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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

[tex]R=(\frac{1}{2}+\frac{1}{2})^-1[/tex]

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.

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# Homework Help: Current Electricity problem

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