In a Wheatstone's network, P = 2 Ω , Q = 2 Ω , R = 2 Ω and S = 3 Ω . Find the resistance (in Ω ) with which S is to be shunted, in order that the bridge gets balanced.

Step-by-step solution

Let a resistance r Ω be shunted with resistance S , so that the bridge is balanced. Let S ' be the resultant resistance of S and r , then In balanced position P Q = R S ′ 2 2 = 2 S ′ ∴ S ′ = 2 Ω Now, 1 S ′ = 1 S + 1 r 1 r = 1 S ′ - 1 S = 1 2 - 1 3 = 3 - 2 6 1 r = 1 6 r = 6 Ω