Circuit Theory ELE 1101: Nodal.
Mesh. Superposition
Course Given by
 M.N.CHRISISTOM
 Dept of Mechanical Engineering
 Emails: cnuwematsiko@tech.mak.ac.ug

dkliskatz@yahoo.com
 Phone: +256-787-258-844


Nodal & Mesh Analyses
Nodal Analysis employs KCL & Ohm’s
Law so as to determine the currents
flowing through resistors (and other
requirements).
 Mesh Analysis employs KVL so as to
determine the currents flowing through
resistors (and other requirements).

Nodal Analysis
Step 1: Determine the earth node
(whose voltage is zero).
 Step 2: Determine & number the other
nodes. Give @ node a voltage value e.g.
v1, v2,..
 Step 3: Where there are voltage sources;
determine some node voltages by
inspection.

Nodal Analysis (cont.)
Step 4: Use KCL to write down
equations for the remaining nodes &
solve them simultaneously.
 Step 5: Complete the solutions required.

Nodal Analysis

Use Nodal Analysis to find current
through R
1
15
4
2
R
Nodal Analysis

Steps 1 & 2
v
i1
1
15
0
i2
i3
4
2
R
Nodal Analysis

Step 3: is not applicable here.
 Steps 4&5: solve for v & iR
 15  i1  i2  i3  0
Use  v  to  det er min e  i1 , i2 , i3
v v v
 15     0
1 4 2
60
 60  4v  v  2v  0  v 
 8.6V
7
v 30
 iR  i3  
 4 .3 A
R 7
Nodal Analysis

Nodal analysis to obtain i, PR
1
+
i
4
2
R
10
15
-
Nodal Analysis

Steps 1,2 & 3: identify all node voltages.
v1
v
1
+
i
2
4
R
10
15
-
By inspection; v1=15
0
Nodal Analysis

Determine v & solve completely
Apply  KCL  at  node  of  v :
v  v1 v v
   10  0
1
4 2
but  v1  15; by  inspection ;
v  15 v v

   10  0
1
4 2
100
 4v  60  v  2v  40  0  v 
7
v 50
i  
 7.1A  PR  i 2 R  102.0W
R 7
Nodal Analysis: 4 nodes

Use Nodal Analysis; find
2
+
97
-
iR
6
1
4
R
3
Nodal Analysis: 4 nodes

Steps 1,2 & 3: identify all node voltages.
2
1
3
2
+
97
-
6
1
4
0
R
3
Nodal Analysis: 4 nodes

Determine v3 & solve completely
v1  97;by  inspection
Apply  KCL  at  node 2 & 3 :
v2  v1 v2  v3 v2

  0.....[Node  2]
2
6
4
v3  v2 v3 v3
   0............[Node  3]
6
1 3
11v2  2v3  582
9v3  v2  0
 v3  6  iR 
6
 2A
3
Mesh Analysis
Step 1: Determine & give current values
to @ mesh.
 Step 2: Where there are current sources;
determine some mesh currents by
inspection.
 Step 3: Use KVL to write the other mesh
equations & solve them simultaneously.

Mesh Analysis (cont.)
Step 4: Determine the resultant current
through @ resistor.
 Step 5: Complete the solutions as
required.

Q1: Mesh Analysis

Use Mesh Analysis to find current
through R (4-ohm resistor)
1
15
4
R
2
Q.1: Mesh Analysis
1
15
mesh
1
4
mesh
2
44
2
mesh
3
Q.1: Mesh Analysis


Steps 1,2,3,4
By  inspection : i1  15
Apply  KVL  to  the  other  2  meshes :
mesh  2;
1(i2  i1 )  4(i2  i3 )  0  i2  15  4i2  4i3  0
 5i2  4i3  15
mesh  3;
4(i3  i2 )  2i3  0  3i3  2i2  0
45
30
15
 i2  , i3 
 iR  i2  i3 
7
7
7
Q.2:Mesh Analysis

