znx
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A battery of e.m.f 6 V is connected across a resistor of resistance 12 ohms. The potential difference across the resistor is 4.5 V.

What is the internal resistance of the battery?

A) 3 ohms

B) 4 ohms

C) 9 ohms

D) 16 ohms

I worked out the current across the resistor which is the same across the battery, by doing I = 4.5/12 = 3/8

As the resistor has an p.d of 4.5 the battery must have a p.d of 1.5, as sum of e.m.f = sum of p.d

So I used E = V + Ir, with the values 6 = 1.5 + 3/8r to get r as 12 ohms, which isn't an answer.
So I then used E = I(R+r), with the values 6 = 3/8(12+r) to get r as 4, which is B)

My question is why has the 1st method not worked and did I even get the right answer using the 2nd method?
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178865
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(Original post by znx)
As the resistor has an p.d of 4.5 the battery must have a p.d of 1.5, as sum of e.m.f = sum of p.d

So I used E = V + Ir, with the values 6 = 1.5 + 3/8r to get r as 12 ohms, which isn't an answer.
The "V" in "E = V + Ir" refers to the potential difference of the external circuit, which in this case is your resistor which has a potential difference of 4.5V.
If you use V = 4.5V, you'll get r = 4 ohms.
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znx
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(Original post by 178865)
The "V" in "E = V + Ir" refers to the potential difference of the external circuit, which in this case is your resistor which has a potential difference of 4.5V.
If you use V = 4.5V, you'll get r = 4 ohms.
oh, thank you
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