In the figure given the value of $X$ resistance will be, when the $p.d.$ between $B$ and $D$ is zero ................... $ohm$
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(c) Potential difference between $B$ and $D$ is zero, it means Wheatstone bridge is in balanced condition

So $\frac{P}{Q} = \frac{R}{S}$

$==>$ $\frac{{21}}{{3 + \frac{{8X}}{{(8 + X)}}}} = \frac{{18}}{6}$

$==>$ $X = 8\,\Omega $

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