$A$ series $LCR$ circuit with $L=0.12\, H$,$C=480\, nF$,$R=23\, \Omega$ is connected to a $230\, V$ variable frequency supply.
$(a)$ What is the source frequency for which current amplitude is maximum? Obtain this maximum value.
$(b)$ What is the source frequency for which average power absorbed by the circuit is maximum? Obtain the value of this maximum power.
$(c)$ For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?
$(d)$ What is the $Q$-factor of the given circuit?

(A) Given: $L = 0.12\, H$,$C = 480\, nF = 480 \times 10^{-9}\, F$,$R = 23\, \Omega$,$V_{rms} = 230\, V$.
Peak voltage $V_0 = \sqrt{2} \times 230 = 325.27\, V$.
$(a)$ Current amplitude is maximum at resonance. Resonance frequency $\omega_R = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.12 \times 480 \times 10^{-9}}} = 4166.67\, rad/s$.
Frequency $f_R = \frac{\omega_R}{2\pi} = \frac{4166.67}{6.28} \approx 663.48\, Hz$.
Maximum current $I_{max} = \frac{V_0}{R} = \frac{325.27}{23} \approx 14.14\, A$.
$(b)$ Average power is maximum at resonance. $f = 663.48\, Hz$.
$P_{max} = I_{rms}^2 R = \left(\frac{I_{max}}{\sqrt{2}}\right)^2 R = \frac{1}{2} I_{max}^2 R = \frac{1}{2} \times (14.14)^2 \times 23 \approx 2299.3\, W$.
$(c)$ Power is half at half-power frequencies $f = f_R \pm \Delta f$.
Bandwidth $2\Delta\omega = \frac{R}{L} = \frac{23}{0.12} = 191.67\, rad/s$.
$\Delta f = \frac{\Delta\omega}{2\pi} = \frac{191.67}{2 \times 2 \times 3.14} \approx 15.26\, Hz$.
Frequencies are $663.48 + 15.26 = 678.74\, Hz$ and $663.48 - 15.26 = 648.22\, Hz$.
Current amplitude $I' = \frac{I_{max}}{\sqrt{2}} = \frac{14.14}{1.414} = 10\, A$.
$(d)$ $Q$-factor $= \frac{\omega_R L}{R} = \frac{4166.67 \times 0.12}{23} \approx 21.74$.