The emf of a cell is E, and its its internal resistance is 1Ω . A resistance of 4Ω is joined to battery in parallel. This is connected in secondary circuit of poetntiometer. The balancing length is 160cm. If 1V cell balances for 100cm of potentiometer wire, the emf of cell E is

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1 V

3 V

2 V

4 V

answer is C.

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Detailed Solution

1v =100cm then 160cm corresponds to 1.6V v=1.6=E-ir and i=E/5