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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$\frac{P}{Q} = \frac{R}{{{S_1} + {S_2}}}$
  • B$\;\frac{P}{Q} = \frac{{2R}}{{{S_1} + {S_2}}}$
  • C$\;\frac{P}{Q} = \frac{{R\left( {{S_1} + {S_2}} \right)}}{{{S_1}{S_2}}}$
  • D$\;\frac{P}{Q} = \frac{{R\left( {{S_1} + {S_2}} \right)}}{{2{S_1}{S_2}}}$
AIEEE 2006, Medium
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