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Meshes and Loops Steps of Mesh Analysis Supermesh Examples

Lecture 6. Mesh Analysis. Meshes and Loops Steps of Mesh Analysis Supermesh Examples. Circuit Analysis – A Systematic Approach. Mesh Analysis is another general method that is almost as powerful as Nodal Analysis.

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Meshes and Loops Steps of Mesh Analysis Supermesh Examples

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

  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. 1kW 1kW + + + V1 Vout 1kW V2 - - - 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. Vout = (V1 + V2)/3

  4. Mesh Analysis: The Recipe • 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. + + V1 - - Step 1: Identifying the Meshes 1kW 1kW 1kW Mesh 1 Mesh 2 V2

  6. Mesh Analysis: The Recipe • 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. + I1 - Step 2: Assigning Mesh Currents 1kW 1kW 1kW + V1 V2 I2 -

  8. Mesh Analysis: The Recipe • 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. VR + - R I1 VR + I2 - R I1 Step 3: Voltages from Mesh Currents VR = I1R VR = (I1 -I2 ) R

  10. 1kW 1kW 1kW + + V1 V2 I1 I2 - - KVL Around Mesh 1 -V1 + I1 1kW + (I1 - I2) 1kW = 0 I1 1kW + (I1 - I2) 1kW = V1

  11. 1kW 1kW 1kW + + V1 V2 I1 I2 - - KVL Around Mesh 2 (I2 - I1) 1kW + I2 1kW + V2 = 0 (I2 - I1) 1kW + I2 1kW = -V2

  12. Mesh Analysis: The Recipe • 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. Step 4: Solve the Equations • Re-organize the equations: • The two equations can be combined into a single matrix/vector equation. I1 + (I1 - I2) 1kW = V1 (1kW + 1kW )I1 - 1kWI2 = V1 (I2 - I1) 1kW + I2 1kW = -V2 - 1kW I1 + (1kW + 1kW ) I2 = -V2

  14. Using MATLAB >> A = [1e3+1e3 -1e3; -1e3 1e3+1e3]; >> v = [7; -4]; >> i = inv(A)*v i = 0.00333 -0.00033 I1 = 3.33mA I2 = -0.33mA Vout = (I1 - I2) 1kW = 3.66V

  15. 2kW 2mA 1kW + 12V 2kW 4mA - I0 Another Example

  16. Identify Mesh’s 2kW Mesh 3 2mA 1kW + 2kW Mesh 1 Mesh 2 12V 4mA - I0

  17. I1 Assign Mesh Currents 2kW 2mA I3 1kW + 2kW 12V 4mA I2 - I0

  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 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?

  19. I1 SuperMesh 2kW The Supermesh does not include this source! The Supermesh surrounds this source! 2mA I3 1kW + 2kW 12V 4mA I2 - I0

  20. KVL Around the Supermesh -12V + I3 2kW + (I3 - I2)1kW + (I1 - I2)2kW = 0 I3 2kW + (I3 - I2)1kW + (I1 - I2)2kW = 12V

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

  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 I1 = 1.2mA I2 = -4mA I3 = -0.8mA I0 = I1 - I2 = 5.2mA

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