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Electrical Circuits I
Emam Fathy
Department of Electrical and Control Engineering
email: [email protected]
http://www.aast.edu/cv.php?disp_unit=346&ser=68525
Lecture 4
Mesh Analysis
Mesh Analysis
• Nodal analysis was developed by applying
KCL at each non-reference node
• Mesh analysis is developed by applying
KVL around loops in the circuit
• Mesh (loop) analysis results in a system of
linear equations which must be solved for
unknown currents
Steps of Mesh Analysis
1. Identify mesh (loops).
2. Assign a current to each mesh.
3. Apply KVL around each loop to get an
equation in terms of the loop currents.
4. Solve the resulting system of linear
equations for the mesh/loop currents.
Example
+
–
Vout
1kW
1kW
1kW
V1 V2
+
–
+
–
• Find Vout
1. Identifying the Meshes
Mesh 2
1kW
1kW
1kW
V1 V2Mesh 1+
–
+
–
2. Assigning Mesh Currents
1kW
1kW
1kW
V1 V2I1 I2
+
–
+
–
3. KVL Around Mesh 1
I1 1kW + (I1 – I2) 1kW = V1
Or I1 (1kW+ 1kW) – I2 1kW = V1
1kW
1kW
1kW
V1 V2I1 I2
+
–
+
–
3. KVL Around Mesh 2
1kW
1kW
1kW
V1 V2I1 I2
+
–
+
–
(I2 – I1) 1kW + I2 1kW = –V2
Or – I1 1kW + I2 (1kW + 1kW) = – V2
Solution
• The two equations can be combined into a
single matrix/vector equation
I1 (1kW+ 1kW) – I2 1kW = V1
– I1 1kW + I2 (1kW + 1kW) = – V2
W+WW
WW+W
2
1
2
1
k1k1k1
k1k1k1
V
V
I
I
4. Solving the Equations
Let: V1 = 7V and V2 = 4V
Results:
I1 = 3.33 mA
I2 = –0.33 mA
Finally
Vout = (I1 – I2) 1kW = 3.66V
Example
1kW
2kW
2kW
12V 4mA
2mA
I0
+
–
Mesh 2
Mesh 3
Mesh 1
1. Identify Meshes
1kW
2kW
2kW
12V 4mA
2mA
I0
+
–
2. Assign Mesh Currents
I1 I2
I31kW
2kW
2kW
12V 4mA
2mA
I0
+
–
The
Supermesh
surrounds
this source!
The
Supermesh
does not
include this
source!
12V 4mA
1kW
2kW
2kW
2mA
I0
I1 I2
I3
+
–
Supermesh
I2 = –4 mA
I1 – I3 = 2 mA
I3 2kW + (I3 - I2)1kW + (I1 - I2)2kW = 12V KVL Around the Supermesh
A Supermesh results when 2 meshes have a current
source in common.
Solution
• The three equations can be combined into
a single matrix/vector equation
W+WWWW
V12
mA2
mA4
1k2k2k1k2k
101
010
3
2
1
I
I
I
I1 = 1.2 mA, I2 = – 4 mA, I3 = – 0.8 mA
I0 = I1 – I2 = 5.2 mA
Example
Find power dissipated in 12W-resistor and 3W-resistor
using mesh analysis.
12 V 8 V
2 W 9 W
12 W
4 W 3 W
Solution:
12 V 8 V
2 W 9 W
12 W
4 W 3 W
I1
I2
KVL I1:
18I1 – 12I2 = 12
KVL I2:
-12I1 + 24I2 = -8
Example
R1
R4
12 V
R2
R3
I1
I3
I2I
s
I3 = -Is
End of Lec
