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A series LCR circuit is connected to a $45 \sin (ωt)$ 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 $ω = 8 \times 10^{4} rad s^{- 1},$ 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

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$P \to 3, Q \to 1, R \to 4, S \to 2$

$P \to 4, Q \to 5, R \to 3, S \to 1$

$P \to 4, Q \to 2, R \to 1, S \to 5$

$P \to 2, Q \to 3, R \to 5, S \to 1$

answer is B.

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Detailed Solution

$\begin{array}{l} ω_{0} = \frac{1}{\sqrt{LC}} \\ C = 2 \times 10^{- 9} F I_{0} = \frac{V_{0}}{R} \\ \frac{5}{100} I_{0} = \frac{V_{0}}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}} ∴ R = \frac{9 \times 10^{3}}{4 \sqrt{399}} Ω \\ R \approx 112.5 Ω ∴ I_{0} = \frac{V_{0}}{R} = 400 mA \\ Q = \frac{1}{R} \sqrt{\frac{L}{C}} = 44.4 Q = \frac{ω_{r} .}{Band width} \Rightarrow Bandwidth = \frac{ω_{r}}{Q} \\ Peak power = \frac{V_{0}^{2}}{R} = 18 watt \end{array}$

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