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Kirchhoff's Rules

  1. Mar 1, 2009 #1
    1. The problem statement, all variables and given/known data
    (a) Determine the currents I1, I2, I3 in the figure. Assume the internal resistance of each batter is r = 1.0 olms

    (b) What is the terminal voltage of the 6.0 V battery?


    I hope you can read my handwriting.

    2. Relevant equations
    V = IR

    3. The attempt at a solution
    I1 + I2 = I3

    I don't understand kirchhoff's rules and how the internal resistance of each batter means.
  2. jcsd
  3. Mar 1, 2009 #2


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    First the way it's drawn it's I2 + I3 = I1

    All it means is that charge has to go somewhere. ∑ I_in = ∑ I_out

    Direction matters.

    The little r's are the internal resistances of the batteries. When they ask for the terminal V of the 6v battery they want you to include the I3*r change in voltage from 6v that you would measure in real life.

    To solve you need to write out the loop equations for the 2 inner loops. Those along with your current conservation give you 3 equations and you have 3 unknown currents. So then you just solve.
  4. Mar 1, 2009 #3

    Okay, can you tell me if my loops are right?

    Top: V1 + V2 = I1r + I1R1 + I2r + I2R2 + I1R3

    Bottom: V3 + I2R2 + I2r = I1r + I3r + I3R5 + V2 + I3R4

    Big: V1 + V3 = I1r + I1R1 + I3R4 + I3r + I3R5 + I1R4
  5. Mar 1, 2009 #4


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    First of all forget the Big loop. The 2 inner loops and the conservation of current is all you need.

    The top and bottom look OK just glancing at them.

    Now use

    I1 = I2 + I3 and you can get it down to 2 equations and 2 unknowns pretty pronto.

    Then solve.
  6. Mar 1, 2009 #5
    Okay, so I figure out the currents. I'm still pretty confused with finding terminal voltage.

    Are you saying that Terminal Voltage = I3r?
  7. Mar 1, 2009 #6


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    No. Terminal voltage is 6v + I3*r. (This is if that is the right direction for I3.)
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