Question

The Wheatstone bridge is shown in figure. What should be the value of $X$ so that Wheatstone bridge will be balanced?

Answer 1

Equivalent resistance of $X\Omega$ and $100\Omega$ in parallel
$R^{\prime }=\frac{x\times 100}{100+x}=\frac{100x}{100+x}$
From Wheatstone bridge
$\frac{100}{10}=\frac{R^{\prime }}{5}$
$10=\frac{100x}{(100+x)\times 5}$
$1=\frac{2x}{100+x}$
$100+x=2x$
$100=2x-x$
$100=x$
Therefore $x=100\Omega$.