1. Lesson 7 Basic Laws of Electric Circuits Mesh Analysis

2. Basic Circuits Mesh Analysis: Basic Concepts:  In formulating mesh analysis we assign a mesh current to each mesh.  Mesh currents are sort of fictitious in that a particular mesh current does not define the current in each branch of the mesh to which it is assigned.

3. Basic Circuits Mesh Analysis: Basic Concepts: Figure 7.2: A circuit for illustrating mesh analysis. Eq 7.1 Around mesh 1:

4. Basic Circuits Mesh Analysis: Basic Concepts: Eq 7.2 Eq 7.3 Eq 7.4

5. Basic Circuits Mesh Analysis: Basic Concepts: We are left with 2 equations: From (7.1) and (7.4) we have, Eq 7.5 Eq 7.6 We can easily solve these equations for I 1 and I 2.

6. Basic Circuits Mesh Analysis: Basic Concepts: The previous equations can be written in matrix form as: Eq (7.7) Eq (7.8)

7. Basic Circuits Mesh Analysis: Example 7.1. Write the mesh equations and solve for the currents I 1, and I 2. Figure 7.2: Circuit for Example 7.1. Mesh 1 4I 1 + 6(I 1 – I 2 ) = 10 - 2 Mesh 2 6(I 2 – I 1 ) + 2I 2 + 7I 2 = 2 + 20 Eq (7.9) Eq (7.10)

8. Basic Circuits Mesh Analysis: Example 7.1, continued. Simplifying Eq (7.9) and (7.10) gives, 10I 1 – 6I 2 = 8 -6I 1 + 15I 2 = 22 Eq (7.11) Eq (7.12) » % A MATLAB Solution » » R = [10 -6;-6 15]; » » V = [8;22]; » » I = inv(R)*V I = 2.2105 2.3509 I 1 = 2.2105 I 2 = 2.3509

9. Basic Circuits Mesh Analysis: Example 7.2 Solve for the mesh currents in the circuit below. Figure 7.3: Circuit for Example 7.2. The plan: Write KVL, clockwise, for each mesh. Look for a pattern in the final equations.

10. Basic Circuits Mesh Analysis: Example 7.2 Mesh 1: 6I 1 + 10(I 1 – I 3 ) + 4(I 1 – I 2 ) = 20 + 10 Mesh 2: 4(I 2 – I 1 ) + 11(I 2 – I 3 ) + 3I 2 = - 10 - 8 Mesh 3: 9I 3 + 11(I 3 – I 2 ) + 10(I 3 – I 1 ) = 12 + 8 Eq (7.13) Eq (7.14) Eq (7.15)

11. Basic Circuits Mesh Analysis: Example 7.2 Clearing Equations (7.13), (7.14) and (7.15) gives, 20I 1 – 4I 2 – 10I 3 = 30 -4I 1 + 18I 2 – 11I 3 = -18 -10I 1 – 11I 2 + 30I 3 = 20 In matrix form: WE NOW MAKE AN IMPORTANT OBSERVATION!! Standard Equation form

12. Basic Circuits Mesh Analysis: Standard form for mesh equations Consider the following: R 11 = of resistance around mesh 1, common to mesh 1 current I 1. R 22 = of resistance around mesh 2, common to mesh 2 current I 2. R 33 = of resistance around mesh 3, common to mesh 3 current I 3.

13. Basic Circuits Mesh Analysis: Standard form for mesh equations R 12 = R 21 = - resistance common between mesh 1 and 2 when I 1 and I 2 are opposite through R 1,R 2. R 13 = R 31 = - resistance common between mesh 1 and 3 when I 1 and I 3 are opposite through R 1,R 3. R 23 = R 32 = - resistance common between mesh 2 and 3 when I 2 and I 3 are opposite through R 2,R 3. = sum of emf around mesh 1 in the direction of I 1. = sum of emf around mesh 2 in the direction of I 2. = sum of emf around mesh 3 in the direction of I 3.

14. Basic Circuits Mesh Analysis: Example 7.3 - Direct method. Use the direct method to write the mesh equations for the following. Figure 7.4: Circuit diagram for Example 7.3. Eq (7.13)

15. Basic Circuits Mesh Analysis: With current sources in the circuit Example 7.4: Consider the following: Figure 7.5: Circuit diagram for Example 7.4. Use the direct method to write the mesh equations.

16. Basic Circuits Mesh Analysis: With current sources in the circuit This case is explained by using an example. Example 7.4: Find the three mesh currents in the circuit below. Figure 7.5: Circuit for Example 7.4. When a current source is present, it will be directly related to one or more of the mesh current. In this case I 2 = -4A.

17. Basic Circuits Mesh Analysis: With current sources in the circuit Example 7.4: Continued. An easy way to handle this case is to remove the current source as shown below. Next, write the mesh equations for the remaining meshes. Note that I 2 is retained for writing the equations through the 5  and 20  resistors.

18. Basic Circuits Mesh Analysis: With current sources in the circuit Example 7.4: Continued. Equation for mesh 1: 10I 1 + (I 1 -I 2 )5 = 10 or 15I 1 – 5I 2 = 10 Equations for mesh 2: 2I 3 + (I 3 -I 2 )20 = 20 or - 20I 2 + 22I 3 = 20 Constraint Equation I 2 = - 4A

19. Basic Circuits Mesh Analysis: With current sources in the circuit Example 7.4: Continued. Express the previous equations in Matrix form: I 1 = -0.667 A I 2 = - 4 A I 3 = - 2.73 A

20. End of Lesson 7 circuits Mesh Analysis