Three resistances P, Q, R each of 2 ,Omega an | vedclass

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(A) Let $X$ be the equivalent resistance of $S$ and $6 \,\Omega$ connected in parallel.For a parallel combination,the equivalent resistance $X$ is giv

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$3$

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(A) Let $X$ be the equivalent resistance of $S$ and $6 \,\Omega$ connected in parallel.For a parallel combination,the equivalent resistance $X$ is given by:$\frac{1}{X} = \frac{1}{S} + \frac{1}{6} \quad \dots(i)$For a balanced Wheatstone bridge,the condition is $\frac{P}{Q} = \frac{R}{X}$.Given $P = Q = R = 2 \,\Omega$,we substitute these values:$\frac{2}{2} = \frac{2}{X} \implies 1 = \frac{2}{X} \implies X = 2 \,\Omega$.Now,substitute $X = 2 \,\Omega$ into equation $(i)$:$\frac{1}{2} = \frac{1}{S} + \frac{1}{6}$$\frac{1}{S} = \frac{1}{2} - \frac{1}{6}$$\frac{1}{S} = \frac{3 - 1}{6} = \frac{2}{6} = \frac{1}{3}$Therefore,$S = 3 \,\Omega$.

Three resistances $P, Q, R$ each of $2 \,\Omega$ and an unknown resistance $S$ form the four arms of a Wheatstone bridge circuit. When a resistance of $6 \,\Omega$ is connected in parallel to $S$,the bridge gets balanced. What is the value of $S$ in $\Omega$?

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