Two cells of emfs `E_(1)` and `E_(2)(E_(1)gtE_(2))` are connected as shown in figure.

When a potentiometer is connected between A and B, the balancing length of the potentiometer wire is 300 cm. On connection the same potentiometer between A and C, the balancing length is 100 cm. The ratio `(E_(1))/(E_(2))` is

Options

A. `3 : 1`
B. `1 : 3`
C. `2 : 3`
D. `3 : 2`

Answer

D

Explanation

`E_(1) prop 300`, `E_(1) - E_(2) prop 100`
`(E_(1))/(E_(1) - E_(2)) = 3` or `E_(1) = 3E_(1) - 3E_(2)`
or `3E_(2) = 2E_(1)` or `(E_(1))/(E_(2)) = (3)/(2)`