In a Wheatstone bridge,P = 9 , Omega,Q = 11 , | vedclass

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(C) For a balanced Wheatstone bridge,the condition is $\frac{P}{Q} = \frac{R}{S'}$,where $S'$ is the equivalent resistance of $S$ and $r$ connected in

Vedclass · Accepted Answer

$26.4 \, \Omega$

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(C) For a balanced Wheatstone bridge,the condition is $\frac{P}{Q} = \frac{R}{S'}$,where $S'$ is the equivalent resistance of $S$ and $r$ connected in parallel.Given $P = 9 \, \Omega$,$Q = 11 \, \Omega$,and $R = 4 \, \Omega$.Substituting these values into the balancing condition: $\frac{9}{11} = \frac{4}{S'}$.Solving for $S'$,we get $S' = \frac{4 	imes 11}{9} = \frac{44}{9} \, \Omega$.Since $S'$ is the parallel combination of $S = 6 \, \Omega$ and $r$,we have $\frac{1}{S'} = \frac{1}{S} + \frac{1}{r}$.Substituting the values: $\frac{9}{44} = \frac{1}{6} + \frac{1}{r}$.Rearranging to solve for $r$: $\frac{1}{r} = \frac{9}{44} - \frac{1}{6} = \frac{54 - 44}{264} = \frac{10}{264} = \frac{5}{132}$.Therefore,$r = \frac{132}{5} = 26.4 \, \Omega$.

In a Wheatstone bridge,$P = 9 \, \Omega$,$Q = 11 \, \Omega$,$R = 4 \, \Omega$,and $S = 6 \, \Omega$. How much resistance $r$ must be connected in parallel to the resistance $S$ to balance the bridge?

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