In the Fig., a long uniform potentiometer wire AB is having a constant potential gradient along its length. The null points for the two primary cells of emf `epsi_1` and ` epsi_2` connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A. Find `(i) epsi_1//epsi_2 ` and (ii) position of null point for the cell `epsi_1`. How is the sensitivity of a potentiometer increased ?

Explanation

(i) If the constant potential gradient be k volt/cm then as per question
`epsi_1 + epsi_2 = k xx 300` ....(i)
`epsi_1 - epsi_2 = k xx 120` ....(ii)
`rArr (epsi_1 + epsi_2)/(epsi_1 - epsi_2) = 300/120 = 5/2 rArr epsi_1/epsi_2 = 7/3`
(ii) If for the cell `epsi_1` the null point be at a distance of 1 from end A, then
`epsi_1 = kl`
Dividing (i) by (iii), we have
`(epsi_1 + epsi_2)/(epsi_1) = 1 + epsi_2/epsi_1 = (300)/(l) " or " 1 + 3/7 = (300)/(l)`
` rArr l = (300 xx 7)/(10) = 210 cm`
The sensitivity of a potentiometer can be increased by (i) increasing the length of the potentiometer wire, and (ii) introducing a high resistance in series of potentiometer.