Thread: HELP with Kirchhof's law equation

Using Kirchhoff’s Laws calculate the following:


a) The three branch currents I₁, I₂, and I_L
b) The power dissipated in the load resistor


R1 = 12
R2 = 14
RL = 7

Explain how you got there if possible. Thanks!

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:

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?

-Dan