Question

With an unknown resistance X in the left gap and a resistance of 30 Ω in the right gap of a meter-bridge, the null point is obtained at 40 cm from the left end of the wire. Calculate the unknown resistance and shift in the position of the null point when resistance in each gap is increased by 15 Ω.

Answer 1

Case I: By first condition,

X = X Ω, R = 30 Ω, 

l_x = 40 cm, 

l_R = 100 − 40 = 60 cm 

For balanced network,

`X/R = l_X/l_R`

`X/30 = 40/60`

∴ X = `30 × 2/3`

X = 20 Ω

Case II: In each gap, resistance is increased by 15 Ω.

`(X + 15)/(R + 15) = l_X/l_R = l_X/(100 - l_X)`

Putting X = 20 and R = 30

`(20 + 15)/(30 + 15) = l_X/(100 - l_X)`

`35/45 = l_X/(100 - l_X)`

`7/9 = l_X/(100 - l_X)`

∴ 7(100 − l_X) = 9l_X

∴ 700 − 7l_X = 9 l_X

∴ 700 = 9 l_X + 7 l_X

∴ 700 = 16 l_X

∴ `l_X = 700/16`

∴ l_X = 43.75 cm

Shift in the position of the null point = 43.75 − 40 = 3.75 cm to the right.