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Finding the Current in a Circuit

  1. Feb 19, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the magnitude of the current in each branch of the circuit shown below, in which B1 = 3.86 V. Specify the direction of each current. (See attached image)

    2. Relevant equations

    Kirchhoff's rule

    3. The attempt at a solution

    I labeled the points in the circuit starting in the lower left hand corner with A, then going up to the first junction B, and so on. Hopefully that makes my explanation make more sense.

    I went ahead and chose E as my junction point (the junction point on the right) to give me the equation:

    I2-I1-I3=0 or I1=I2-I3 (1)

    Then I looked and the bottom loop, starting in the lower left hand corner going up. I got:

    -I2R2-I3R3+E1=0 (2)

    I next looked at the loop starting at point B and going up. I got:

    -E2+I1R1+I2R2=0 (3)

    I substituted equation 1 into equation 3 (giving me equation 4). I then multiplied both sides of equation 4 by 75 and both sides of equation 3 by 22 in order to get matching coefficients on the I3 term. I then subtracted those two equations to eliminate I3, giving me:


    I2=99.59mA which was incorrect.

    I spent a lot of time on this and can't figure out where I went wrong, ant help is appreciated, and I will clarify anything that doesn't make sense. Thanks.

    Attached Files:

  2. jcsd
  3. Feb 19, 2008 #2
    Any tips?

    It's due in an hour and I still can't figure out where I've gone wrong :(

    I've done it a few times from the beginning and still have gotten the same answer...
  4. Feb 19, 2008 #3
    You should get three equations for the three unknowns. It looks like you have two, though the loop descriptions are ambiguous (going up?). There are three loops you can go around, all of which you know from Kirchoff's Voltage law must sum to zero. I'll give you two of them. Let me name the currents I1, I2, and I3 from bottom to top, and have them all go right (the signs will work themselves out later), from KVL I get:

    Loop one is going around the bottom half of the circuit starting at the battery:

    Loop two is going around the top half of the circuit starting at the 56Ω resistor:

    Can you do the third one? Are you familiar with linear algebra (some highschool algebra classes cover it)?
  5. Feb 19, 2008 #4
    I was trying to use Kirchhoffs current rule for my first equation, where at the junction point on the right in the middle I2-I1-I3=0. But I see where I could use the third loop. Is -EB+75I1-22I3+5=0 correct?

    I am kind of confused where you wrote -EB+56I1+75I2, doesn't the 75 ohm resistor correspond to I1?

    Thanks for the help so far.
  6. Feb 19, 2008 #5
    Oh, whoops I accidentally switched the resistors. It should be


    Yes your equation is correct. Now you just have to go through the algebra, which you can do with substitution at this point, but it may take awhile.

    I'll give you one of the answers in a sec that may help you, though you really ought to learn about procrastination. :p
  7. Feb 19, 2008 #6
    Okay, you should get 62mA for the top current, but if I made a mistake you are in trouble as you only have 15 mins.
  8. Feb 20, 2008 #7
    The bottom ought to be 3mA and the mid ought to be 65mA.
    Last edited: Feb 20, 2008
  9. Feb 20, 2008 #8
    I've been learning about that since high school :redface: lol.

    Thank you for the help, did you get the other two values by plugging the value you got for the middle current into the previous equations?

    Also, you said that substitution may take a while...do you have another technique for solving the three linear equations? This whole process with currents and circuits really takes me a while to do so anything to speed me up a little will definitely help on exams and such..

    Thank you again for the help.
  10. Feb 20, 2008 #9
    I actually used a technique called mesh current analysis which went a little faster. But yeah, the idea is that once you get one of the answers you can just put it into the other answers. A quick way to solve systems of equations is with linear algebra techniques, Gauss Elimination probably being the best. To learn them requires a couple pages of writing I'd rather not do, but you could probably look it up.
  11. Feb 20, 2008 #10
    Gaussian Elimination is a good one. You can also use Kramer's rule, or if you have a calculator that can do matrix, that would be the easiest method.
  12. Feb 20, 2008 #11
    What website is this on
  13. Feb 20, 2008 #12
    Thanks, I'll look that up. And ya I do have a calculator so maybe I'll try to figure out the matrices, though I haven't dealt with them for a while...

    Also, my teacher decided to extend it a couple days so apparently I wasn't the only one procrastinating :biggrin:

    The original question? It's on webassign.
  14. Feb 20, 2008 #13


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    Science Advisor

    I think you mean Cramer's Rule. :wink:


  15. Feb 20, 2008 #14
    What's the deal with Cramer? /Seinfeld voice

    Do the kids these days know about Seinfeld?

    Anyway, a site about mesh current:

    You should also learn about node voltage if you have a couple more weeks of circuits:

    A site that shows Guassian Elimination (my favorite) though Cramer's rule is probably better for straight numerics:

    Linear algebra is about solving two things Ax=b and Av=Ev, and Gauss elimination gives you the big tool to solve the first guy.
  16. Feb 16, 2009 #15
    I actually have the same question i'm working on and went through the steps to solve this problem. I got the top branch right and the bottom keeps turning up to be wrong. Was 3 right for you?
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