Question

A potentiometer wire of length L and a resistance r are
connected in series with a battery of e.m.f. E₀ and a resistance
r₁. 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: [2015 RS]
(a)E₀r/(r + r₁) . l/L
(b) E₀I/L
(c) LE₀r/(r + r₁)ˡ(d) LE₀r/Ir₁

Answer 1

ANSWER

Current in the circuit having E

0

is: i=

r+r

1

E

0

Voltage dropped in potentiometer wire of length L is: ir=

r+r

1

E

0

r

Since the balanced point is obtained at length l of the potentiometer wire, the emf of the cell E has to match the voltage dropped in length l of the wire.

∴E=

r+r

1

E

0

r

×(

L

l

)