Two cells of emfs `E_(1)` and `(E_(2) (E_(1) gt E_(2))` are connected as shows in Fig. `6.45`.

When a potentiometer is connected between `A` and `B`, the balancing length of the potentiometer wire is `300 cm`. On connecting 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

When potentiometer is connected between A and B, then it measures only `E_(1)` and when connected between A and C, then it measures `E_(1)-E_(2)`
`therefore(E_(1))/(E_(1)-E_(2))=(l_(1))/(l_(2))`
or `(E_(1)-E_(2))/(E_(1))=(l_(2))/(l_(1))` or `1-(E_(2))/(E_(1))=(100)/(300)`
`implies(E_(2))/(E_(1))=1-(1)/(3)implies(E_(2))/(E_(1))=(2)/(3)`
`implies(E_(1))/(E_(2))=(3)/(2)`