logging in or signing up Mesh Analysis with problems kvvbapiraju Download Post to : URL : Related Presentations : Let's Connect Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Copy embed code: Embed: Flash iPad Dynamic Copy Does not support media & animations Automatically changes to Flash or non-Flash embed WordPress Embed Customize Embed URL: Copy Thumbnail: Copy The presentation is successfully added In Your Favorites. Views: 9466 Category: Science & Tech.. License: All Rights Reserved Like it (11) Dislike it (1) Added: August 27, 2009 This Presentation is Public Favorites: 1 Presentation Description No description available. Comments Posting comment... By: dhaval.kavathia (34 month(s) ago) Dear sir, PLEASE send me this ppt on my email id- email@example.com. Ver useful for me. Saving..... Post Reply Close Saving..... Edit Comment Close Premium member Presentation Transcript Slide 1: 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: Slide 4: 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. Slide 13: 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 You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.