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Kirchoff's law

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

    I have to find the current through each resistor in this: http://i7.photobucket.com/albums/y297/SpaceTrekkie/Physics/Untitled.jpg circuit. All the resistors labeled R1 are 100ohms, the one labeled R2 is 1000 ohms and E1 = 10V E2= 6

    3. The attempt at a solution

    I figured out that there are two loops, one in the upper portion and one below, but I am not sure how to set up the equations because of the E2 (6V) battery is in the parallel portion.
    I don't need the numerical answer, I just want to see how to set it up.
  2. jcsd
  3. Apr 1, 2008 #2
    Well it's fine that the battery is there, in fact that is great because you automatically know the potential difference at the battery.
  4. Apr 2, 2008 #3
    But how does it affect the current comming into that parallel part when the current "splits" at the junction? Since it is flowing in the opposite direction.
  5. Apr 2, 2008 #4
    I don't understand what you are asking. Current can go through a battery. How is this any different than if you were to have one battery feeding into a parallel set of resistance?

    You may be thinking about this too hard. For now simply get your equations by summing the voltages around the loop. Then find your currents. With more time and practice your questions will start to be answered.
  6. Apr 2, 2008 #5
    For the loop rule, just follow a line in a loop, if something falls onto the line include it in your equations

    For the junction rule, the current has to be conserved. Current "flowing" in the direction opposite the battery could simple be interpreted as charging the battery. (Although, doesn't really make sense why one would want to charge a battery with a battery)
  7. Apr 2, 2008 #6
    Okay, thanks all, I think I get it now. Basically I go around the loop starting at the positve side of the 10V batter, and just go around like normal, calculating the potential drops at each resistor, then when I get to the 6V batter, since that is oriented the opposite way subtract off the 6V from whatever the potential is when it gets there, I dont have to do anything weird with the direction or currents or anything.
  8. Apr 2, 2008 #7
    Yes, that's correct.
  9. Apr 2, 2008 #8
    If I were you I would use superposition principle. Calculate the current from each source separetly (the other one is shorted) and then simply add the currents.
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