In a Wheatstone's bridge, three resistances P, Q and R connected in three arms and fourth arm is

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3.Current Electricity

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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)

Solution

$\frac{P}{Q}=\frac{R}{S} \text { where } S=\frac{S_{1} S_{2}}{S_{1}+S_{2}}$

Standard 12

Physics

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