In a Wheatstone's bridge, there resistances $$P, Q$$ 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 bridge to be balanced will be

A

$$\dfrac{P}{Q} = \dfrac{R}{S_1 + S_2}$$

B

$$\dfrac{P}{Q} = \dfrac{2R}{S_1 + S_2}$$

C

$$\dfrac{P}{Q} = \dfrac{R(S_1 + S_2)}{S_1 S_2}$$

D

$$\dfrac{P}{Q} = \dfrac{R(S_1 + S_2)}{2 S_1 S_2}$$