- #1

Couperin

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The question is:

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.

**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.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.

Last edited: