In a Wheatstone network,P=2 Omega, Q=2 Omega, | vedclass

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(A) Given: $P=2 \Omega, Q=2 \Omega, R=2 \Omega, S=3 \Omega$.The condition for a balanced Wheatstone bridge is $\frac{P}{Q} = \frac{R}{S_{eq}}$,where $

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$6$

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(A) Given: $P=2 \Omega, Q=2 \Omega, R=2 \Omega, S=3 \Omega$.The condition for a balanced Wheatstone bridge is $\frac{P}{Q} = \frac{R}{S_{eq}}$,where $S_{eq}$ is the equivalent resistance of $S$ shunted with $X$.The equivalent resistance $S_{eq}$ when $S$ is shunted with $X$ is given by $S_{eq} = \frac{S \cdot X}{S+X}$.Substituting the values into the balance equation: $\frac{2}{2} = \frac{2}{\left(\frac{3X}{3+X}\right)}$.This simplifies to $1 = \frac{2(3+X)}{3X}$.Therefore,$3X = 6 + 2X$,which gives $X = 6 \Omega$.Thus,$S$ must be shunted with a resistance of $6 \Omega$ to balance the bridge.

In a Wheatstone network,$P=2 \Omega, Q=2 \Omega, R=2 \Omega$,and $S=3 \Omega$. The resistance with which $S$ is to be shunted so that the bridge may be balanced is: (in $\Omega$)

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