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

Answer

C

Explanation

Current through the potentionmeter wire is
`1 = (E_(0))/(r+r_(1))`
Potential difference across potentiometer wire
`V = Ir = (E_(0)r)/(r+r_(1))`
Potential gradient across potentiometer wire
`K =(V)/(L)=((E_(0)r)/(r+r_(1)))(1)/(L)`
Now `E = Kl = ((E_(0)r)/(r+r_(1)))(l)/(L)`