DIT, Kevin St.
Electric Circuit Theory DT287/2
Mesh Analysis
Summary
In this section we use the format approach for solving circuit problems using mesh analysis. This technique generates a set of equations, which can be solved using determinant algebra and the application of Cramers rule. We will assign the SAME DIRECTION for each current flow in each branch of a particular circuit. We can then write the loop equation by inspection. Let the sum of the impedances in the loop be positive and the mutual or transfer impedance, be negative . The mutual or transfer impedance is that impedance which is common to two loops. KVL states that the sum of the e.f.m`s in a closed loop must be equal to the sum of the potential differences. If the e.m.f is in the same direction as the current direction, it is given a positive e.m.f. If it is in the opposite direction to the current direction, it is assigned a negative sign.
Consider the following example of a Tee network fed from two DC sources in figure 29.
Figure 29: Tee network.
The loop equations are written applying Kirchoff’s voltage law
E
1
=
I R
1
+
I R
3
−
I R
3
(1)
And
−
E
2
= −
I
1
R
3
+
I
2
R
2
+
I
2
R
3
(2)
Note the negative sign in front of E
2
because the assigned current is in the opposite direction to the rise in potential. But by inspection
E
1
=
I
1
( R
1
+
R
3
)
−
I
2
R
3
(3)
−
E
2
= −
I
1
R
3
+
I
2
( R
2
+
R
3
) (4)
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DIT, Kevin St.
Electric Circuit Theory DT287/2
Here the sum of the self-impedances is R
1
+ R
2 and the mutual impedance is –
R
3
. Expressed in matrix form:


−
E
1
E
2


=


I
I
1
2




( R
1
(
−
+
R
3
R
2
)
)
( R
(
−
R
3
2
+
)
R
3
)

 (5)
Use the following rules to can write the equations by inspection:
♦
Convert all the current sources to voltage sources
♦
Note the location of the major node,
♦
Calculate the number of loops using the equation N = ( b - n ) + 1 where N = number of independent loops, b = number of branches, n = number of major nodes,
♦
Select all loops and draw a line around each loop in a clockwise direction
(you could go anti-clockwise) representing each current. Each loop must contain one impedance (or generator) which is not in any other loop (this ensures the equations are independent), and
♦
Apply KVL to the loop using the following rule:
♦
A positive sign is assigned to each e.m.f. if the rising voltage is in the same direction as the assigned current or negative if in the opposite direction.
Exercise
Consider the circuit shown in figure 30.
Loop 1
Loop 2
Loop 3
Figure 30: Typical circuit under investigation.
50
∠
30 o = i
1
( 14
+ j )
− i
2
( j 4 )
− i
3
( 8 )
0
= − i
1
( j 4 )
+ i
2
( 7
+ j 3 )
− i
3
(
− j )
−
50
∠
30 o = − i
1
( 8 )
− i
2
(
− j )
+ i
3
( 6
+ j 3 )
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DIT, Kevin St.
Electric Circuit Theory
We will solve for the individual currents at a later stage.
DT287/2
Determinants
We have to consider determinant algebra in order to solve for the currents in the above matrix. A matrix consists of elements arranged in rows and columns.
The first letter of the subscript of each element tells you the row and the second subscript tells you the column in which the element is contained. For example, consider a 3 by 3 matrix.
A = 


 a
11 a
21 a
31 a
12 a
22 a
32 a a
13
23 a
33




(6)
Determinant of a square Matrix
The determinant of a square matrix A , is calculated as:
A
= a
11 a
22
− a
21 a
12
(7)
Co-factors and Minors :
If the determinant of a matrix is bigger than 2 x 2 then cofactors must be used.
The minor of an element A ij
, of a determinant of order n, is the determinant of order ( n – 1) obtained by deleting the row and column that contains the given element. The minor of an element A i j
is denoted as:
(8)
A
11
=
M
11
= a
22 a
23
a
32 a
33
For the element a
12 the minor is
M
12
= a
11 a
13
− a
31 a
33
And for the element a
13
M
13
= a
21 a
22
− a
31 a
32
The signed minor is
(
−
1 )
2
+ j
Mij
= cofactor
(9)
(10)
(11)
(12)
Example
For i = 1 , and j = 1 , the cofactor is: (
−
1 )
2
Mij
=
Mij
For i = 1 , and j = 2, the cofactor is: (
−
1 )
3
Mij
= −
Mij
Copyright Paul Tobin 48

DIT, Kevin St.
Electric Circuit Theory DT287/2
Using the formula we generate the following matrix with the sign of each element given as:




+
−
+
−
+
−
+
−
+




(13)
The value of a determinant:
The value of a determinant A of order n is the sum of the n products obtained by multiplying each element of a chosen row or column by its co-factor.
−
A
12
∆
12
+
A
22
∆
22
−
A
32
∆
32
(14)
Evaluate the following matrix around the 3 5 0 row.


1


2
3
4
1
5
7
(
−
6 )
0




=
3[-(4)(-6) - (1)(7)] - 5[(1)(-6) - (2)(7)] - 0 = 7
Note the brackets around the negative value. This is a good practice.
Example 1
Solve for the currents in (5)
I
1
=


−
E
1
E
2


( R
1
(
−
+
R
3
R
2
)
(
)
R
(
2
−
R
3
+
)
R
3
)


( R
(
−
R
3
2
+
)
R
3
)


I
2
=


( R
1
(
−
+
R
3
R
2
)
)


( R
1
(
−
+
R
3
R
2
)
)
−
E
E
1
2


( R
(
−
R
3
2
+
)
R
3
)


Example 2
Calculate the value of current i
2
in the circuit in figure 31 using
1) Thévenin's Theorem
2) Superposition Theorem
3) Mesh analysis
Copyright Paul Tobin 49

DIT, Kevin St.
Electric Circuit Theory
R
1
=10
Ω
, R
2
=5
Ω
, R
3
= 15
Ω
, X
1
= j10
Ω
, X
2
= j 5
Ω
,
E
1
=
10
∠
60 o and E
2
=
5
∠
30 o
.
DT287/2
Figure 31 :
Solution
The Thévenin impedance Z
TH
= jX
= − j 5 . Remember an ideal voltage source has a zero ohm source impedance and may be replaced by a short circuit.
Figure 32: Application of Thévenin's theorem.
The Thévenin voltage is:
V
TH
=
E
1
+
E
2
Convert each voltage vector in polar form to rectangular form and then add each component i.e.
E
1
E
1
=
10
∠
60 o
.
=
10
∠
60 o
.
=
E
1
( cos
+
=
10 ( cos60
+ j sin
) j sin60 ) and E
2 and E
2
=
5
∠
30 o
=
5
∠
30 o
=
E
2
( cos
+
=
5 ( cos 30
+ j sin
) j sin 30 )
The Thévenin equivalent circuit, with the load reinserted, is shown in figure
33.
Copyright Paul Tobin 50

DIT, Kevin St.
Electric Circuit Theory DT287/2
Figure 33: Thévenin equivalent circuit with load.
Note: In figure 33, maximum power is transferred to the load when the load is the complex conjugate of the Thévenin impedance.
Copyright Paul Tobin 51