Solving Circuit Problems with Kirchhoff's Rules

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The discussion revolves around solving a circuit problem using Kirchhoff's Rules to determine the currents I1, I2, and I3. The user initially struggles with the number of loops needed and where to apply Kirchhoff's junction rule, ultimately establishing that I1 = I2 + I3 at junction point b. After deriving equations for the loops, the user calculates I3 and arrives at 0.18A, which contradicts the answer key that states I3 should be 0.055A. Multiple participants agree with the user's calculation, suggesting the answer key may be incorrect, and they emphasize the importance of verifying the problem setup and battery orientation. The conversation concludes with a consensus that the user's calculations appear accurate, casting doubt on the answer key's validity.
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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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