1.

In a potentiometer experiment, the balancing point with a cell is at a length 240 cm. On shunting the cell with a resistance of 2 Omega, the balancing length becomes 120 cm. The internal resistance of the cell is

Answer»

`1 Omega`
`4 Omega`
`0.5 Omega`
`2 Omega`

Solution :(d): E = `kl_(1)` ...(i)
`E-ir=E-(ER)/(2+r) =kl_(2) rArr E = kl_(2) [(2+r)/(2)]`...(ii)
From eqn. (i) and (ii), `(2+r)/(2) = (l_(1))/(l_(2)) = (240)/(120)`
`rArr r= 2OMEGA`


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