Two primary cells of emfs `E_(1)` and `E_(2)` are connected to the potentiometer wire AB as shown in the figure . If the balancing lengths for the two combinations of the cells are 250 cm and 400 cm , find the ratio of `E_(1)` and `E_(2)` .

Explanation

Here , `E_(1) - E_(2) = K xx 250 " " ... (i) `
`E_(1) + E_(2) = K xx 400 = 400 K " " ... (ii)`
`therefore` K is potential gradient
On adding the equations (i) and (ii)
`2 E_(1) = 650 K`
`E_(1) = 325 K`
On subtracting equations (i) from (ii) we get
`2 E_(2) = 150 K`
`E_(2) = 75K`
Hence , `" " (E_(1))/(E_(2)) = (325K)/(75K) = (13)/(3) = 4.33`.