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Question

A cell balances against a length of 200cm on a potentiometer wire, when it is shunted by a resistance of 8Ω. The balancing length reduces by 40cm, when it is shunted by a resistance of 4Ω. Calculate the balancing length when the cell is in open circuit. Also calculate the internal resistance of the cell.

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Solution

Let balancing length be l1 and internal resistance be 'r'.
r=R(l1l2l2)
r=8(l1200200) ....(1)
r=4(l1160160) ....(2)
From equations (1) and (2)
8(l1200200)=4(l1160160)
l120010=l116016
16l1=3200+10l11600
l1=266.66cm
r=8(266.66200200)
r=2.66Ω

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