How do I solve 2 simultaneous equations using KVL?

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Discussion Overview

The discussion revolves around solving two simultaneous equations using Kirchhoff's Voltage Law (KVL). Participants are exploring the process of deriving equations for mesh currents in a circuit and addressing discrepancies between their results and those from a reference example.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents their equations for mesh currents, expressing confusion over discrepancies with a reference example.
  • Another participant suggests verifying whether the reference answers satisfy the participant's equations.
  • There is a discussion about the signs of voltage in the equations, with one participant noting a difference in the sign of a voltage term compared to the reference.
  • Participants clarify that the equations presented are identical to those in the reference example, indicating that sign changes do not affect the validity of the equations.
  • A later reply outlines a method for solving the simultaneous equations by manipulating them to eliminate one variable.

Areas of Agreement / Disagreement

Participants generally agree on the form of the equations but disagree on the interpretation of the signs and the method of arriving at the mesh currents. The discussion remains unresolved regarding the specific values of the currents derived from the equations.

Contextual Notes

There are unresolved issues regarding the assumptions made about the signs of voltage in the equations, and the steps taken to solve the simultaneous equations have not been fully detailed.

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So I'm trying to get a better understanding of KVL by looking up practice problem and I'm having some difficulties. I'm attempting to do this problem
loopcurrentmethod.gif

my first step is to get an equation for Ia (32v-10ia+8ib = 0) an Ib (12ib+20v-8ia = 0), adn then solve for either b or a first and plug into the other equation, but this is where the problem occurs. on the site http://fourier.eng.hmc.edu/e84/lectures/ch2/node2.html it got a current for each mesh to be something completely different. I was wondering how they got 4 and 1 for the currents?
 
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Have you checked to see whether their answers satisfy your equations?
 
NascentOxygen said:
Have you checked to see whether their answers satisfy your equations?
What do you mean? That is a pratice example so I'm assuming their answers and equations are correct and if mine doesn't match up then I'm doing something wrong
 
Do yours "match up" or not?
 
NascentOxygen said:
Do yours "match up" or not?
at first, no and the only thing that didnt match up was that i got my 20v to be the opposite sign of what they had
 
Your equations are identical with theirs. You can multiply everything on both sides by –1 and the equation still holds.
 
NascentOxygen said:
Your equations are identical with theirs. You can multiply everything on both sides by –1 and the equation still holds.
no not what I mean, on the B side, i have all the other sign matching up except the 20v so multiply by -1 won't make it correct, but that's not my question I just want to know how they got their currents for each of the meshes
 
If these are your equations:
32v-10ia+8ib = 0 and
12ib+20v-8ia = 0
then they are indeed identical with those in the worked example. All signs correspond.

You are asking how to solve 2 simultaneous equations?

If we use yours, which I'll write out again neatly:
32 - 10ia + 8ib = 0 ... (i)

12ib + 20 - 8ia = 0 ... (ii)

Multiply eqn (i) by 4
Multiply eqn (ii) by -5

Then add the corresponding sides of both new equations, and the resultant term for ia has a coefficient of 0, i.e., ia disappears from the equation, and with only one unknown in the equation you can solve for it to find the value for ib.
 

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