In a Wheatstone bridge,three resistors P,Q,an | vedclass

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(B) For a balanced Wheatstone bridge,the ratio of resistances in adjacent arms must be equal: $\frac{P}{Q} = \frac{R}{S_{eq}}$,where $S_{eq}$ is the e

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) For a balanced Wheatstone bridge,the ratio of resistances in adjacent arms must be equal: $\frac{P}{Q} = \frac{R}{S_{eq}}$,where $S_{eq}$ is the equivalent resistance of the fourth arm.Since $S_1$ and $S_2$ are connected in parallel,their equivalent resistance is given by $S_{eq} = \frac{S_1 S_2}{S_1 + S_2}$.Substituting this into the balance condition,we get: $\frac{P}{Q} = \frac{R}{(\frac{S_1 S_2}{S_1 + S_2})}$.Simplifying this,we obtain: $\frac{P}{Q} = \frac{R(S_1 + S_2)}{S_1 S_2}$.

In a Wheatstone bridge,three resistors $P$,$Q$,and $R$ are connected in three arms,and the fourth arm is formed by the parallel combination of two resistors $S_1$ and $S_2$. The condition for the bridge to be balanced is:

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