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 :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)`

A potentiometer wire of Length `L` and a resistance `r` are connected

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 balanced at a length `l` of the potentiometer wire. The e.m.f. `E` will be given by :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» Correct Answer - c 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)`

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