The length of a wire of a potentiometer is 100 cm, and the e.m.f. of its standard cell is E volt. It is employed to measure the e.m.f. of a battery whose internal resistance is `0.5 Omega`. If the balance point is obtained at I = 30 cm from the positive end, the e.m.f. of the battery is .
where i is the current in the potentiometer wire.

Options

A. `(30 E)/(100)`
B. `(30 E)/(100.5)`
C. `(30 E)/((100 - 0.5))`
D. `(30 (E - 0.5 i))/(100)`, where `i` is the current in the potentiometer

Answer

A

Explanation

(a) From the principle of potentiometer `V prop l`
`implies (V)/(E) = (l)/(L)`, Where `V = ` emf of battery, `E = ` emf of standard cell, `L`= Length of potentiometer wire
`V = (El)/(L) = (30 E)/(100)`