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

Answer

C

Explanation

`(E_(0) rl)/((r+ r_(1))L)`
Pote ntial gradient `sigma = (Ir)/(L)`
But I `= (E_(0))/(r + r_(1))`
`therefore sigma = (E_(0)r)/((r + r_(1))L)`
`therefore` emf E = `sigma l = (E_(0)rl)/((r+ r_(1))L)`