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Loop Equation in Circuit

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data

    What is the magnitude i in the figure below, if all resistances are 4 ohms, and all batteries are ideal with an emf of 12.4 V?

    http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c27/pict_27_70.gif

    2. Relevant equations

    3. The attempt at a solution

    I realize I need to set up a loop equation. I decided to start in the lower left hand corner, turn left at the first "branch", go through the parallel resistors, through the unknown current, then turn left again, through the resistor, then left again, through the battery, and back to my starting point. I realize this completely disregards the rest of the circuit, but I do need a simple loop equation because as soon as I incorporate additional currents, I have too many unknowns and it can't be solved. My answer of 0.886 A through the given resistor is incorrect, and I have checked my loop equation multiple times.

    The picture looks daunting, but I feel like it's very do-able; I just need a loop that will work. Can anyone point me in the right direction?
     
  2. jcsd
  3. Oct 5, 2009 #2
    A single loop equation is no use. You don't know if the same current is going through all the resistances in the loop. You need to replace all parallel and series resistances you can find. Some of the points of the circuit can be shown to have the same potential. For example the top left and bottom left corners. You can connect those with a wire.

    If you ever get 2 of the voltage sources in parallel, you can replace them with a single voltage source of the same voltage if plus is connected to plus and minus to minus.
    If they short each other something must have gone wrong.
     
  4. Oct 5, 2009 #3
    It turns out that there is indeed a single loop that you can use. If your loop only follows wires and voltage sources, except for the chain of 4 resistors on the bottom left, than you can work out the voltage across those resistors, because the voltage sources always have 12.4 V across them, no matter what.

    make a copy of the image, and erase all of the resistors, except for the 4 in the bottom left corner. you'll have one loop left.
     
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