In a Wheatstone bridge,three resistors $P$,$Q$,and $R$ are connected in three arms,and the fourth arm is formed by the parallel combination of two resistors $S_1$ and $S_2$. The condition for the bridge to be balanced is:
- A$\frac{P}{Q} = \frac{2R}{S_1 + S_2}$
- B$\frac{P}{Q} = \frac{R(S_1 + S_2)}{S_1 S_2}$
- C$\frac{P}{Q} = \frac{R(S_1 + S_2)}{2S_1 S_2}$
- D$\frac{P}{Q} = \frac{R}{S_1 + S_2}$
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