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A potentiometer PQ is set up to comapre ...

A potentiometer PQ is set up to comapre two resistances as shown in the figure. The ammeter A in the circuit reads 1.0 A when two way key `K_(3)` is open. The balance point is at a length `l_(2)` cm from P when two way key `K_(3)` is plugged in between 2 and 1, while the balance point is at a length `l_(2)` cm from P when key `K_(3)` is plugged in between 3 and 1. The ratio of the two resistance `(R_(1))/(R_(2))`, is found to be-

A

`(l_(1))/(l_(1)+l_(2))`

B

`(l_(2))/(l_(1)-l_(2))`

C

`(l_(1))/(l_(1)-l_(2))`

D

`(l_(1))/(l_(2)-l_(1))`

Text Solution

Verified by Experts

The correct Answer is:
D
`I =1A, phi` = potential gradient
`phi l_(1) =l (R_(1)) rArr phi l_(1) =R_(1)" …..(1)"`
Again, `phi l_(2)= R_(1) +R_(2)" …..(2)"`
Divide (2) by (1) `rArr (l_(1))/(l_(2)) =(R_(1)+R_(2))/(R_(1)) rArr (R_(2))/(R_(1)) =(l_(2)-l_(1))/(l_(1))" "therefore (R_(1))/(R_(2)) =(l_(1))/(l_(2)-l_(1))`
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