Question

In a meter-bridge experiment with a resistance `R_(1)` in left gap and a resistance `X` in a right gap. null point is obtained at `40 cm` from the left emf. With a resistance `R_(2)` in the left gap, the null point is obtainned at `50 cm` from left hand. Find the position of the left gap is containing `R_(1)` and `R_(2)` (i) in series and (ii) in parallel.

Answer 1

Correct Answer - (i)` 62.5 cm `,(ii)`28.6 cm`
`(R_(1))/(X) = (40)/(60) = (2)/(3) or R_(1) = (2X)/(3)`….(i)
`(R_(2))/(X) = (50)/(50) = 1 or R_(2) = X `…..(i)
when `R_(1) and R_(2)` are in series the effective resistance `= R_(1) + R_(2) = (2X)/(3) + X = (5X)/(3)`
Then , `(5X//3)/(X) = (1)/((100 -1)) or 5(100 - 1) = 3 l `
`or 1 = 5 xx 100//8 = 62.5 cm `
(ii) when `R_(1) and R_(2) ` are in parallel th eeffective resisrtence is
`R_(p) = R_(1) R_(2) (R_(1) + R_(2)) = (2X//3 xx X)/((2X//3+X)) = (2)/(5) X`
Then `(2X//5)/(X) = (1)/((100 - 1)) or 2 (100 - 1) = 5 l `
or ` l = (200)/(7) = 28.6 cm`