Solving Circuit Problems with Kirchhoff's Rules

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Homework Help Overview

The discussion revolves around solving a circuit problem using Kirchhoff's Rules, specifically focusing on determining the currents I1, I2, and I3 in a circuit with multiple batteries and resistances. The original poster describes their approach and the circuit diagram they have drawn, which includes assumptions about battery resistance and terminal voltage.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply Kirchhoff's Rules but questions the number of loops needed and where to apply the junction rule. Participants discuss the necessity of three equations for three unknowns and the relationship between the loop equations.

Discussion Status

Participants are actively engaging in the problem, with some providing feedback on the original poster's equations. There is a divergence in results regarding the value of I3, leading to uncertainty about the correctness of the answer key. Multiple interpretations of the problem setup are being explored.

Contextual Notes

There is mention of potential issues with the orientation of the batteries and the accuracy of the answer key, which remains unverified by all participants. The original poster has expressed confidence in their setup after multiple checks.

joej
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okay I got a question asking me to determine the currents in I1, I2, and I3 in a circuit assuming that the resistance of each battery is 1 ohm (I drew a separate 1ohm resistor after every battery to represent this) and the terminal voltage of the 6V battery

now I have drawn a diagram of the circuit and labeled every part of it, I am assuming I have to use Kirchhoff's Rules to solve this problem

now here is where I get stuck, do I make 3 loops since there are 3 batteries or will 2 loops be fine
(using the diagram loop 1 will probably be 'abefa' and loop 2 will be 'abcdefa', do I need loop 3 'ebcde')

also at which point should I apply Kirchhoff's junction rule?
 

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You have 3 unknowns, so you need three equations. One of these is the junction rule for the three currents (at either of the two junctions...you'll see that they both give you the same equation). The other two will be two loop equations.

If you write the third loop equation, you'll find that it is actually just a combination of the two previous ones, and so, gives you nothing new.
 
okay still seems I'm screwing up somewhere...

junction is at point b on my graph so... I1 = I2+I3

loop 1 is 'bcdefab' => +12-9(I1)-15(I3)+6-19(I3) = 0

loop 2 is 'befab' => +11(I2)-12-15(I3)+6-19(I3) = 0

I1 = 18-34(I3) / 9 ==> I1 = 2-3.8(I3)

I2 = -6-34(I3) / -11 ==> I2 = 0.55+3.1(I3)


I3 = I1 - I2

I3 = (2-3.8(I3))-(0.55+3.1(I3)) ==> I3 = 1.45 - 6.9(I3)


I'm guessing I'm screwing up somewhere in the veginning with either the loops of the junction cause that's th only place I can think of where I migth be going wrong unless I'm not noticing something
 
What makes you think you're wrong ? Everything looks okay to me, so far.

You've got : I3 = 1.45 - 6.9(I3)

Or, 7.9(I3) = 1.45 ==>I3 = (1.45/7.9) A
 
that gives me

I3 = 0.18A

the answer key tells me I am wrong as it says I3 should be 0.055A

so yeah basically I'm stuck and not quite sure where I'm going wrong
 
When I solved it, i got i3 = 0.185 A. I believe the answer key is wrong.
 
not that I don't trust you but is there someone else that could confirm this as well, I would like to have a few confirmations that they key is actually wrong before I decide to go against it, thnx ;)
 
joej said:
not that I don't trust you but is there someone else that could confirm this as well, I would like to have a few confirmations that they key is actually wrong before I decide to go against it, thnx ;)

I get 0.18A for I3 also. Make sure that you've got the question right... You might have gotten the battery orientation wrong or something.
 
okay tripple checked everything, I got nothing wrong so it seems that the answer key is bogus, thnx for the help
 

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