In a Wheatstone's bridge, three resistance P, Q and R connected in the three arms and the fourth arm is formed by two resistances S₁ and S₂ connected in parallel. The condition for the bridge to be balanced will be (a) P/Q = (2R)/(S₁+S₂) (b) P/Q = (R(S₁+S₂))/(S₁S₂) (c) P/Q = (R(S₁+S₂))/(2S₁S₂) (d) P/Q = (R)/(S₁+S₂)

Answer: (b). Fourth arm: S = (S₁ S₂)/(S₁ + S₂) Balance: P/Q = R/S = (R(S₁+S₂))/(S₁ S₂) So the answer is (b).

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Wheatstone Bridge & Metre Bridge

In a Wheatstone's bridge, three resistance $P, Q$ and $R$ connected in the three arms and the fourth arm is formed by two resistances $S_1$ and $S_2$ connected in parallel. The condition for the bridge to be balanced will be

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Answer: (b)

Fourth arm: $S = \dfrac{S_1 S_2}{S_1 + S_2}$
Balance: $\dfrac{P}{Q} = \dfrac{R}{S} = \dfrac{R(S_1+S_2)}{S_1 S_2}$
So the answer is (b).

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