# Maximum power from a battery with source resistance

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1. Jan 6, 2016

### JordanHood

1. The problem statement, all variables and given/known data
A 5V battery rated at 1800mAh has a soucre resistance of 10 Ohms. What is the maximum power that can be extracted from the battery?

2. Relevant equations

3. The attempt at a solution
I understand that had the battery not had any source resistance then the power out would have been equal to Energy x Time giving us P=VIT, but i do no understand what to do when there is a source resistance. Do I calculate another current that is passing through the battery? using V=IR to get a current of 0.5A? If so then what does the 1800mAh mean? I have then tried putting this current into P=I^2R but this means im totally ignoring one of the values in the question which means I cant be right

EDIT
Apologies I didnt realise this was a multiple choice question, the possible answers are
a) 625mW
b) 1.25mW
c) 2.5 mJ
d) 11.664 MJ

Last edited: Jan 6, 2016
2. Jan 6, 2016

### cnh1995

I don't see the significance of 1800mAh in this problem. Do you know the maximum power transfer theorem in electrical circuits?

3. Jan 6, 2016

### JordanHood

Is this when the source resistance and load resistance are equal?

4. Jan 6, 2016

### cnh1995

Yes. But I just realized that the theorem will give the value of load resistance for maximum power dissipation in the load, while the question asks about the maximum power in the entire circuit.

5. Jan 6, 2016

### cnh1995

I believe 1800mAh means the battery can supply 1800mA current for 1hr.
Edit: I think the proper meaning of this is that the battery can supply 1mA current for 1800 hrs.

Last edited: Jan 6, 2016
6. Jan 6, 2016

### JordanHood

So do I need to calculate the current per second?

7. Jan 6, 2016

### cnh1995

No. Seeing the options, I think this problem does belong to the maximum power transfer theorem. After all, it's the load that extracts power from the battery. So when we say 'power extracted from the battery', it means the power consumed by the load from the battery.

Last edited: Jan 6, 2016
8. Jan 6, 2016

### JordanHood

Ok so I set the internal resistance of the battery to 10 Ohms and work from there?

9. Jan 6, 2016

### cnh1995

I think so.

10. Jan 6, 2016

### JordanHood

That gives me 625mW, which is one of the answers, so assuming that is right. Bit concerned about not using the 1800mAh at all though, possibly there to confuse you?

11. Jan 6, 2016

### Staff: Mentor

No. Energy delivered = power x time

But you are not asked about energy, the question involves maximizing output power.

12. Jan 6, 2016

### cnh1995

I think so. I haven't seen mAh rating being used for calculating maximum power so far.