A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Ω 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) Here,
Inductance, L = 0.12 H 

$$\begin{gathered} \mathrm{Resistance}, \mathrm{R} = 23 \mathrm{\Omega } \\ \mathrm{Capacitance}, \mathrm{C} = 480 \mathrm{nF} = 480 \times {10}^{-9}\mathrm{F} \end{gathered}$$
                               $$\begin{gathered} \mathrm{Rms} \mathrm{voltage}, {\mathrm{E}}_{\mathrm{v}} = 230 \mathrm{volt} \\ \mathrm{Peak} \mathrm{voltage},{\mathrm{E}}_0 = \sqrt{2} {\mathrm{E}}_{\mathrm{v}} = \sqrt{2} \times 230 \mathrm{volt}. \end{gathered}$$ 

$\mathrm{Maximum} \mathrm{current}, {\mathrm{I}}_0 = \frac{{\mathrm{E}}_0}{\sqrt{{\mathrm{R}}^2+{\left(\mathrm{\omega L} - {\displaystyle \frac{1}{\mathrm{\omega C}}}\right)}^2}}$ 
I₀ would be maximum, when
$$\begin{gathered} {\mathrm{\omega }}_{\mathrm{r}} = \mathrm{\omega } = \frac{1}{\sqrt{\mathrm{LC}}} \\ =\frac{1}{\sqrt{0.12 \times 480 \times {10}^{-9}}} \\ = 4166.7 \mathrm{rad} {\mathrm{s}}^{-1} \\ \Rightarrow {\mathrm{I}}_0 = \frac{{\mathrm{E}}_0}{\mathrm{R}} \\ = \frac{\sqrt{2} \times 230}{23} = 14.14 \mathrm{amp}. \end{gathered}$$ 

(b) Average power absorbed by the circuit is maximum, when I = I₀
          $$\begin{gathered} {\mathrm{P}}_{\mathrm{av}} = \frac{1}{2}{\mathrm{I}}_0^2\mathrm{R} \\ = \frac{1}{2}(14.14{)}^2 \times 23 \\ = 2299.3 \mathrm{watt} \end{gathered}$$ 

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

                        $\boxed{\mathrm{\omega } = {\mathrm{\omega }}_{\mathrm{r}}\pm ∆\mathrm{\omega }}$ 
When,  
            $$\begin{gathered} \boxed{∆\mathrm{\omega } = \frac{\mathrm{R}}{2\mathrm{L}}} \\ =\frac{23}{2 \times 0.12} =95.83 \mathrm{rad} {\mathrm{s}}^{-1} \end{gathered}$$ 

∴ angular frequencies at which power transferred is half = ω_r ± ∆ω
= 4166.7 ± 95.83 = 4262.3 and 4070.87 rad s^–1

Current amplitude at these frequencies is
                    $\frac{{\mathrm{I}}_0}{\sqrt{2}} = \frac{14.14}{1.414} = 10 \mathrm{A}.$ 

(d)  Q-factor  = $\frac{{\mathrm{\omega }}_{\mathrm{r}}\mathrm{L}}{\mathrm{R}} = \frac{4166.7 \times 0.12}{23} = 21.74.$