Mesh Circuit Analysis: Solving KCL Equations for i1, i2, and i3 | Homework Help

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Discussion Overview

The discussion revolves around solving Kirchhoff's Current Law (KCL) equations for a mesh circuit analysis problem. Participants are seeking clarification on their approach to the equations and the correctness of their solutions, as well as resources for further learning.

Discussion Character

  • Homework-related, Technical explanation, Debate/contested

Main Points Raised

  • One participant presents a set of KCL equations derived from a circuit diagram and asks if their approach is correct.
  • Another participant suggests that the approach looks okay but advises checking the sign on the i3 term in one of the equations.
  • There is a discussion about the labeling of loops and current directions in the circuit diagram to enhance clarity.
  • A participant expresses confusion regarding the sign of the term related to the 5-ohm resistor in their second equation, questioning whether it should be negative.
  • Another participant confirms that the term should indeed be negative and points out the specific line in the equations where the sign needs correction.
  • One participant acknowledges their misunderstanding and expresses gratitude for the assistance received.

Areas of Agreement / Disagreement

Participants generally agree on the structure of the equations but disagree on the correctness of the sign in one of the terms. The discussion remains unresolved regarding the participant's understanding of the KCL application.

Contextual Notes

There are indications of missing assumptions regarding the circuit's configuration and the specific definitions of current directions, which may affect the interpretation of the equations.

Who May Find This Useful

Students studying circuit analysis, particularly those learning about mesh analysis and KCL, may find this discussion beneficial.

dwn
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Homework Statement



I have attached an image of the problem:

Homework Equations



KCL Equation

The Attempt at a Solution



0 = 2i11 + 3(i1-i2)
212 = 3(i2-i1) + 5(i2-i3)
-122 = 3i3 + 5(i3-i2)

5i1 - 3i2 = 0
-3i1 +8i2 + 5i3 = 212
-5i2 + 8i3 = -122

Is this the correct approach?

Also, does anybody have a good reference to learn this stuff? I'm taking this course online and I'm drowning. Thanks!
 

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dwn said:

Homework Statement



I have attached an image of the problem:

Homework Equations



KCL Equation


The Attempt at a Solution



0 = 2i11 + 3(i1-i2)
212 = 3(i2-i1) + 5(i2-i3)
-122 = 3i3 + 5(i3-i2)

5i1 - 3i2 = 0
-3i1 +8i2 + 5i3 = 212
-5i2 + 8i3 = -122

Is this the correct approach?
The approach looks okay. Check the sign on the i3 term for your second equation. Otherwise looks okay to me.

You should label the loops and current directions on the circuit diagram and associate your equations with the labels so it's clear which loop each equation belongs to. Granted it's pretty clear in this relatively simple case with only three loops, but it's good to get in the habit of making your work easy to interpret for those who are just seeing it for the first time.

Also, does anybody have a good reference to learn this stuff? I'm taking this course online and I'm drowning. Thanks!
You can probably find many worked examples on the net by searching for "mesh analysis examples". For that matter there are plenty here on the Physics Forums. Take a look at the "Related Discussions" section at the bottom of your thread. Also investigate the forum search facilities.
 
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Thank you for your help and I will make sure to clearly present everything in the future.

I'm not sure I quite understand why the second equation is incorrect..? If the less simplified equation above is correct, which as I understand it, you subtract the current adjacent to the resistor of interest (5 ohm) from the current loop you are analyzing, shouldn't it be a negative 5i3 ::: 212 = 3(i2-i1) + 5(i2-i3)
 
dwn said:
Thank you for your help and I will make sure to clearly present everything in the future.

I'm not sure I quite understand why the second equation is incorrect..? If the less simplified equation above is correct, which as I understand it, you subtract the current adjacent to the resistor of interest (5 ohm) from the current loop you are analyzing, shouldn't it be a negative 5i3 ::: 212 = 3(i2-i1) + 5(i2-i3)

Yes, it should be -5i3. That's the point I was making. Look at the relevant section from your first post...
5i1 - 3i2 = 0
-3i1 +8i2 + 5i3 = 212 <---------
-5i2 + 8i3 = -122

The 5i3 in the indicated line should be negative, right?
 
Ooops...I thought you had changed it and that's why you used the red...my bad! I think I'm going cross eyed!
Thanks again for the help gneill.
 

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