the node at which the three resistors are 
connected.
Therefore, the voltage across 13 
W
must be 2i
x
+
3
× 
1.5i
x
=
6.5i
x 
V.
Fig. 2.6-4 
Circuit for 
Example 2.6-3
13 
Ω
2 
Ω
4 
Ω
3 
Ω
4 A
i
x
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KVL and KCL in Operational Amplifier Circuits 
2.23
Therefore, the current through 13 
W
must be 6.5i
x
/13 
=
0.5i
x
A from the top terminal to the bottom 
terminal. Now, apply KCL at the current source node to get 1.5i
x
 
+ 
0.5i
x
=
4.
\
i
x
=
2A.
example: 2.6-4
Find the ratio 
v
v
o
s
in the circuit shown in Fig. 2.6-5. 
v
0
v
S
0.002 
v
0
20 k
Ω
40 k
Ω
10 k
Ω
100
+
+
+
–
–
–
100 k
Ω
i
x
i
x
Fig. 2.6-5 
Circuit for Example 2.6-4
Solution
The current that flows through 100 k
W
at output side is 100i
x
from the bottom terminal to the top 
terminal. Therefore, v
i
o
x
= −
10
7
V. Therefore, the voltage generated by the VCVS at input side is 
-
0.002
× 
10
7 
i
x
=
-
2
× 
10
4
i
x
V with polarity as shown in Fig. 2.6-5.
Therefore, voltage across 40 k
W
resistance is 
=
20 10
2 10
0
3
4
×
− ×
=
i
i
x
x
V.
Therefore, current through 40 k
W
resistance is 
=
0 A.
Applying KVL in the first mesh, we get, voltage drop across 10k
W
=
v
s
V.
Therefore, current through 10 k
W
resistance 
=
10
-
4 
v
s
A
Current through 40 k
W
is zero. Therefore, by KCL, i
x
=
10
-
4 
v
s
A.
We know that v
i
o
x
= −
10
7
V.
v
i
v
v
v
v
o
x
s
s
o
s
= −
= −
×
= −
∴
= −
−
10
10
10
1000
1000
7
7
4