A series LCR circuit is connected to a 45 sin($$\omega t$$) Volt source. The resonant angular frequency of the circuit is $$10^5$$ rad s$$^{-1}$$ and current amplitude at resonance is $$I_0$$. When the angular frequency of the source is $$\omega = 8 \times 10^4$$ rad s$$^{-1}$$, the current amplitude in the circuit is 0.05 $$I_0$$. If $$L = 50$$ mH, match each entry in List-I with an appropriate value from List-II and choose the correct option.

List-I		List-II
(P)	$$I_0$$ in mA	(1)	44.4
(Q)	The quality factor of the circuit	(2)	18
(R)	The bandwidth of the circuit in rad s$$^{-1}$$	(3)	400
(S)	The peak power dissipated at resonance in Watt	(4)	2250
		(5)	500

The applied voltage is $$v = 45 \sin \left( \omega t \right)\;{\rm V}$$, so its peak (amplitude) value is $$V_0 = 45\;{\rm V}$$.

For a series LCR circuit the current amplitude at any angular frequency $$\omega$$ is
$$I(\omega) = \frac{V_0}{Z},\qquad Z = \sqrt{\,R^{2} + \left( \omega L - \frac{1}{\omega C} \right)^{2}}.$$

At resonance $$\omega_0 = \frac{1}{\sqrt{LC}} = 10^{5}\;{\rm rad\,s^{-1}}$$, the impedance is purely resistive, so
$$I_0 = I(\omega_0) = \frac{V_0}{R} \qquad -(1)$$

Using $$\omega_0^2 = \dfrac{1}{LC}\;$$ we get the capacitance
$$C = \frac{1}{L\omega_0^{2}}.$$

At $$\omega = 8\times10^{4}\;{\rm rad\,s^{-1}}$$ the current amplitude becomes $$I = 0.05\,I_0$$. Hence
$$\frac{I}{I_0} = \frac{R}{\sqrt{\,R^{2} + \left(\omega L - \frac{1}{\omega C}\right)^{2}}} = 0.05.$$ Squaring and rearranging, $$R^{2} = \frac{0.0025}{0.9975}\left(\omega L - \frac{1}{\omega C}\right)^{2}$$ $$\Longrightarrow\; R = 0.05006\left|\omega L - \frac{1}{\omega C}\right|.$$(2)

Now $$\omega L - \frac{1}{\omega C} = \,L\left(\omega - \frac{\omega_0^{2}}{\omega}\right) = L\frac{\omega^{2}-\omega_0^{2}}{\omega}.$$ For $$\omega = 8\times10^{4}\;{\rm rad\,s^{-1}}$$, $$\omega_0 = 10^{5}\;{\rm rad\,s^{-1}}$$ and $$L = 50\;{\rm mH}=0.05\;{\rm H}$$, $$\left|\omega^{2}-\omega_0^{2}\right| = \left| (8\times10^{4})^{2}-(10^{5})^{2}\right| = \left|6.4\times10^{9}-1.0\times10^{10}\right| = 3.6\times10^{9},$$ $$\left|\omega L - \frac{1}{\omega C}\right| = 0.05 \times \frac{3.6\times10^{9}}{8\times10^{4}} = 0.05 \times 4.5\times10^{4} = 2.25\times10^{3}\;\Omega.$$ Thus from (2) $$R = 0.05006 \times 2250 \approx 1.13\times10^{2}\;\Omega \approx 113\;\Omega.$$

Using $$I_0 = \dfrac{V_0}{R}$$ from (1): $$I_0 = \frac{45}{113}\,{\rm A} \approx 0.40\;{\rm A} = 400\;{\rm mA}.$$ So, P $$\to$$ 400 mA corresponding to List-II (3).

The quality factor is $$Q = \frac{\omega_0 L}{R} = \frac{10^{5}\times0.05}{113} \approx 44.4.$$ Thus, Q $$\to$$ 44.4 corresponding to List-II (1).

The bandwidth is $$\Delta\omega = \frac{\omega_0}{Q} = \frac{10^{5}}{44.4} \approx 2.25\times10^{3}\;{\rm rad\,s^{-1}}.$$ Hence, R $$\to$$ 2250 rad s$$^{-1}$$ corresponding to List-II (4).

At resonance the voltage and current are in phase, so the instantaneous power is $$p = v\,i.$$ Its maximum (peak) value is $$P_{\text{peak}} = V_0\,I_0 = 45 \times 0.40 = 18\;{\rm W}.$$ Therefore, S $$\to$$ 18 W corresponding to List-II (2).

Collecting the matches:
P $$\to$$ 3, Q $$\to$$ 1, R $$\to$$ 4, S $$\to$$ 2.

Hence the correct option is Option B: P $$\to$$ 3, Q $$\to$$ 1, R $$\to$$ 4, S $$\to$$ 2.