In a Wheatstone's bridge,three resistances P, | vedclass

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(C) The balanced condition for a Wheatstone bridge is given by $\frac{P}{Q} = \frac{R}{S}$,where $S$ is the equivalent resistance of the fourth arm.Si

Vedclass · Accepted Answer

$\frac{P}{Q} = \frac{R(S_1 + S_2)}{S_1 S_2}$

Vedclass · Answer

(C) The balanced condition for a Wheatstone bridge is given by $\frac{P}{Q} = \frac{R}{S}$,where $S$ is the equivalent resistance of the fourth arm.Since $S_1$ and $S_2$ are connected in parallel in the fourth arm,their equivalent resistance $S$ is given by $\frac{1}{S} = \frac{1}{S_1} + \frac{1}{S_2} = \frac{S_1 + S_2}{S_1 S_2}$.Therefore,$S = \frac{S_1 S_2}{S_1 + S_2}$.Substituting this value into the bridge balance condition,we get $\frac{P}{Q} = \frac{R}{\left( \frac{S_1 S_2}{S_1 + S_2} \right)}$.This simplifies to $\frac{P}{Q} = \frac{R(S_1 + S_2)}{S_1 S_2}$.

In a Wheatstone's bridge,three resistances $P, Q$ and $R$ are 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

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