Chapter 2 Circuit analysis
Topics to be covered:
Nodal and Mesh Analysis
Chapter No. From Text Book : 2.1 to 2.3
No. of lectures : 03
ELECTRICAL SCIENCES (EEE F111) by Dr. Jagadish Nayak , BITS Pilani, Dubai Campus
Circuit analysis techniques
 To Study Nodal and Mesh analysis
• we make use of Ohms law, KVL and KCL
In Nodal analysis,
y , A set of simultaneous equations
q
are written,, in
which , the variables are voltages.
In mesh analysis, A set of simultaneous equations are written, in
which , the variables are current (loop analysis).
ELECTRICAL SCIENCES (EEE F111) by Dr. Jagadish Nayak , BITS Pilani, Dubai Campus
Nodal Analysis
c
a
va
+
+
va
vc
-
vc
Node Voltage designation g
Voltage between non‐
reference node
reference node va +
vac
-
vc
By KVL ,between node c and a
node c and a vca=vc-va
Reference node By KVL ,between node a and c vac=va-vc
ELECTRICAL SCIENCES (EEE F111) by Dr. Jagadish Nayak , BITS Pilani, Dubai Campus
Nodal Analysis Example
Consider a circuit shown below
v1
3Ω
v2
i3
1Ω
6A
i1
4Ω
12A
i2
Arbitrarily choose the current directions
v1 v1  v2

6
1
3
Let the node voltage be labeled as v1 and v2
Applying KCL at node v2
Apply KVL at node v1 , we get i1+i3=6
By Ohms law i1=v1/1 and i3=(v1-v2)/3
 4v1  v2  18
i2+i3=6 by Ohms law i2=-v2/2
v v v
 2  1 2  12  2v1  5v2  72
2
3
ELECTRICAL SCIENCES (EEE F111) by Dr. Jagadish Nayak , BITS Pilani, Dubai Campus
Nodal Analysis Example
Using the simultaneous equations 4v1  v2  18 and
2v1  5v2  72
We get , v2=-14V
= 14V and v1=1V and corresponding currents i1=1A , i2=7A and i3=5A
Let us consider another example
v1
3V
+
-
2Ω
1Ω
v2
v3
i4
i3
i1
6Ω
i2
1
v2 A
4
8Ω
ELECTRICAL SCIENCES (EEE F111) by Dr. Jagadish Nayak , BITS Pilani, Dubai Campus
Nodal Analysis Example
v1
3V
2Ω
+
-
1Ω
v2
v3
i4
i3
i1
6Ω
i2
1
v2 A
4
8Ω
Apply KCL at node v1,
There is a 3V source between node v1 and reference node, we can directly
write v1=3V and there are two unknowns to be solved now v1 and v2
ELECTRICAL SCIENCES (EEE F111) by Dr. Jagadish Nayak , BITS Pilani, Dubai Campus
Nodal Analysis Example
v1
3V
2Ω
+
-
1Ω
v2
v3
i4
i3
i1
6Ω
i2
1
v2 A
4
8Ω
Apply KCL at node v2,
i1  i2  i3  0
Apply Ohms Law
v1  v2  v2 v3  v2


 0   10v2  6v3  9
2
6
1
ELECTRICAL SCIENCES (EEE F111) by Dr. Jagadish Nayak , BITS Pilani, Dubai Campus
Nodal Analysis Example
v1
3V
2Ω
1Ω
v2
i4
i3
i1
+
-
v3
6Ω
i2
1
v2 A
4
8Ω
Apply KCL at node v3,
1
v2  i3  i4  0
4
Apply Ohms Law
v v v
1
v2  3 2  3  0   6v2  9v3  0
4
1
8
ELECTRICAL SCIENCES (EEE F111) by Dr. Jagadish Nayak , BITS Pilani, Dubai Campus
Nodal Analysis Example
Using simultaneous equations
 10v2  6v3  9 and  6v2  9v3  0
We get v2=1.5V
1 5V and v3=1V
1V
Also we can calculate the currents
i1  0.75 A i2  0.25 A
i3  0.5 A
i4  0.125 A
ELECTRICAL SCIENCES (EEE F111) by Dr. Jagadish Nayak , BITS Pilani, Dubai Campus
Drill Exercise
For the circuit shown below , find v1, v2 and v3. What is the current directed
down through the dependent source? What is the resistance loading the
independent voltage source.
source
v1
3V
+
-
2Ω
1Ω
v2
v3
i3
i1
6Ω
i2
i
+
-
2v 2 V
i4
8Ω
ELECTRICAL SCIENCES (EEE F111) by Dr. Jagadish Nayak , BITS Pilani, Dubai Campus
Example ( Voltage sources between non- reference nodes)
Consider following circuit diagram.
v1
24V
+
-
4V
2Ω
v2
+
i1
6Ω
i2
-
i
v3
i3
3A
8Ω
ELECTRICAL SCIENCES (EEE F111) by Dr. Jagadish Nayak , BITS Pilani, Dubai Campus
Example ( Voltage sources between non- reference nodes)
v1
24V
+
-
4V
2Ω
v2
+
i1
6Ω
-
v3
i
i2
i3
3A
8Ω
By Looking at the circuit we can write v1=24V
Applying KCL at node v2 , we get i1+i2= i
By Ohms Law i1 
24  v2
v
and i2  2
2
6
ELECTRICAL SCIENCES (EEE F111) by Dr. Jagadish Nayak , BITS Pilani, Dubai Campus