The length of a wire of 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 l = 30 cm from the positive end, the e.m.f. of the battery is:

Options

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

Answer

D

Explanation

In a potentiometer the fall of potential across any portion of the wire is directly proportional to the length of that portion,
i.e., `VproplrArrV/E=1/L`
where V = e.m.f of battery, and L = length of the potentiometer wire,
E = e.m.f of standard cell and l = balance point length.
`thereforeV=(El)/L=(Exx30)/100=(30E)/100`