Question
In a Wheatstone's bridge, three 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 the bridge to be balanced will be
A.
$$\frac{P}{Q} = \frac{{2R}}{{{S_1} + {S_2}}}$$
B.
$$\frac{P}{Q} = \frac{{R\left( {{S_1} + {S_2}} \right)}}{{{S_1}{S_2}}}$$
C.
$$\frac{P}{Q} = \frac{{R\left( {{S_1} + {S_2}} \right)}}{{2{S_1}{S_2}}}$$
D.
$$\frac{P}{Q} = \frac{R}{{{S_1} + {S_2}}}$$
Answer :
$$\frac{P}{Q} = \frac{{R\left( {{S_1} + {S_2}} \right)}}{{{S_1}{S_2}}}$$
Solution :
$$\frac{P}{Q} = \frac{R}{S}\,{\text{where}}\,S = \frac{{{S_1}{S_2}}}{{{S_1} + {S_2}}}$$