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A potentiometer has a wire of length 100 cm and the emf of the standard cell used is E volt. It is employed to measure the emf of a battery whose internal resistance is $0.5 Ω,$ If the balance length obtained is 30cm from the positive end, the emf of the battery is..

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$\frac{30 E}{99.5}$

$\frac{30 (E - 0.5 i)}{100};$ (i =current across potentiometer wire)

$\frac{30 E}{100}$

$\frac{30 E}{100.5}$

answer is D.

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Detailed Solution

Let V be the potential across balance point from one end of wire.
Hence according to the principle of potentiometer $V \propto l$

Also if a cell of emf E is employed in the circuit between the ends of potentiometer wire of length L then $E \propto L$

Therefore $\frac{V}{E} = \frac{l}{L}$

$V = \frac{l}{L} E = \frac{30}{100} E = \frac{30 E}{100}$

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