Question

The resistance of a potentiometer wire is 92 and
its length is 9 m. An unknown resistance and 2V
cell is connected in series with it. What is the
value of resistance so that the potential gradient
will be 1uV/mm?​

Answer 1

Answer:

Length of the potentiometer wire, I=9 m

Resistance of the potentiometer wire, R=92 Ω

Emf of the accumulator connected in series with the wire, E=2V

A resistance box in series is connected to the potentiometer wire.

Required potential gradient, k=1*10^-6 V/mm=1 uV/mm

Potential drop along the potentiometer wire,

V=k.I= 1*10^-6*9*1000=0.009

Therefore, current through the potentiometer wire,

I= V/R = 0.009/92 = 97.8 uA

Let R'' be the resistance of the resistance box then,

Using the relation

I= E/R - V/R' = 97.8uA

2/92 - 0.009/R' = 97.8 u

0.0217391304 - 0.009/R' = 97.8 u

21739.13 u - 97.8 u = 0.009/R'

21,641.3304 u

R ′= 4.15 ohm