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. `(LE_(0) r)/(Ir_(1))`
C. `(E_(0) r)/((r + r_(1))) (I)/(L)`
D. `(E_(0) I)/(L)`

Answer

C

Explanation

(c ) Current in potentiometer wire is
`I = (E_(0))/(r + r_(1))`
Volt drop across potentiometer wire
`V_(0) = (E_(0) r)/((r + r_(1)))`
So, `E = KI = (V_(0) I)/(L)`
`E = (E_(0) rI)/((r + r_(1)) L)`