Soln. (a) At resonant frequency, the current amplitude is maximum. f = $\frac{1}{2\pi\sqrt{LC}}$ = $\frac{1}{2\pi\sqrt{0.12 \times 480 \times 10^{-9}}}$ = 663 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 A (b) Maximum power loss at resonant frequency, P = $E_v I_v$ cos ϕ P = $\frac{E_v E_v}{R}$ cos 0° = $\frac{E_v^2}{R}$ = $\frac{(230)^2}{23}$ = 2300 W (c) Let at an angular frequency, the source power is half the power at resonant frequency. P = $E_v I_v$ cos ϕ $\frac{1}{2} \left[ \frac{E_v^2}{R} \right]$ = $\frac{E_v E_v}{Z} \frac{R}{Z}$ $Z^2$ = $2R^2, R^2$ + $(X_L - X_C)^2$ = $2R^2$ $X_L - X_C$ = R $\omega_L$ - $\frac{1}{\omega C}$ = R or $\omega^2 - \frac{1}{LC}$ = $\frac{R}{L}$ ω where resonant angular frequency, $\omega_r$ = $\frac{1}{LC}$ = $\frac{1}{0.12 \times 480 \times 10^{-9}}$ so , $\omega^2 - \omega_r^2$ = ± $\frac{R}{L}$ ω two quadratic equations can be formed $\omega^2 - \frac{R}{l}$ ω - $\omega_r^2$ = 0 and $\omega^2 + \frac{R}{L}$ ω - $\omega_r^2$ = 0 On solving, we get, $\omega_1$ = $\frac{R}{2L} + \left[ \omega_r^2 + \frac{R^2}{4L^2} \right]^{1/2}$ = $\omega_r + \Delta \omega$ and $\omega_2$ = - $\frac{R}{2L} + \left[ \omega^2 + \frac{R^2}{4L^2} \right]^{1/2}$ = $\omega_r$ - Δω Now, $\omega_1 - \omega_2$ = R/L [$\omega_r$ + Δω] – [$\omega_r$ – Δω] = R/L or Δω = R/L Δω = $\frac{R}{2L}$ bandwidth of angular frequency So, band width of frequency Δf = $\frac{\Delta \omega}{2\pi}$ = $\frac{R}{4\pi L}$ = $\frac{23}{4 \times 3.14 \times 0.12}$ = 15.26 Hz Hence the two frequencies for half power $f_Z$ = $f_2$ – Δf and $f_1$ = $f_r$ + Δf $f_Z$ = 663 – 15.26 = 647.74 Hz and $f_1$ = 663 + 15.26 = 678.26 Hz At these frequencies the current amplitude is I = $\frac{I_0}{\sqrt{2}}$ = 10 A (d) Q factor, Q = $\frac{1}{R} \sqrt{\frac{L}{C}}$ ⇒ Q = $\frac{1}{23} \sqrt{\frac{0.12}{480 \times 10^{-9}}}$ = 21.7