Use Mesh Analysis to find i & PR
2
i
1
4
+
R
10
15
-
Q.2 Mesh Analysis
2
1
4
+
15
mesh
mesh
mesh
10
1
2
3
-
Q.2: Mesh Analysis


Steps 1,2,3,4
By  inspection : i1  15
Apply  KVL  to  the  other  2  meshes :
mesh  2;
1(i2  i1 )  4(i2  i3 )  0  i2  15  4i2  4i3  0
 5i2  4i3  15
mesh  3;
4(i3  i2 )  2i3  10  0  3i3  2i2  5
25
5
20
 i2  , i3   iR  i2  i3  , PR  32.7W
7
7
7
Principle of Superposition
“The response of a linear circuit, acted
upon by several independent sources, is
the algebraic sum of the responses when
each source acts alone”.
 The Principle is useful either in analyses
where it is difficult to use the above
methods easily or when the individual
effects of sources are needed.

Exercise on Superposition

In the circuits Q1,Q2 & Q3 use any
method to determine the current through
R.
Superposition : Q1

Use Superposition to find i &
PR
2
i
1
4
+
R
10
15
-
Superposition : Q1

Step 1: Let current source act alone:
 Find i1 Nodal analysis:
2
v1
i1
1
4
R
15
0
Superposition : Q1

Step 2: Let the voltage source act alone:
 Find i2 by Nodal Analysis
2
v2
i2
1
4
+
R
10
-
0

By Nodal Analysis:
v1 v1 v1
60
v1 15
   15  0  v1 
 i1  
1 4 2
7
4 7
v2 v2 v2  10
20
v2 5
 
 0  v2 
 i2  
1 4
2
7
4 7
 By  Superposit ion. Pr inciple
20
1600
2
i  i1  i2 
 PR  i R 
 32.7W
7
49
Superposition : Q2

Use Superposition to find i & PR
1
i
6
+
27
-
R
3
4
2
15
Superposition : Q2

Step 1: Let voltage source act alone.
 Use Nodal Analysis to find iR1
v1
v
1
iR1
6
+
27
-
4
3
0
2
Superposition : Q2

Use Nodal analysis:
By  inspection ; v1  27
By  nodal  analysis ;
v  v1 v v v v
    0
1
6 3 4 2
12(v  27)  2v  4v  3v  6v  0
 v  12
v 12
 iR1  
 4A
R 3
Superposition : Q2

Step 2: Let current source act alone.
 Use Nodal Analysis to find iR 2
v2
1
iR 2
6
R
4
3
2
15
0
Superposition : Q2

Use Nodal analysis:
v2 v2 v2 v2 v2
     15  0
1
6
3
4
2
20
20
 v2 
 iR 2 
 2.2 A
3
9
 By  Superposit ion  Pr inciple;
20
iR  iR1  iR 2  4 
 6.2 A
9
2
 PR  iR  R  116.1W
Q3: Superposition

By Superposition Principle find PR
4
2
+
R
10
-
3
12
Q3: Superposition

Step 1: Let current source act alone
4
2
i1
R
By current division: i1 = 24/5= 4.8A
3
12
Q3: Superposition

Step 2: let voltage source act alone
2
4
i2
+
R
10
-
Apply KVL: i2 = 10/(2+3) = 2A
3
Q3: Superposition

Apply Superposition Principle
 iR  i1  i2  4.8  2  6.8 A
2


 PR  iR R  138.72W
Note;
PR  PR1  PR 2  i1  R  i2  R
2
2
 69.12  12  81.12W
[ Because  of  i 2  power  is  not  linear ]
Q4: Superposition

Use the Superposition Principle to
determine the current in R.
3
+
26
-
4
39
R
2
13
Q4: Superposition

Let @ source act alone. Nodal Analysis:
Voltage  source  alone :
v1  26 v1 v1
   0  v1  8  i1  2
3
4 2
39 A  current  source  alone :
v2 v2 v2
   39  0  v2  36  i2  9
3 4 2
13 A  current  source  alone :
v3 v3 v3
   13  0  v3  12  i3  3
3 4 2
 i  i1  i2  i3  2  9  3  4  4 A  upwards .