AC Nodal and Mesh Analysis
Discussion D11.1
Chapter 4
4/10/2006
1
AC Nodal Analysis
2
How did you write nodal equations
by inspection?
v1
2A
i1
v3
v2
R2
R1
i3
R3
R4
is
i5
R5
3
Writing the Nodal Equations by Inspection
v1
2A
i1
v3
v2
R2
R1
i3
 G1  G2

 G2
 0

R3
is
R4
G2
G2  G3  G4
G3
i5
R5
Gv  i
 v1   2 
   
G3  v2    0 
  
G3  G5 
 v3   is 
0
•The matrix G is symmetric, gkj = gjk and all of the off-diagonal terms
are negative or zero.
The gkk terms are the sum of all conductances connected to node k.
The gkj terms are the negative sum of the conductances connected to
BOTH node k and node j.
The ik (the kth component of the vector i) = the algebraic sum of the
independent currents connected to node k, with currents entering the
4
node taken as positive.
Example with resistors
v1
1
1S
2A
2
1/2 S
v2
3
1/3 S
4
1/4 S
v3
1A
1/2 S
2
1
0  v1   2 
1  1 2

   

1
1

1
4

1
3

1
3

 v2    0 
 0
  
1 3
1 3  1 2 

 v3   1 
1
0  v1   2 
1.5

   

1
1.583

0.333

 v2    0 
 0 0.333 0.833  v   1 

 3   
5
For steady-state AC circuits we can
use the same method of writing nodal
equations by inspection if we replace
resistances with impedances and
conductances with admittances.
Let's look at an example.
6
Problem 4.31 in text
V1
j2
V2
40 A
20 A

-j1
Change impedances to admittances
V1
j2
V2
-j/2 S

S
20 A
40 A
-j1
j1 S
7
V1
j2
V2
-j/2 S
20 A

S
-j1
j1 S
1  j / 2

 j/2
1  j 0.5

 j 0.5
40 A
j / 2   V1   2 
 V    
j / 2   2   4 
j 0.5   V1   2 
 V    
j 0.5   2   4 
 V1   4.24345 

 
 V2   5.831121 
8
Matlab Solution
1  j 0.5

 j 0.5
j 0.5   V1   2 
 V    
j 0.5   2   4 
 V1   4.24345 

 
 V2   5.831121 
9
Nodal Analysis for Circuits Containing Voltage Sources
That Can’t be Transformed to Current Sources
• If a voltage source is connected between
two nodes, assume temporarily that the
current through the voltage source is
known and write the equations by
inspection.
10
Problem 4.33 in text
Note: V2 = 10
V1
-j4
V2
j/4 S
645 A

1/2 S
I0
+
j2
-j/2 S
100 V
AC
I2
assume I2
1/ 2  j / 4  j / 4   V1   645 

  V    I 
 j / 4  2  
 j/4
2 
1/ 2  j / 4  j / 4  V1   645 

    I 
 j / 4  10  
 j/4
2 
11
1/ 2  j / 4  j / 4  V1   645 

    I 
 j / 4  10  
 j/4
2 
1/ 2  j / 4 V1  j 2.5  645
  j / 4 V1  j 2.5  I2
1/ 2  j / 4 0   V1   645  j 2.5 

