Question

In the figure a long uniform potentiometer wire AB is having a constant potential gradient along its length. The null points for the two primary cells of emfs `epsilon_(1) " and " epsilon_(2)` connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A. Find (i) `epsilon_(1)//epsilon_(2)` and (ii) position of null point for the cell `epsilon_(1)`.
How is the sensitivity of a potentiometer increased ?

Answer 1

(i) Let K be the potential gradient in volt/cm of the potentimeter wire. Then
`epsilon_(1) - epsilon_(2) = 120 K and epsilon_(1)+epsilon_(2) = 300K`
`:. epsilon_(1)= 210 K and epsilon_(2) =90K`
Hence, `epsilon_(1)/epsilon_(2) = (210K)/(90K) = 7/3`
(ii) If cell of emf `epsilon_(1)` is balanced across length `l_(1)` on potentiometer wire, then
`epsilon_(1) = 210 K= l_(1) K or l_(1)=210cm`
The sensitivity of a potentiometer wire can be increased by decreasing its potential gradient.The same can be achieved either by increasing the length of potentiometer wire or by increasing the resistance in series with the driving cell.