In the Wheatstone's bridge shown,P = 2,Omega, | vedclass

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(D) For a Wheatstone bridge to be balanced,the ratio of resistances in the arms must satisfy the condition $\frac{P}{Q} = \frac{S'}{R}$,where $S'$ is

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$8$

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(D) For a Wheatstone bridge to be balanced,the ratio of resistances in the arms must satisfy the condition $\frac{P}{Q} = \frac{S'}{R}$,where $S'$ is the effective resistance of the arm containing $S$ with a shunt $r$ connected in parallel.Given $P = 2\,\Omega$,$Q = 3\,\Omega$,$R = 6\,\Omega$,and $S = 8\,\Omega$.The effective resistance $S'$ of $S$ and $r$ in parallel is given by $S' = \frac{S \cdot r}{S + r} = \frac{8r}{8 + r}$.Substituting the values into the balance condition:$\frac{2}{3} = \frac{8r / (8 + r)}{6}$$\frac{2}{3} = \frac{8r}{6(8 + r)}$$12(8 + r) = 24r$$96 + 12r = 24r$$12r = 96$$r = 8\,\Omega$.Thus,the required shunt resistance is $8\,\Omega$.

In the Wheatstone's bridge shown,$P = 2\,\Omega$,$Q = 3\,\Omega$,$R = 6\,\Omega$ and $S = 8\,\Omega$. In order to obtain balance,the shunt resistance across '$S$' must be ............. $\Omega$.

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