Question

Two resistances X and Y in the two gaps of a metre bridge give a null point dividing the wire in the ratio 2 : 3. If each resistance is increased by 30 Ω, the new null point divides the wire in the ratio 5 : 6, calculate each resistance.

Answer 1

From the data, we have in the first case,

\(\cfrac XY\) = \(\cfrac{I_x}{I_y}\) = \(\cfrac 23\)

∴ 3X = 2Y ……….. (1)

and in the second case, \(\cfrac{X+30}{Y+30}\)

= \(\cfrac{I_X+30}{I_Y+30}\) = \(\cfrac56\)

∴ 6X + 180 = 5Y + 150 

∴ 6X – 5Y = -30 

∴ 6X = 5Y – 30 

∴ 2(3X) = 5Y – 30 …………. (2)

Substituting the value of 3X from Eq. (1) in Eq. (2),

we get, 

2(2Y) = 5Y – 30 

∴ Y = 30 Ω 

∴ X = \(\cfrac23\)Y = 20 Ω

Answer 2

hope so this will help you

SORRY FOR CALCULATION MISTAKE 

X= 20

Y= 30