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The circuit shown hereis used to compare...

The circuit shown hereis used to compare the EMFs of two cells `E_(1)` and `E_(2)(E_(1)gtE_(2))`. The null point is at C when the galvanometer is connected to `E_(1)`. When the galvanometer is connected to `E_(2)`, the null point will be

A

To the left of C

B

To the right of C

C

At C itself

D

Nowhere on AB

Text Solution

Verified by Experts

The correct Answer is:
B
We know that, in potentiometer
`(E_(1))/(l_(1)) =(E_(2))/(l_(2))= ` Constant
In Case 1,
Let `AC=l_(1) " "` (When galvonometer is connected to `E_(1)`)
In Case 2, `E_(2)` is connected with galvonometer
`:.` By equation (i),
`E_(1)l_(2)=E_(2)l_(1)=` Constant
Given, `E_(2) gt E_(1)`
`:.l_(2) gt l_(1)`
`:.l_(2) gt AC`
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