Capacitance, $C=\frac{10^{-3}}{2 \pi} \mathrm{F}$ Inductance, $L=\left(\frac{100}{\pi}\right) \mathrm{mH}=\frac{100}{\pi} \times 10^{-3} \mathrm{H}$
Resistance, $R=10 \Omega$
Voltage of source $=220 \mathrm{volt}$
Frequency of source, $f=50 \mathrm{~Hz}$
Phase angle $\phi$ is given by
$$
\tan \phi=\frac{X_L-X_C}{R}
$$
where, $X_L=$ inductive impedance $=\omega L$
$$
\begin{aligned}
& =2 \pi f L \\
\therefore \quad X_L & =2 \pi \times 50 \times \frac{100}{\pi} \times 10^{-3} \\
& =10 \Omega
\end{aligned}
$$
$$
(\because \omega=2 \pi f)
$$
and
$$
\begin{aligned}
X_C & =\text { Capactive impedance } \\
& =\frac{1}{\omega C}=\frac{1}{2 \pi f C}
\end{aligned}
$$
$$
=\frac{1 \times 2 \pi}{2 \pi \times 50 \times 10^{-3}} \Omega=\frac{100}{5}=20 \Omega
$$
$\therefore$ Phase angle,
$$
\tan \phi=\frac{X_C-X_L}{R}=\frac{20-10}{10}=1 \text { or } \phi=45^{\circ}
$$