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Current in 10 ohm Resistor

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the current in the 10.0 ohm resistor in the drawing (V1 = 24.0 V and R1 = 15.0 Ohm)

    Diagram: 20_82alt.gif

    2. Relevant equations
    Kirchoffs loop rule

    3. The attempt at a solution

    well do i have to solve for both parts of the circuit if i just want to find the current at the 10.0 ohm resistor or can i just do one side?

    could i do like 10 - R1*I1 - 10*(I1-I2) + 10 = 0?

    well actually then i need to find I2 right? or how can i solve it, like am i just looking for I1-I2? or what im just confused at this point
  2. jcsd
  3. Feb 13, 2012 #2


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    You pretty much answered your own question. Using only one loop leaves you with two unknowns in one equation. You also need to do the other loop.
  4. Feb 13, 2012 #3
    Yeah, you need another loop. Since when do rappers take physics? :p
  5. Feb 13, 2012 #4
    how do you set up the 2nd loop?

    V1 + 10 - 10*(I2-I1)? - 5.0*I2 ??

    or is the bolded part +10*(I2-I1) or is it I2 + I1 ? or what? and does the rest of that equation look right?
  6. Feb 13, 2012 #5
    Yeah, you've got it right. Solve the simultaneous eqns.
  7. Feb 13, 2012 #6
    well for the first loop the +10 from the first voltage would cancel out with the one to the right of it correct?

    and am i looking for I2-I1 or I1-I2 when looking for the current in the 10 ohm resistor?
  8. Feb 13, 2012 #7
    Alright can someone double check this?

    so i did for the first loop
    10 - 15I1 - 10I1 + 10I2 - 10 = 0
    25I1 = 10I2

    then 2nd loop did
    24 + 10 - 10(I2-I1) - 5I2 = 0
    simplified it to => 34 = 15I2-10I1 and then to => 27.5*I1 = 34

    plugged I1 in for I2 from the first equation and ended up with
    I1 = 1.236
    I2 = 3.09

    Plugged it in now to 10 ohms * (I2 - I1) = 10*(3.09-1.236) = 18.55 amps

    is that the right answer? and i only did I2-I1 because I2 was bigger but im not sure it should matter? i dont know lol im confused on this part

    edit: i def. did something wrong lol it should NOT be that high
    Last edited: Feb 13, 2012
  9. Feb 13, 2012 #8


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    If you're looking for the current through the 10Ω resistor, why did you multiply current by resistance? --- That gives the voltage drop across the 10Ω resistor.
  10. Feb 13, 2012 #9
    wait oh yeah, but the Current through that point I2-I1 or I1-I2?
  11. Feb 13, 2012 #10


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    It's possible to do it either way.

    Which way makes sense to you and why?
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