In Wheatstone's bridge $P = 9\, ohm$, $Q = 11\, ohm$, $R = 4\,ohm$ and $S = 6\,ohm$. How much resistance must be put in parallel to the resistance $S$ to balance the bridge ............... $ohm$
Medium
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$\frac{P}{Q} = \frac{R}{{S'}}$ (For balancing bridge)
$ \Rightarrow $ $S' = \frac{{4 \times 11}}{9} = \frac{{44}}{9}$
$ \Rightarrow $ $\frac{1}{{S'}} = \frac{1}{r} + \frac{1}{6}$
$ \Rightarrow $ $\frac{9}{{44}} - \frac{1}{6} = \frac{1}{r}$
$ \Rightarrow $ $r = \frac{{132}}{5} = 26.4\,\Omega $
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