Match Parameters of Series LCR Circuit Driven by Sinusoidal Voltage

To find the matching entries, we analyze the series $LCR$ circuit driven by the voltage $V(t) = 45 \sin(\omega t)\text{ V}$.

1. Calculation of Resistance ($R$) and Capacitance ($C$)

The given parameters are:

• Voltage amplitude, $V_0 = 45\text{ V}$
• Resonant angular frequency, $\omega_0 = 10^5\text{ rad s}^{-1}$
• Inductance, $L = 50\text{ mH} = 0.05\text{ H}$

At resonance: $\omega_0 = \frac{1}{\sqrt{LC}} \implies C = \frac{1}{\omega_0^2 L} = \frac{1}{(10^5)^2 \times 0.05} = 2 \times 10^{-9}\text{ F}$

At source frequency $\omega = 8 \times 10^4\text{ rad s}^{-1}$:

• Inductive reactance: $X_L = \omega L = (8 \times 10^4) \times 0.05 = 4000\ \Omega$
• Capacitive reactance: $X_C = \frac{1}{\omega C} = \frac{1}{(8 \times 10^4) \times (2 \times 10^{-9})} = 6250\ \Omega$
• Difference in reactance: $|X_L - X_C| = |4000 - 6250| = 2250\ \Omega$

The current amplitude at frequency $\omega$ is given as: $I = \frac{V_0}{\sqrt{R^2 + (X_L - X_C)^2}} = 0.05 I_0 = \frac{1}{20} \left(\frac{V_0}{R}\right)$

Equating the impedances: $\sqrt{R^2 + (2250)^2} = 20R$ $R^2 + 2250^2 = 400 R^2 \implies 399 R^2 = 2250^2$ $R = \frac{2250}{\sqrt{399}} \approx \frac{2250}{19.975} \approx 112.64\ \Omega \approx 112.5\ \Omega$

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2. Evaluation of List-I Items

(P) Current amplitude at resonance ($I_0$) in mA: $I_0 = \frac{V_0}{R} \approx \frac{45}{112.5}\text{ A} = 0.4\text{ A} = 400\text{ mA}$ $\implies (P) \rightarrow (3)$

(Q) Quality factor ($Q$) of the circuit: $Q = \frac{\omega_0 L}{R} = \frac{10^5 \times 0.05}{112.5} = \frac{5000}{112.5} \approx 44.44$ $\implies (Q) \rightarrow (1)$

(R) Bandwidth of the circuit in $\text{rad s}^{-1}$: $\text{Bandwidth} = \frac{\omega_0}{Q} = \frac{R}{L} \approx \frac{112.5}{0.05} = 2250\text{ rad s}^{-1}$ $\implies (R) \rightarrow (4)$

(S) Peak power dissipated at resonance in Watt: At resonance, the phase angle $\phi = 0$, so: $P_{\text{peak}} = V_0 I_0 = 45 \times 0.4 = 18\text{ W}$ $\implies (S) \rightarrow (2)$

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Conclusion

Matching the results:

• $P \rightarrow 3$
• $Q \rightarrow 1$
• $R \rightarrow 4$
• $S \rightarrow 2$

This corresponds to Option B.