Transcript 7. Loop or mesh analysis

Lecture 6. Mesh Analysis
• Meshes and Loops
• Steps of Mesh Analysis
• Supermesh
• Examples
1
Circuit Analysis – A Systematic Approach
• Mesh Analysis is another general method that is almost as
powerful as Nodal Analysis.
- Nodal analysis was developed by applying KCL at each nonreference node.
- Loop analysis is developed by applying KVL around loops in the
circuit.
- Loop analysis results in a system of linear equations which must
be solved for unknown currents.
2
Learning by Examples: A Summing Circuit
• The output voltage V of this circuit is proportional to the sum of the
two input voltages V1 and V2.
• This circuit could be useful in audio applications or in
instrumentation.
• The output of this circuit would probably be connected to an
amplifier.
1kW
V1
+
-
1kW
+
Vout
+
1kW
-
V2
-
Vout = (V1 + V2)/3
3
Mesh Analysis: The Recipe
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.
4
Step 1: Identifying the Meshes
1kW
V1
+
-
Mesh 1
1kW
1kW
Mesh 2
+
-
V2
5
Mesh Analysis: The Recipe
1. Identify mesh (loops).
2. Assign a current to each mesh.
3. Apply KVL around each mesh to get an
equation in terms of the mesh currents.
4. Solve the resulting system of linear
equations.
6
Step 2: Assigning Mesh Currents
1kW
V1
1kW
+
-
1kW
I1
+
I2
-
V2
7
Mesh Analysis: The Recipe
1. Identify mesh (loops).
2. Assign a current to each mesh.
3. Apply KVL around each mesh to get an
equation in terms of the mesh currents.
4. Solve the resulting system of linear
equations.
8
Step 3: Voltages from Mesh Currents
+ VR
R
+
-
I1
VR = I1 R
VR
I2
-
R
I1
VR = (I1 - I2 ) R
9
KVL Around Mesh 1
1kW
V1
1kW
+
-
1kW
I1
+
I2
-
V2
-V1 + I1 1kW + (I1 - I2) 1kW = 0
I1 1kW + (I1 - I2) 1kW = V1
10
KVL Around Mesh 2
1kW
V1
1kW
+
-
1kW
I1
+
I2
-
V2
(I2 - I1) 1kW + I2 1kW + V2 = 0
(I2 - I1) 1kW + I2 1kW = -V2
11
Mesh Analysis: The Recipe
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.
12
Step 4: Solve the Equations
• Re-organize the equations:
I1 + (I1 - I2) 1kW = V1
(I2 - I1) 1kW + I2 1kW = -V2
(1kW + 1kW )I1 - 1kW I2 = V1
- 1kW I1 + (1kW + 1kW ) I2 = -V2
• The two equations can be combined into a single matrix/vector
equation.
 1kW   I1   V1 
1kW  1kW




  1kW

I

V
1
k
W

1
k
W

 2   2 
13
Using MATLAB
>> A = [1e3+1e3 -1e3;
-1e3 1e3+1e3];
>> v = [7; -4];
>> i = inv(A)*v
i =
I1 = 3.33mA
0.00333
I2 = -0.33mA
-0.00033
Vout = (I1 - I2) 1kW = 3.66V
14
Another Example
2kW
2mA
1kW
+
12V
-
2kW
I0
4mA
15
Identify Mesh’s
2kW
2mA Mesh 3
+
12V
-
Mesh 1
1kW
2kW
Mesh 2
I0
4mA
16
Assign Mesh Currents
2kW
2mA
I3
+
12V
-
2kW
I2
I1
1kW
4mA
I0
17
How to Deal with Current Sources
• The current sources in this circuit will have whatever voltage is
necessary to make the current correct.
• We can’t use KVL around the loop because we don’t know the
voltage. What to do?
• The 4mA current source sets I2:
I2 = -4mA
• The 2mA current source sets a constraint on I1 and I3:
I1 - I3 = 2mA
• We have two equations and three unknowns. Where is the third
equation?
18
SuperMesh
2kW
The
Supermesh
surrounds this
source!
2mA
I3
+
12V
-
2kW
I2
I1
1kW
The
Supermesh
does not
include this
source!
4mA
I0
19
KVL Around the Supermesh
-12V + I3 2kW + (I3 - I2)1kW + (I1 - I2)2kW = 0
I3 2kW + (I3 - I2)1kW + (I1 - I2)2kW = 12V
20
Solve the Equations
• The three equations can be combined into a single matrix/vector
equation.
1
0
 0
  I 1   4mA 
 1
  I    2mA 
0

1

 2  

2kW  1kW  2kW 2kW  1kW  I 3   12V 
21
Solve Using MATLAB
>> A = [0 1 0; 1 0 -1;
2e3 -1e3-2e3 2e3+1e3];
>> v = [-4e-3; 2e-3; 12];
>> i = inv(A)*v
i =0.0012
-0.0040
-0.0008
I1 = 1.2mA
I2 = -4mA
I3 = -0.8mA
I0 = I1 - I2 = 5.2mA
22