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Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1.

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Presentation on theme: "Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1."— Presentation transcript:

1 Meshes and Loops Steps of Mesh Analysis Supermesh Examples Lecture 6. Mesh Analysis 1

2 2 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 non- reference 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.

3 3 Learning by Examples: A Summing Circuit The output voltage V of this circuit is proportional to the sum of the two input voltages V 1 and V 2. This circuit could be useful in audio applications or in instrumentation. The output of this circuit would probably be connected to an amplifier. + - V out 1k  V1V1 + - V2V2 + - V out = (V 1 + V 2 )/3

4 4 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.

5 5 Mesh 2 1k  Step 1: Identifying the Meshes V1V1 + - V2V2 + - Mesh 1

6 6 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.

7 7 1k  Step 2: Assigning Mesh Currents V1V1 + - V2V2 + - I1I1 I2I2

8 8 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.

9 9 Step 3: Voltages from Mesh Currents R I1I1 +- VRVR V R = I 1 R R I1I1 +- VRVR I2I2 V R = (I 1 - I 2 ) R

10 10 1k  V1V1 + - V2V2 + - I1I1 I2I2 KVL Around Mesh 1 -V 1 + I 1 1k  + (I 1 - I 2 ) 1k  = 0 I 1 1k  + (I 1 - I 2 ) 1k  = V 1

11 11 1k  V1V1 + - V2V2 + - I1I1 I2I2 KVL Around Mesh 2 (I 2 - I 1 ) 1k  + I 2 1k  + V 2 = 0 (I 2 - I 1 ) 1k  + I 2 1k  = -V 2

12 12 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.

13 13 Step 4: Solve the Equations The two equations can be combined into a single matrix/vector equation. I 1 + (I 1 - I 2 ) 1k  = V 1 (I 2 - I 1 ) 1k  + I 2 1k  = -V 2 Re-organize the equations: - 1k  I 1 + (1k  + 1k  ) I 2 = -V 2 (1k  + 1k  )I 1 - 1k  I 2 = V 1

14 14 Using MATLAB >> A = [1e3+1e3 -1e3; -1e3 1e3+1e3]; >> v = [7; -4]; >> i = inv(A)*v i = 0.00333 -0.00033 I 1 = 3.33mA I 2 = -0.33mA V out = (I 1 - I 2 ) 1k  = 3.66V

15 15 Another Example 1k  2k  12V + - 4mA 2mA I0I0

16 16 Mesh 2 Mesh 3 Mesh 1 Identify Mesh’s 1k  2k  12V + - 4mA 2mA I0I0

17 17 Assign Mesh Currents 1k  2k  12V + - 4mA 2mA I0I0 I1I1 I2I2 I3I3

18 18 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 I 2 : I 2 = -4mA The 2mA current source sets a constraint on I 1 and I 3 : I 1 - I 3 = 2mA We have two equations and three unknowns. Where is the third equation?

19 19 1k  2k  12V + - 4mA 2mA I0I0 I1I1 I2I2 I3I3 The Supermesh surrounds this source! The Supermesh does not include this source! SuperMesh

20 20 KVL Around the Supermesh -12V + I 3 2k  + (I 3 - I 2 )1k  + (I 1 - I 2 )2k  = 0 I 3 2k  + (I 3 - I 2 )1k  + (I 1 - I 2 )2k  = 12V

21 21 Solve the Equations The three equations can be combined into a single matrix/vector equation.

22 22 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 I 1 = 1.2mA I 2 = -4mA I 3 = -0.8mA I 0 = I 1 - I 2 = 5.2mA


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