Calculating Resistance in a Circuit

[Cornfused]

Homework Statement

A resistor with resistance R is connected to a battery that has emf 12.0 V and internal resistance r=0.40 ohms. For what two values of R will the power dissipated in the resistor be 80.0 W?

P=80 W
r=0.40 ohms
EMF=12V
Find R

Homework Equations

P=VI, = I^2R = V^2/R
V= EMF - Ir
V=IR

The Attempt at a Solution

I need to find resistance R of the resistor. But I can't seem to find the current I, which I need to find V, so that I can find R. I have tried rearranging the equations all manner of ways and can't seem to figure it out.

The other issue is how the question asks to find the two values of R. I assume that for there to be two values that somehow R has to be squared so that you have the same number, just positive and negative. That is my best guess at the moment.

Thanks for the help.

 

[Cornfused]

I was reworking it and wanted to know if this was valid.

P=EMF*I - I^2r

I used I^2r = P to get

P = EMF*I - P ----> P+P = EMF*I ----> 2P/EMF = I

I know P = I^2R but wasn't sure if I could do the same with P = I^2r

 

[gneill]

First find an expression for the current in the circuit. It's a series circuit consisting of the 12V EMF, internal resistance 0.40 Ohms, and resistance R, so what is the current?

Next plug that expression for the current into your expression for dissipated power, P = I²*R. Solve for R.