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

Explanation

Let K be the potential gradient of the given potentiometer wire.
When potentiometer is connected between A and B, then `epsilon_(1) = K xx 300`
When potentiometer is connected between A and C, then `epsilon_(1) - epsilon_(2) = K xx 100`
`:. (epsilon_(1)-epsilon_(2))/epsilon_(1) = 100/300 = 1/3 or 1 - epsilon_(2)/epsilon_(1) = 1/3`
or `epsilon_(2)/epsilon_(1) = 1 - 1/3 = 2/3 or epsilon_(1)/epsilon_(2)=3/2`