A series LCR circuit is connected to a Volt source. The resonant angular frequency of the circuit is and current amplitude at resonance is . When the angular frequency of the source is the current amplitude in the circuit is . If , match each entry in

Question

A series LCR circuit is connected to a $45 \sin (ω t)$ 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 $ω = 8 \times 10^{4} rad s^{- 1},$ the current amplitude in the circuit is $0.05 I_{0}$ . If $L = 50 mH$ , match each entry in List-I with appropriate value from List-II and choose the correct option. [2023]

	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

	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

Smart Achievers · Accepted Answer

(2)

As, $V = V_{0} \sin ω t$ and the resonant angular frequency is

$ω_{0} = \frac{1}{\sqrt{L C}} \Rightarrow ω_{0} = 10^{5} rad/s = \frac{1}{\sqrt{L C}} \Rightarrow C = \frac{1}{L ω_{0}^{2}} = \frac{1}{5 \times 10^{- 2} \times 10^{10}} = 2 \times 10^{- 9} F$

$[∵ L = 50 mH = 5 \times 10^{- 2} H]$

$I_{0} = \frac{V_{0}}{R} = \frac{45}{R}$ $[∵ V = 45 \sin ω t is given]$

$ω = 8 \times 10^{4} rad/s = 0.8 ω_{0}$

$I = 0.05 I_{0} = \frac{I_{0}}{20} \Rightarrow Z = 20 R$

$X_{L} = L ω = 8 \times 10^{4} \times 5 \times 10^{- 2} = 4 k Ω$

$X_{C} = \frac{1}{C ω} = \frac{1}{8 \times 10^{4} \times 2 \times 10^{- 9}} = \frac{1}{16} \times 10^{5} Ω = \frac{25}{4} k Ω$

$Z^{2} = R^{2} + {(X_{C} - X_{L})}^{2}$ or, $400 R^{2} = R^{2} + {(\frac{9}{4} k Ω)}^{2}$

$∴$ $R = \frac{\frac{9}{4} k Ω}{\sqrt{399}} = \frac{9}{80} k Ω = \frac{900}{8} Ω$

$I_{0} = \frac{V_{0}}{R} = \frac{45 \times 8}{900} = \frac{8}{20} A \approx 0.4 A = 400 mA$

Quality factor,

$Q = \frac{1}{R} \sqrt{\frac{L}{C}} = \frac{8}{900} \sqrt{\frac{5 \times 10^{- 2}}{2 \times 10^{- 9}}} = \frac{8}{900} \sqrt{25 \times 10^{6}}$

$= \frac{8}{900} \times 5000 = 44.4$

$Q = \frac{ω_{0}}{Δ ω}$

$∴$ $Bandwidth, Δ ω = \frac{ω_{0}}{Q} = \frac{10^{5}}{44.4} = 2250.0 rad/s$

Peak power dissipated, $P_{\max} = I_{0}^{2} R = \frac{45^{2}}{R^{2}} \times R = \frac{45^{2}}{R}$

$= \frac{45^{2}}{900} \times 8 = 18.4 W \approx 18 W$

Solution

1 P → 2, Q → 3, R → 5, S → 1

2 P → 3, Q → 1, R → 4, S → 2

3 P → 4, Q → 5, R → 3, S → 1

4 P → 4, Q → 2, R → 1, S → 5

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