A series LCR circuit with L = 0.12 H,

Inductance, L = 0.12 H

Capacitance, C = 480 nF = 480 × 10−9 F

Resistance, R = 23 Ω

Supply voltage, V = 230 V

Peak voltage is given as:

$V_{0}=\sqrt{2} \times 230=325.22 \mathrm{~V}$

(a) Current flowing in the circuit is given by the relation, $\quad I_{0}=\frac{V_{0}}{\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}}$

Where,

I0 = maximum at resonance

At resonance, we have

$\omega_{R} L-\frac{1}{\omega_{R} C}=0$

Where,

ωR = Resonance angular frequency

$\therefore \omega_{R}=\frac{1}{\sqrt{L C}}$

$=\frac{1}{\sqrt{0.12 \times 480 \times 10^{-9}}}=4166.67 \mathrm{rad} / \mathrm{s}$

$\therefore$ Resonant frequency, $v_{R}=\frac{\omega_{R}}{2 \pi}=\frac{4166.67}{2 \times 3.14}=663.48 \mathrm{~Hz}$

And, maximum current $\left(I_{0}\right)_{\mathrm{Max}}=\frac{V_{0}}{R}=\frac{325.22}{23}=14.14 \mathrm{~A}$

(b) Maximum average power absorbed by the circuit is given as:

$\left(P_{\mathrm{av}}\right)_{\mathrm{Max}}=\frac{1}{2}\left(I_{0}\right)_{\mathrm{Max}}^{2} R$

$=\frac{1}{2} \times(14.14)^{2} \times 23=2299.3 \mathrm{~W}$

Hence, resonant frequency $\left(v_{R}\right)$ is $663.48 \mathrm{~Hz}$.

(c) The power transferred to the circuit is half the power at resonant frequency.

Frequencies at which power transferred is half, $=\omega_{R} \pm \Delta \omega$

$=2 \pi\left(v_{R} \pm \Delta v\right)$

Where,

$\Delta \omega=\frac{R}{2 L}$

$=\frac{23}{2 \times 0.12}=95.83 \mathrm{rad} / \mathrm{s}$

Hence, change in frequency, $\Delta v=\frac{1}{2 \pi} \Delta \omega=\frac{95.83}{2 \pi}=15.26 \mathrm{~Hz}$

$\therefore v_{R}+\Delta v=663.48+15.26=678.74 \mathrm{~Hz}$

And, $v_{R}-\Delta v=663.48-15.26=648.22 \mathrm{~Hz}$

Hence, at 648.22 Hz and 678.74 Hz frequencies, the power transferred is half.

At these frequencies, current amplitude can be given as:

$I^{\prime}=\frac{1}{\sqrt{2}} \times\left(I_{0}\right)_{\mathrm{Max}}$

$=\frac{14.14}{\sqrt{2}}=10 \mathrm{~A}$

(d) $Q$-factor of the given circuit can be obtained using the relation, $Q=\frac{\omega_{R} L}{R}$

$=\frac{4166.67 \times 0.12}{23}=21.74$

Hence, the Q-factor of the given circuit is 21.74.