# Power disspated in a circuit

1. May 1, 2012

### Pi Face

http://imgur.com/WO1zq

I tried the loop rule method but I couldn't get it to work since I had 2 unknowns (currrent and R1/R2)
I know that for resistors in series, the current through both are the same. Also, R1 should equal R2, beecase they need to dissipiate the same power. This means the equivalent resistance of R1 and R2 is just half of either one (R1/2). So, the total equivalent resistance is 2k+R_1/2.
I also know that the voltage across two resistors in parallel is the same, so V across R1 and R2 are equal, meaning the current through both are equal as well. Using the junction rule, if I1 is the current through the first resistor Rb, then the current through R1 and R2 is I_1/2 for each.

I think I know a fair amount about the circuit but I'm not sure how to combine all of it into something sensible...

2. May 2, 2012

### ehild

You have found out a lot, just go ahead! You know that half of the current flowing through the 2kΩ resistor flows through R1. How is the power related to the resistance and current?

ehild

3. May 2, 2012

### Pi Face

P=I^2*R right?
but R is still R1 or R2, and I don't know the current numerically. I know it's half of I1, but what's I1? The only way I can think of finding it is I=V/Req, where V is 10 and Req is 2k+R_1/2, which for R1 I don't know either and leads me in a circle

4. May 2, 2012

### ehild

It is right. If I is the current through the 2kΩ resistor, then P=2000I2.
It is the same as the power on the unknown R1 with I/2 current: P=R1 (I/2)2, is not it? Compare the equations.

ehild

5. May 2, 2012

### Pi Face

ahh I see, I don't actually need to find the value of I

P=2000I^2=R*(I/2)^2=R*(1/4)I^2

meaning R must be 4 times 2k, or 8kΩ. And both R1 and R2 are equal so 8k for both, which I think is what the solutions said

cool, thanks!