Question

In a simple meter-bridge circuit, the both gaps are bridge by coils P and Q having the smaller resistance. A balance is obtained when the jockey key makes contact at a point of the bridge wire 40 cm from the P end. On shunting the coil Q with a resistance of 50 Ω the balance point is moved through 10 cm. What are the resistance of P and Q ? 

(A) [(100) / 3]Ω, [(100) / 2]Ω respectively 

(B) [(50) / 3]Ω, [(50) / 2]Ω respectively 

(C) [(25) / 3]Ω, [(25) / 2]Ω respectively 

(D) [(75) / 3]Ω, [(75) / 2]Ω respectively

Answer 1

The correct option is (B) [(50) / 3]Ω, [(50) / 2]Ω respectively.

Explanation:

Initially, jockey key at 40cm from P end i.e. (100 – 40) = 60cm from Q end 

∴ (P/Q) = (40 / 60) = (2/3) (1) 

Now Q is shunted with 50Ω. 

∴ New resistance Q' = [(50Q) / (50 + Q)] and new balance point is moved by 10 cm. 

Hence (P / Q') = (50 / 50) = 1 

∴ P = Q' i.e. P = [(50Q) / (50 + Q)] 

∴ (P/Q) = [(50) / (50 + Q)] 

from (1), 

(2/3) = [(50) / (50 + Q)] 

∴ 100 + 2Q = 150 

2Q = 50 

Q = 25Ω 

and P = (2/3)Q = (2/3) × 25 = (50 / 3)Ω