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+1 vote
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in Physics by (58.2k points)

Describe briefly, with the help of a circuit diagram, how a potentiometer is used to determine the internal resistance of a cell.

1 Answer

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by (66.1k points)
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Best answer

Initially key K2 is off

Then at balancing length l1

e = Kl1 ------------------ (i)

Now key K2 is made on

V = Kl2 --------------- (ii)

So, ε/V = l1/l2 ......(iii)

where internal resistance r is

r =  (ε/V-1)R

⇒ r = (l1/l1 -1)R.

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