Kirchhoff's Rules Homework Solutions for Circuit Problem

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Homework Help Overview

The discussion revolves around a circuit problem involving Kirchhoff's Rules, specifically analyzing a circuit with three loops and three unknown currents. The circuit includes resistances of 8 Ohms, 6 Ohms, and 4 Ohms, along with a 4V and a 12V battery.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of equations based on Kirchhoff's Rules, with some questioning the independence of the loops and the correctness of the equations formed. There is mention of simplifying the problem by reducing the number of currents used.

Discussion Status

Some participants have provided guidance on simplifying the problem by focusing on two independent loops instead of three. The original poster expresses uncertainty but indicates progress after adjusting their approach, while others share their own challenges with similar problems.

Contextual Notes

There is a mention of confusion regarding the flow of current through specific resistors and the need for clarity on the equations derived from the circuit. The original poster also notes a sense of frustration with the problem, indicating it is perceived as simpler than it has turned out to be.

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Homework Statement


Sorry for edit, accidentally posted.

Circuit is show in my attachment. I have 3 loops with three unknown currents.

8 Ohms of resistance down the left side.
6 Ohms of resistance and a 4V battery down the center.
4 Ohms of resistance and a 12V battery down the right side.

I honestly have no idea if I'm even approaching this correctly.

Homework Equations



Kirchhoff's Rules

The Attempt at a Solution



See attachment:

Loop A (Big Loop)
-8I1 +12V +4I1 +4I3 = 0
-4I1 +4I3 = -12V

Loop B (Left)
-8I1 +4V -6I1 +6I2 = 0
-14I1 + 6I2 = -4V

Loop C (Right)
-6I3 - 6I2 - 4V + 12V - 4I3 = 0
-10I3 - 6I2 = -8V

combine in a matrix
I1 I2 I3 V
a -4 0 -4 -12
b -14 6 0 -4
c 0 -6 -10 8

rref()

I1 = 1.083
I2 = 1.861
I3 = -1.917

Im not entirely sure, but I think these are voltages for the loops, so from there I would set
V= IR for each of the loops to find the current, but this produces the wrong answer. This is the easiest circuit in my textbook, and I have been searching the web for problems which is only making matters worse I think.

Any help would be greatly appreciated.
 

Attachments

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Last edited:
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You don't have three independent loops. Only two are independent, equation for the third loop is a linear combination of two others. Third equation that you need to solve is about currents in the upper node.

But second opinion won't hurt, could be I am missing something. It happens when I stick my head outside of Chemistry forum

--
methods
 
I don't think it is right. For instance, the -8I1 +12V +4I1 +4I3 = 0
should be -8I1 +12V + 4I3 = 0 because I1 does not flow through the 4 ohm resistor.

Recommend you just use two currents. Say, I1 as you have it down the left side and I2 in place of your I3 up on the right side. Then the current down the center is I2 - I1. This gives you a system of two equations to solve so quite a bit simpler. Just do the left and right loops to get them.
 
I'm still not sure that I know what I'm doing, but sticking to two loops helped; I have the correct answers now. This is such a headache, you guy have no idea how long I've been stuck on that simple little problem, onto the next one.

Thanks for the help.
 

Attachments

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    scan0002.jpg
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Super! Glad to see you using that matrix method for solving linear systems.
I had a lot of trouble with that problem, too. I find it hard to get the sign of each term right.
 

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