Solve Mesh Currents: Find i1, i2 & i3

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The discussion focuses on solving mesh currents i1, i2, and i3 using Kirchhoff's Voltage Law (KVL). The user presents their KVL equations for each mesh but questions the inclusion of a -10 in the equation for i1. Responses emphasize the importance of correctly identifying the polarity of voltage sources and suggest solving the first two equations before substituting into the third. Clarification on voltage source polarities is deemed essential for accurate equation formulation. The thread highlights the procedural approach to solving mesh current problems in circuit analysis.
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Homework Statement



Find the mesh currents i1, i2 and i3

Homework Equations





The Attempt at a Solution



I did it this way

KVL at i1: 12 + 1i2 + 1i1 - 1I2 + 10 = 0

KVL at i2: 10 + 1i2 - 1i1 + 1i2 + 1i2 - 1i3 = 0

KVL at i3: 1i3 - 1i2 + 1i3 + 12 = 0

I was wondering though for KVL at i1, should i add a -10 to go with the 10?? Thanks.
 

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I would take a 2nd look at the polarity of your voltage sources on your mesh at i1. Remember you are looking clockwise.
 
What's the polarity on each of the voltage sources? You can't really write any equations without knowing them...or at least arbitrarily assigning them. The way you have your equations written, though, I would take Ouabache's advice...
 
THIS WILL BE DONE BY ,FIRST,SOLVING EQUATION 1 AND 2,THEN SUBSTITUTING THE VALUE IN THE 3RD EQUATION ...THIS PROCEDURE MAY HELP U,I HAVE DONE THIS TYPE OF QUESTION TODAY,,IT MIGHT WORK,,,DO REPLY ME IF WORKED...THNX

..ATHAR
engineer_dave said:

Homework Statement



Find the mesh currents i1, i2 and i3

Homework Equations





The Attempt at a Solution



I did it this way

KVL at i1: 12 + 1i2 + 1i1 - 1I2 + 10 = 0

KVL at i2: 10 + 1i2 - 1i1 + 1i2 + 1i2 - 1i3 = 0

KVL at i3: 1i3 - 1i2 + 1i3 + 12 = 0

I was wondering though for KVL at i1, should i add a -10 to go with the 10?? Thanks.
 

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