# In a Wheatstone's bridge their resistances P, Q and R connected in the three arms and the fourth arm is formed by the two resistances $S_1\: and \: S_2$ connected in parallel. The condition for bridge to be balanced will be
$\begin {array} {1 1} (A)\;\large\frac{P}{Q}= \large\frac{R}{S_1+S_2} & \quad (B)\;\large\frac{P}{Q}= \large\frac{2R}{S_1+S_2} \\ (C)\;\large\frac{P}{Q}= \large\frac{R(S_1+S_2)}{S_1S_2} & \quad (D)\;\large\frac{P}{Q}= \large\frac{R(S_1+S_2)}{2S_1S_2} \end {array}$