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Thread: HELP with Kirchhof's law equation

  1. #1
    Mar 2013

    HELP with Kirchhof's law equation

    Using Kirchhoff’s Laws calculate the following:

    a) The three branch currents I1, I2, and IL
    b) The power dissipated in the load resistor
    HELP with Kirchhof's law equation-photo-04-03-2013-03-19-51-pm.jpg

    R1 = 12
    R2 = 14
    RL = 7

    Explain how you got there if possible. Thanks!
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  2. #2
    Forum Admin topsquark's Avatar
    Jan 2006
    Wellsville, NY

    Re: HELP with Kirchhof's law equation

    For starters the only time you have a problem like this without capacitors (or inductors) you will not have a differential equation.

    So you have the junction rule: $\displaystyle I_1 + I_L = I_2$

    You didn't put a unit on the 19 (tsk tsk). I'm going to assume that it's the potential difference across the source. Now we need to talk about potentials. There is a loop around the whole thing and I'm going to assume the current is clockwise. (So we have a loop through the source and around the circuit going through the 12 (Ohm) and the 14 (Ohm) resistors.) You will need one more loop: Let's make it go across the bridge...that is a clockwise current going through the 12 (Ohm) and 7 (Ohm) resistors. (What is the potential across the bridge?) Note that your 7 (Ohm) resistor has the current going the "wrong" way, What do you do with the potential across the 7 (Ohm) resistor in this case?

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