# Mesh Analysis with problems

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Mesh Analysis

### Slide 2:

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.

### Slide 3:

A circuit for illustrating mesh analysis. Eq 1 Around mesh 1:

Eq 2 Eq 3 Eq 4

### Slide 5:

We are left with 2 equations: From (1) and (4) we have, Eq 5 Eq 6 We can easily solve these equations for I1 and I2.

### Slide 6:

The previous equations can be written in matrix form as: Eq (7) Eq (8)

### Slide 7:

Mesh Analysis: Example 1. Write the mesh equations and solve for the currents I1, and I2. Circuit for Example 1. Mesh 1 4I1 + 6(I1 – I2) = 10 - 2 Mesh 2 6(I2 – I1) + 2I2 + 7I2 = 2 + 20 Eq (9) Eq (10)

### Slide 8:

Mesh Analysis: Example 1, continued. Simplifying Eq (9) and (10) gives, 10I1 – 6I2 = 8 -6I1 + 15I2 = 22 Eq (11) Eq (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

### Slide 9:

Mesh Analysis: Example 2 Solve for the mesh currents in the circuit below. Circuit for Example 2. The plan: Write KVL, clockwise, for each mesh. Look for a pattern in the final equations.

### Slide 10:

Mesh Analysis: Example 2 Mesh 1: 6I1 + 10(I1 – I3) + 4(I1 – I2) = 20 + 10 Mesh 2: 4(I2 – I1) + 11(I2 – I3) + 3I2 = - 10 - 8 Mesh 3: 9I3 + 11(I3 – I2) + 10(I3 – I1) = 12 + 8 Eq (13) Eq (14) Eq (15)

### Slide 11:

Mesh Analysis: Example 7.2 Clearing Equations (13), (14) and (15) gives, 20I1 – 4I2 – 10I3 = 30 -4I1 + 18I2 – 11I3 = -18 -10I1 – 11I2 + 30I3 = 20 In matrix form: WE NOW MAKE AN IMPORTANT OBSERVATION!! Standard Equation form

### Slide 12:

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.

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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.

### Slide 14:

Mesh Analysis: Example 3 - Direct method. Use the direct method to write the mesh equations for the following. Circuit diagram for Example 3. Eq (13)

### Slide 15:

Mesh Analysis: With current sources in the circuit Example 4: Consider the following: Circuit diagram for Example 4. Use the direct method to write the mesh equations.

### Slide 16:

Mesh Analysis: With current sources in the circuit This case is explained by using an example. Example 4: Find the three mesh currents in the circuit below. Circuit for Example 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.

### Slide 17:

Mesh Analysis: With current sources in the circuit Example 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.

### Slide 18:

Mesh Analysis: With current sources in the circuit Example 4: Continued. Equation for mesh 1: 10I1 + (I1-I2)5 = 10 or 15I1 – 5I2 = 10 Equations for mesh 2: 2I3 + (I3-I2)20 = 20 or - 20I2 + 22I3 = 20 Constraint Equation I2 = - 4A

### Slide 19:

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

### Slide 20:

End of Lesson circuits Mesh Analysis