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
Answer: (A) $30\ \Omega$
Balance: $\dfrac{P}{Q} = \dfrac{R}{S} \Rightarrow S = \dfrac{QR}{P} = \dfrac{20 \times 15}{10} = 30\ \Omega$.
Solution by Sreeraj P, M.Sc Physics