Explanation

(b) Given, $L = 0.04\ H,R = 12\Omega,\ V = 220\ $volt and $f = 50\ Hz$
The value of current
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I = \frac{V}{Z}\ $
Or $\ \ \ \ \ \ I = \frac{V}{\sqrt{R^{2} + (\omega L)^{2}}}$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ I = \frac{V}{\sqrt{R^{2} + (2\pi fL)^{2}}}$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I = \frac{220}{\sqrt{144 + (2\pi \times 50 \times 0.04)^{2}}}$
Or $\ \ \ \ \ \ \ \ I = 12.7\ A\ $