In a Wheatstone bridge, P = 10 Ω, Q = 20 Ω and R = 15 Ω. For balance, the fourth resistance S must be

(A) 30 Ω. Balance: (P)/(Q) = (R)/(S) S = (QR)/(P) = (20 × 15)/(10) = 30 Ω.

Current Electricity question for NEET / JEE Main, with solution

Q 12-03-036NEETJEE MainMedium

In a Wheatstone bridge, $P = 10\ \Omega$, $Q = 20\ \Omega$ and $R = 15\ \Omega$. For balance, the fourth resistance $S$ must be