In a Wheatstone bridge,three resistances P, Q | vedclass

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(C) For a balanced Wheatstone bridge,the condition is given by $\frac{P}{Q} = \frac{R}{S}$,where $S$ is the resistance in the fourth arm.In this case,

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) For a balanced Wheatstone bridge,the condition is given by $\frac{P}{Q} = \frac{R}{S}$,where $S$ is the resistance in the fourth arm.In this case,the fourth arm consists of two resistances $S_1$ and $S_2$ connected in parallel.The equivalent resistance $S$ of two resistors in parallel 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 of $S$ into the balance condition $\frac{P}{Q} = \frac{R}{S}$,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 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 is

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