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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

$\begin {array} {1 1} (A)\;\large\frac{P}{Q}= \large\frac{R}{S_1+S_2} & \quad (B)\;\large\frac{P}{Q}= \large\frac{2R}{S_1+S_2} \\ (C)\;\large\frac{P}{Q}= \large\frac{R(S_1+S_2)}{S_1S_2} & \quad (D)\;\large\frac{P}{Q}= \large\frac{R(S_1+S_2)}{2S_1S_2} \end {array}$


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In the standard formula for the wheatstone bridge, replace the fourth resistance S by the equivalent of
$ S_1\: and \: S_2$ in parellel
$S= S_1S_2 /(S_1+S_2)$
Ans : (C)


answered Feb 15, 2014 by thanvigandhi_1
edited Sep 10, 2014 by thagee.vedartham

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