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Assessment Problem # 4.12
Mesh-Current Method
Special Case
Current Source in a branch
ECE 201 Circuit Theory I
1
Use the mesh-current method to find the
power dissipated in the 1 Ω resistor.
R4
2 Ohm
I1
R2
2 Ohm
2A
V1
10 V
R1
2 Ohm
V2
6V
R3
1 Ohm
ECE 201 Circuit Theory I
2
Identify the mesh currents
R4
2 Ohm
I3
I1
R2
2 Ohm
2A
V1
10 V
R1
2 Ohm
I1
I2
V2
6V
R3
1 Ohm
ECE 201 Circuit Theory I
3
Assign the voltage polarities
Identify the voltage @ the 2A source
R4
+
-
2 Ohm
I3
I1
R2
+
2A
V
V1
10 V
I1
-
2 Ohm
+
+
-
+
-
R1
2 Ohm
V2
6V
I2
+
-
R3
+
1 Ohm
ECE 201 Circuit Theory I
4
Write the mesh-current equations (1)
R4
+
-
2 Ohm
I3
I1
R2
+
2A
-
V
+
V1
10 V
I1
-
V + 2I1 – 2I2 = 10
2 Ohm
+
+
-
R1
2 Ohm
V2
6V
I2
+
-
R3
+
1 Ohm
ECE 201 Circuit Theory I
5
Write the mesh-current equations (2)
R4
+
-
2 Ohm
I3
I1
+
2A
-
V
+
+
V1
10 V
I1
-
V + 2I1 – 2I2 = 10
R2
2 Ohm
+
-
R1
2 Ohm
I2
V2
6V
+
-2I1 +5I2 -2I3 = 6
- R3 +
1 Ohm
ECE 201 Circuit Theory I
6
Write the mesh-current equations (3)
R4
+
-
2 Ohm
I3
-V -2I2 +4I3 = 0
-
I1
+
2A
-
V
+
+
V1
10 V
I1
-
V + 2I1 – 2I2 = 10
R2
2 Ohm
+
-
R1
2 Ohm
I2
V2
6V
+
-2I1 +5I2 -2I3 = 6
- R3 +
1 Ohm
ECE 201 Circuit Theory I
7
Eliminate V
from the equations for meshes 1 and 3
V+2I1  2I 2  10
-V-2I 2  4I3  0
Adding,
2I1  4I 2  4I3  10
Also,
2I1  5I 2  2I3  6
I1  I3  2
ECE 201 Circuit Theory I
8
Write the equations in matrix form
 6 4   I1  18 

 I    
 4 5   2   2 
Solving with a TI-89,
[6,-4;-4,5],[18;2])ENTER
ECE 201 Circuit Theory I
9
Solving
• Result looks like
7
 
6
I1 = 7A
I2 = 6A
Calculate I3
I3 =I1 – 2 = 5A
ECE 201 Circuit Theory I
10
Find the power dissipated
in the 1Ω resistor
• P1Ω = (I2)2R = (6)2(1) = 36 W
ECE 201 Circuit Theory I
11