# Current Electricity problem

1. Nov 12, 2006

### Couperin

The question is:

A battery drives a current of 3.0 A round a circuit consisting of two 2.0 ohms resistors in parallel. When these resistors are connected in series, the current changes to 1.2A. Calculate:

a) the e.m.f of the battery

and

b) the internal resistance of the battery.

Here's how far I've got:

I calculated the effective parallel resistance to 1.0 ohms by doing

$$R=(\frac{1}{2}+\frac{1}{2})^-1$$

The potential difference in each case is 3V and 4.8V

So I've put them in a simultaneous equation:

Let E be the e.m.f, and r be the internal resistance:

E = 3r + 3
E = 1.2r + 4.8

Which can be written as

3r + 3 = 1.2r + 4.8

But I can't figure out how to isolate the r values!

Any help would be greatly appreciated.

Last edited: Nov 12, 2006
2. Nov 12, 2006

### rsk

Once you get that far it's just a question of getting your rs on one side and your numbers on the toher....but I don't get teh same equations.

Apologies - I've mixed up my series and paralle - it should be exactly as you said.

3r + 3 = 1.2r + 4.8

So

3r - 1.2r = 4.8 - 3

then subs back to find E

Last edited: Nov 12, 2006
3. Nov 12, 2006

### Couperin

Aha! Why on earth didn't I think of that... :(

Anyway, I got the answer now! Thanks!

3r - 1.2r = 4.8 -3
1.8r = 1.8
r=1

So EMF = 6 and internal resistance = 1

Thanks!