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Lesson 7

Basic Laws of Electric Circuits. Mesh Analysis. Lesson 7. 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

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Lesson 7

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  1. Basic Laws of Electric Circuits Mesh Analysis Lesson 7

  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. Around mesh 1: Eq 7.1

  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 I1 and I2.

  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 I1, and I2. Figure 7.2: Circuit for Example 7.1. Eq (7.9) 4I1 + 6(I1 – I2) = 10 - 2 Mesh 1 Eq (7.10) Mesh 2 6(I2 – I1) + 2I2 + 7I2 = 2 + 20

  8. Basic Circuits Mesh Analysis: Example 7.1, continued. Simplifying Eq (7.9) and (7.10) gives, 10I1 – 6I2 = 8 -6I1 + 15I2 = 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 I1 = 2.2105 I2 = 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: 6I1 + 10(I1 – I3) + 4(I1 – I2) = 20 + 10 Eq (7.13) Mesh 2: 4(I2 – I1) + 11(I2 – I3) + 3I2 = - 10 - 8 Eq (7.14) Mesh 3: 9I3 + 11(I3 – I2) + 10(I3 – I1) = 12 + 8 Eq (7.15)

  11. Basic Circuits Mesh Analysis: Example 7.2 Clearing Equations (7.13), (7.14) and (7.15) gives, Standard Equation form In matrix form: 20I1 – 4I2 – 10I3 = 30 -4I1 + 18I2 – 11I3 = -18 -10I1 – 11I2 + 30I3 = 20 WE NOW MAKE AN IMPORTANT OBSERVATION!!

  12. Basic Circuits Mesh Analysis:Standard form for mesh equations Consider the following: R11 = of resistance around mesh 1, common to mesh 1 current I1. R22 = of resistance around mesh 2, common to mesh 2 current I2. R33 = of resistance around mesh 3, common to mesh 3 current I3.

  13. Basic Circuits Mesh Analysis:Standard form for mesh equations R12 = R21 = - resistance common between mesh 1 and 2 when I1 and I2 are opposite through R1,R2. R13 = R31 = - resistance common between mesh 1 and 3 when I1 and I3 are opposite through R1,R3. R23 = R32 = - resistance common between mesh 2 and 3 when I2 and I3 are opposite through R2,R3. = sum of emf around mesh 1 in the direction of I1. = sum of emf around mesh 2 in the direction of I2. = sum of emf around mesh 3 in the direction of I3.

  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 I2 = -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: 10I1 + (I1-I2)5 = 10 or 15I1 – 5I2 = 10 Equations for mesh 2: Constraint Equation 2I3 + (I3-I2)20 = 20 I2 = - 4A or - 20I2 + 22I3 = 20

  19. Basic Circuits Mesh Analysis: With current sources in the circuit Example 7.4: Continued. Express the previous equations in Matrix form: I1 = -0.667 A I2 = - 4 A I3 = - 2.73 A

  20. circuits End of Lesson 7 Mesh Analysis

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