Given,

Inductance, 𝐿 = 0.12 H

Capacitance, 𝐶 = 480 nF

rms voltage, 𝑉 = 230 V

Resistance, 𝑅 = 23 Ω

(a) At resonance the impedance of the circuit will be minimum, resulting in
the maximum current in the circuit. This frequency is given by-

$\begin{array}{l}{\omega }_r=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{0.12\times 480\times {10}^{-9}}}=4167rad{s}^{-1}\\ f_r=\frac{{\omega }_r}{2\pi }=663s^{-1}\end{array}$

The maximum current is given by,

$I_m=\sqrt{2}\times \frac{V}{R}=\sqrt{2}\times \frac{230}{23}=10\sqrt{2}A$

(b) The maximum average power is given by,

$P_{\text{ avg },\max }=\frac{1}{2}{I}_m^{2}R=\frac{1}{2}\times (10\sqrt{2}{)}^2\times 23=2300W$

(c) The two angular frequencies for which the power transferred to the circuit
is half the power at the resonant frequency,

$\begin{array}{l}\omega ={\omega }_r\pm \mathrm{\Delta }\omega \\ \text{ Where, }\\ \mathrm{\Delta }\omega =\frac{R}{2L}=\frac{23}{2\times 0.12}=95.8rad{s}^{-1}\\ \mathrm{\Delta }f=\frac{\mathrm{\Delta }\omega }{2\pi }=15.2Hz\end{array}$

Hence, at frequencies 648 Hz and 678 Hz the power absorbed is half the
peak power.

The current amplitude at these frequencies can be given as,

$I^{\mathrm{\prime }}=\frac{I_m}{\sqrt{2}}=\frac{10\sqrt{2}}{\sqrt{2}}=10$

(d) Q-factor of the circuit is given by

$Q=\frac{{\omega }_{r}L}{R}=\frac{4167\times 0.12}{23}=21.7$