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In a Wheatstone's bridge, three resistances P, Q and R are 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 :


( A ) $\frac{P}{Q} = \frac{R(S_1 + S_2)}{S_1S_2}$
( B ) $\frac{P}{Q} = \frac{R(S_1 + S_2)}{2S_1S_2}$
( C ) $\frac{P}{Q} = \frac{R}{S_1+S_2}$
( D ) $\frac{P}{Q} = \frac{2R}{S_1+S_2}$

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