1.

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 :

Answer»

`(LE_(0)r)/((r+r_(1))l)`
`(LE_(0)r)/(lr_(1))`
`(E_(0) RL)/((r+ r_(1))L)`
`(E_(0)l)/(L)`

Solution :`(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)`


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