A potentiometer wire of Length `L` and a resistance `r` are connected in series with a battery of e.m.f. `E_(0)` and a resistance `r_(1)`. An unknown e.m.f. `E` is balanced at a length `l` of the potentiometer wire. The e.m.f. `E` will be given by :

Options

A. `(E_(0)l)/(L)`
B. `(LE_(0)r)/((r+r_(1))l)`
C. `(LE_(0)r)/(lr_(1))`
D. `(E_(0)r)/((r+r_(1))).(l)/(L)`

Answer

D

Explanation

The current through the potentiometer wire is
`I=(E_(0))/((r+r_(1))`
and the potential difference across the wire is
`V=Ir=(E_(0)r)/((r+r_(1)))`
The potential gradient along the potentiometer wire is
`k=(V)/(L)=(E_(0)r)/((r+r_(1))L)`
As the unknown e.m.f. E is balanced against length l of the potentiometer wire,
`therefore E=kl=(E_(0)r)/((r+r_(1)))(l)/(L)`