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 - 0.5i))/(100)`

Answer

A

Explanation

Using the principle of potentiometer, `V prop l.` So
`(V)/(E) = (l)/(L)` or `V = (l)/(L) E = (30)/(100) E = (30E)/(100)`