Question

A coil of inductance $0.50H$ and resistance $100\Omega$ is connected to a $240V,50Hz$ ac supply. The maximum current in the coil is given as $\frac{x}{100}A$. Find $x$.

Answer 1

$V=240\sqrt{2}sin\left(2\pi 50t\right)V\omega =2\pi 50rad/secinphasor,I_{max}=\frac{240\sqrt{2}}{\sqrt{R^2+L^2{\omega }^2}}=1.823A$
$x$ = 182