 I   


j
/
4
1
j
2.5

 2  

 0.5  j 0.25 0   V1   4.243  j 6.743 

 I   


j
0.25
1
j
2.5

 2  

AX = B
1
X=A B
 V1   14.2531.25 
 

I
5.846

108.4

 2 
12
Problem 4.33 in text
Note: V2 = 10
V1
-j4
V2
j/4 S
645 A

1/2 S
+
j2
-j/2 S
I0
100 V
AC
I2
assume I2
 V1   14.2531.25 
 

I
5.846

108.4

 2 
V1
I0 
 7.12531.25
2
13
Matlab Solution
1/ 2  j / 4 0   V1   645  j 2.5 

 I   

1 2  
j 2.5
 j/4

 V1   14.2531.25 
 

I
5.846

108.4

 2 
14
AC Mesh Analysis
15
How did you write mesh equations
by inspection?
R2
R1
+
DC
Vs2
i1
+
R3
-
v3
+
+ v2 -
v1 -
v5 -
+
v7
-
R7
+ v6 -
R5
i3
i2
R6
DC
Vs1
i4
+
v4
-
R4
- v +
8
R8
16
Writing the Mesh Equations by Inspection
R2
R1
+
DC
Vs2
i1
+
R3
-
v3
+
v5 -
+
v7
-
R7
i2
+ v6 -
R5
i3
Ri = v
+ v2 -
v1 -
R6
DC
Vs1
i4
- v +
8
+
v4
-
R4
 R1  R5  R7

 R7


 R5

0

 R7
R2  R6  R7
0
 R6
 R5
0
R3  R5
0
0
 i1   Vs2 
   0 
 R6

 i2   
 i3   Vs1 
0

  


R4  R6  R8   i4 
 Vs1 
R8
•The matrix R is symmetric, rkj = rjk and all of the off-diagonal terms
are negative or zero.
The rkk terms are the sum of all resistances in mesh k.
The rkj terms are the negative sum of the resistances common to
BOTH mesh k and mesh j.
The vk (the kth component of the vector v) = the algebraic sum of the
independent voltages in mesh k, with voltage rises taken as positive.
17
Example with resistors
DC
4V
Ri = v
R3
3
R1
R2
2
3
i1
4
i2
R7
1
2
R5
R6
i3
DC
2V
i4
4
1
R8
4
1
0   i1   4 
 2  4 1

i   

4
3

2

4
0

2

 2    0 
 1
0
3 1
0   i3   2 

   
0

2
0
2

4

1

  i4   2 
18
R4
For steady-state AC circuits we can
use the same method of writing mesh
equations by inspection if we replace
resistances with impedances and
conductances with admittances.
Let's look at an example.
19
Problem 4.38 in text:
Find I1 and I2

j1
+
I1
AC
120 V

-
-j1
+
I2
60 A
AC
-
 2  j1 1  j1  I1   12 

I    
 1  j1 1  j 0   2   6 
 I1   3.79518.43 

 
 I 2   2.683  116.6 
20
Matlab Solution
 2  j1 1  j1  I1   12 

I    
 1  j1 1  j 0   2   6 
 I1   3.79518.43 

 
I
 2   2.683  116.6 
21
What happens if we have independent
current sources in the circuit?
1. Assume temporarily that the voltage across each
current source is known and write the mesh
equations in the same way we did for circuits
with only independent voltage sources.
2. Express the current of each independent current
source in terms of the mesh currents and replace
one of the mesh currents in the equations.
3. Rewrite the equations with all unknown mesh
currents and voltages on the left hand side of the
equality and all known voltages on the r.h.s of
the equality.
22
Problem 4.40 in text: Find I0

Assume you know V2
-j1
+
j2
+
AC
60 V
-
I2
I1

I0
20 A V2
-
Note I2 = -2
 3  j1 2  j 2   I1   6 

  I    V 
 2  j 2 2  j 2  2   2 
 3  j1 2  j 2  I1   6 

    V 
 2  j 2 2  j 2  2   2 
23
 3  j1 2  j 2  I1   6 

    V 
 2  j 2 2  j 2  2   2 
3  j1 I1  4  j 4  6
 2  j 2 I1  4  j 4  V2
 3  j1 0   I1   2  j 4 

 V   

 2  j 2 1   2   4  j 4 
AX = B
1
X=A B
 I1  1.414  81.87 
X 

 V2   7.37612.53 
24
Matlab Solution
 3  j1 0   I1   2  j 4 

 V   


2

j
2
1
4

j
4

 2  

 I1  1.414  81.87 

 
 V2   7.37612.53 
25
Problem 4.40 in text: Find I0

Assume you know V2
-j1
+
j2
+
AC
60 V
-
I2
I1

I0
20 A V2
-
Note I2 = -2
 I1  1.414  81.87 

 
V
 2   7.37612.53 
I 0  I1  I 2  I1  2
I 0  1.414  81.87  20  2.608  32.47
26