Explanation

(b) The phase angle $(\theta)$ between I and V is given by
$\tan{\theta = \frac{X_{L} - X_{C}}{R}}$ …(i)
Where, $X_{L} = 2\pi fL$
$= 2\pi \times 50 \times \left\lbrack \frac{200}{\pi} \times 10^{- 3} \right\rbrack$
$= 20\ \Omega$
$X_{L} = \frac{1}{2\pi fC}$
$= \frac{1 \times \pi}{2\pi \times 50 \times 10^{- 3}}$
=10$\ \Omega$
and $R = 10\ \Omega$
Substituting values of $X_{L}$, $X_{C\ }$and R in eq.(i), we get
$\tan{\theta = \frac{20 - 10}{10}} = 1$
$\Rightarrow \ \ \ \ \ \ \tan{\theta = \tan\frac{\pi}{4}}$
$\therefore\ \ \ \theta = \frac{\pi}{4}$
The phase angle of the circuit is$\ \frac{\pi}{4}.$