Using Kirchhoff’s Laws calculate the following:

a) The three branch currents I_{1}, I_{2}, and I_{L}

b) The power dissipated in the load resistor

Attachment 27353

R1 = 12

R2 = 14

RL = 7

Explain how you got there if possible. Thanks!

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- Mar 4th 2013, 07:23 AMjamietreeHELP with Kirchhof's law equation
Using Kirchhoff’s Laws calculate the following:

a) The three branch currents I_{1}, I_{2}, and I_{L}

b) The power dissipated in the load resistor

Attachment 27353

R1 = 12

R2 = 14

RL = 7

Explain how you got there if possible. Thanks! - Mar 4th 2013, 02:00 PMtopsquarkRe: 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:

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