Electrical circuits

1. Nov 20, 2008

dimpledur

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

Suppose that a battery of emf E and internal resistance r is being recharged: another emf sends a current I through the battery. Why rate is electrical energy being converted into chemical energy?

2. Relevant equations

P=EI-I^2*r
V=E-Ir

I know that since it is being recharged, the current flows in the reverse direction (positive terminal --> negative terminal), however, I'm not sure how to go about doing this.

3. The attempt at a solution

2. Nov 20, 2008

Feldoh

You can look at power as the rate at which energy is transferred so that is exactly what the question is asking for. You know the voltage the current and the resistance, and you have an equation..

3. Nov 20, 2008

dimpledur

Since it is going in opposite direction, would it be:
P=I^2r-EI ???

4. Nov 20, 2008

Feldoh

Regardless of which way a current is going it's still going to dissipate energy into heat energy correct? However we're still increasing the energy of the actual battery so how can it be negative? In your equation for have for just the battery P = -EI, you're not dissipating energy from doing this you're gaining some energy!

5. Nov 20, 2008

dimpledur

Okay, thanks. So really, the question is only asking for the equation that is already giving me in my text book?

It doesn't seem right for the answer to be P=EI-I^2*r

6. Nov 20, 2008

dimpledur

I don't get how that can be the answer when the second part of the question asks:

What is the power supplied by the recharging circuit to the battery?

7. Nov 20, 2008

dimpledur

Wait, I don't think the internal resistance of the first battery matters when it is being recharged, does it?

So we have a resistor then the positive terminal and then the negative terminal. This battery is then attached to another emf (it doesn't state that this battery has an internal resistance, so we assume its an ideal battery?). So wouldn't the answer simply be:

P = EI because the internal resistance on the one battery has no effect?