Lecture 4
Node/Loop Analysis
Although they are very
important concepts,
series/parallel equivalents and
the current/voltage division
principles are not sufficient to
solve all circuits.
Node Voltage Analysis
Node Voltage Analysis
v1  v s
v2  v1 v2 v2  v3


0
R2
R4
R3
v3  v1 v3 v3  v2


0
R1
R5
R3
Example 2.6
v1 v1  v 2

 is  0
R1
R2
v 2  v1 v2 v 2  v3


0
R2
R3
R4
v3 v3  v2

 is
R5
R4
Exercise 2.6
v1  v2 v1  v3

 ia
R2
R1
v2  v1 v2 v2  v3


0
R2
R3
R4
v3  v2 v3  v1 v3


 ib  0
R4
R1
R5
Exercise 2.8
Exercise 2.8
V1  10 V1 V1  V2
 
 0  5V1  50  2V1  V1  V2  0  8V1  V2  50
2
5
10
V2  10 V2  V1 V2


 0  V2  10  V2  V1  2V2  0  V1  4V2  10
10
10
5
8V1  V2  50
 8V1  32V2  80
31V2  130
130
31
 4.19
V2 
V1  4V2  10
 6.77
Exercise 2.9
Exercise 2.9
v1  v3 v1 v1  v2
 
 0  v1  v3  4v1  2v1  2v2  0
20
5
10
v2  v1 v2  v3

 10  0  v2  v1  2v2  2v3  100  0
10
5
v3 v3  v1 v3  v2


 0  2v3  v3  v1  4v3  4v2  0
10
20
5
7v1  2v2  v3  0
 v1  3v2  2v3  100
 v1  4v2  7v3  0
Supernodes
Circuits with Voltage Sources
We obtain dependent
equations if we use all of the
nodes in a network to write
KCL equations.
v1 v1   15 v2 v2   15



0
R2
R1
R4
R3
Use KVL for a second independent
equation:
v2  v1  10V
Exercise 2.11
KVL:
 v1  10  v2  0
KCL for supernode at 10V source:
v1 v1  v3 v2 v 3


1
R1
R2
R3
KCL for node 3:
v3  v1 v3  v 2
v3


0
R2
R3
R4
KCL for reference node:
v1 v3

1
R1 R4
Node-Voltage Analysis with a
Dependent Source
First, we write KCL equations at
each node, including the current
of the controlled source just as if
it were an ordinary current
source.
Example 2.9
v1  v 2
 is  2i x
R1
v2  v1 v2 v2 v 3


0
R1
R2
R3
v3  v 2 v3

 2i x  0
R3
R4
Example 2.9
Next, we find an expression for the
controlling variable ix in terms of the
node voltages.
v3 v 2
ix 
R3
Example 2.9
Substitution yields:
v3 v 2
v1  v2
 is  2
R1
R3
v2  v1 v2 v2 v 3


0
R1
R2
R3
v3  v 2 v3
v3  v 2

2
0
R3
R4
R3
Node-Voltage Analysis
1. Select a reference node and assign
variables for the unknown node
voltages. If the reference node is
chosen at one end of an
independent voltage source, one
node voltage is known at the start,
and fewer need to be computed.
2. Write network equations. First, use KCL
to write current equations for nodes and
supernodes. Write as many current
equations as you can without using all of
the nodes. Then if you do not have enough
equations because of voltage sources
connected between nodes, use KVL to
write additional equations.
3. If the circuit contains dependent
sources, find expressions for the
controlling variables in terms of the
node voltages. Substitute into the
network equations, and obtain
equations having only the node
voltages as unknowns.
4. Put the equations into standard form and
solve for the node voltages.
5. Use the values found for the node
voltages to calculate any other currents or
voltages of interest.
Exercise 2.12(a)
v2  v1  10V  v2  10  v1
v1
v2
v1
v1 v2

 1  v1  2(10  v1 )  10
10 5
3v1  10
10
20
v1    v2 
3
3
v2 4
ia 
  1.33 A
5 3
Exercise 2.12(b)
v1  25 v1 v1  v2
 
0
10
10
20
v2 v2  v1 v2  25


0
20
20
5
v1  13 .79V
v2  18 .97V
v1  v2
ib 
20
 0.259 A
v1
v2
Exercise 2.13(a)
v1
v1  10 v1
  2i x  2v1  10  10 i x
5
5
10  v1
ix 
 2v1  20  10 i x
5
20 i x  10
i x  0.5 A
Exercise 2.13(b)
v1 v1  2i y

3 0
v2
5
2
v2 v2  2i y

3  0
5
10
v2
iy 
 i y  2.31 A
5
v1