(a) At natural frequency the current amplitude is maximum.
$$
\begin{aligned}
F &=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \pi \sqrt{0.12 \times 480 \times 10^{-6}}} \\
&=663 \mathrm{~Hz} \\
I_v &=\frac{E_v}{R}, I_0=I_v \sqrt{2}=\frac{E_v \sqrt{2}}{R}=\frac{230 \sqrt{2}}{23}=14.14 \mathrm{~A}
\end{aligned}
$$
(b) Maximum power loss at resonant frequency, $\mathrm{P}=E_{\mathrm{v}} I_{\mathrm{v}} \cos \phi$
$\mathrm{P}=\mathrm{E}_{\mathrm{v}} \frac{E_v}{R} \cos 0^{\circ}=\frac{E_v^2}{R}=\frac{(230)^2}{23}=2300 \mathrm{~W}$
(c) Let at an angular frequency the source power is half the power at resonant frequency.
$$
\begin{aligned}
&P=E_{\mathrm{v}} I_{\mathrm{v}} \cos \phi \\
&\frac{1}{2}\left[\frac{E_v^2}{R}\right]=\frac{E_v E_v}{Z} \frac{R}{Z} \\
&Z^2=2 R^2 \\
&R^2+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^2=2 \mathrm{R}^2 \\
&X_{\mathrm{L}}-X_{\mathrm{C}}=\mathrm{R} \\
&\omega_L-\frac{1}{\omega C}=R \text { or } \omega^2-\frac{1}{L C}=\frac{R}{L} \omega
\end{aligned}
$$
where resonant angular frequency
$$
\omega_r=\frac{1}{L C}=\frac{1}{0.12 \times 480 \times 10^{-6}}
$$
so, $\omega^2-w_r^2=\pm \frac{R}{L} \omega$
two quadratic equations can be formed
$\omega^2-\frac{R}{L} \omega-\omega r^2=0$
and $\omega^2+\frac{R}{L} \omega-\omega_r^2=0$
On solving, we get
$\omega_1=\frac{R}{2 L}+\left[\omega_r^2+\frac{R^2}{4 L^2}\right]^{\frac{1}{2}}=\omega_{\mathrm{r}}+\Delta \omega$ and
$$
\omega_2=-\frac{R}{2 L}+\left[\omega^2+\frac{R^2}{4 L^2}\right]^{\frac{1}{2}}=\omega_{\mathrm{r}}-\Delta \omega
$$

Now, $\omega_1-\omega_2=\frac{R}{L}$
$$
\left[\omega_r+\Delta \omega\right]-\left[\omega_r-\Delta \omega\right]=\frac{R}{L}=\Delta \omega=\frac{R}{L}
$$
$\Delta \omega=\frac{R}{2 L}$ bandwidth of angular frequency
So, band width of frequency
$$
\begin{aligned}
&\Delta f=\frac{\Delta \omega}{2 \pi}=\frac{R}{4 \pi L}=\frac{23}{4 \times 3.14 \times 0.12} \\
&\Delta f=15.26 \mathrm{~Hz}
\end{aligned}
$$
Hence the two frequencies for half power
$$
\begin{aligned}
&f_{\mathrm{z}}=f_2-\Delta f \text { and } f_1=f_{\mathrm{r}}+\Delta f \\
&f_{\mathrm{z}}=663-15.26=647.74 \mathrm{~Hz} \\
&\text { and } f_1=663+15.26=678.26 \mathrm{~Hz}
\end{aligned}
$$
At these frequencies the current amplitude is
$$
I=\frac{I_0}{\sqrt{2}}=10 \mathrm{~A}
$$
(d) $Q$ factor, $Q=\frac{1}{R} \sqrt{\frac{L}{C}}$
$$
Q=\frac{1}{23} \sqrt{\frac{0.12}{480 \times 10^{-6}}}=21.7
$$