# Finding the maximum power delivered from a baterry

1. Jan 17, 2009

### pentazoid

1. The problem statement, all variables and given/known data

A battery of emf [insert symbol for emf here] and internal resistance is hooked up to a variable "load" resistance R. If you want to deliever the maximum possible power to the load, what resistance should you choose?(You can't change the emf and r, of course).

2. Relevant equations

P=I^2*R=V^2/R
Possibly V=[insert symbol for emf here]-Ir

3. The attempt at a solution

I think what I should do is differentiate the Power with respect to the variable load R and sint dP/dR =0

P=V^2/R=([insert symbol for emf here]-Ir)^2/R

2. Jan 17, 2009

### tiny-tim

Hi pentazoid!

(what's "sint"? )
Looks good!

3. Jan 17, 2009

### pentazoid

4. Jan 17, 2009

### nrqed

This is correct but the current I depends on R as well so this not yet in a useful form.

I think it's simpler to use $$P = R I^2$$ with $$I = {\cal E} /(R+r)$$. Now differentiate wrt R, set to zero and solve for R.

5. Nov 15, 2011

### Marioqwe

I'm trying to solve this problem but I'm having a hard time figuring out why I = (emf)/(r+R). Isn't V = emf when the conductivity is infinite? How is it infinite (or very big) in this case if the resistance is r.

6. Nov 15, 2011

### cepheid

Staff Emeritus
To be honest, I don't know what you're talking about.

Kirchoff's voltage law (KVL) says that the sum of the voltages around a closed loop in a circuit has to equal zero (this is just the conservation of energy, when you think about it). In other words, the voltage drop across the two resistors has to be equal to the voltage across the battery terminals (which is just the emf). Everything is in series, so current through the circuit is the total voltage over the total resistance, hence I = ε/(r+R).