A series LCR circuit is connected to a $45 \sin (\omega t)$ Volt source. The resonant angular frequency of the circuit is $10^{5} \mathrm{rad} \mathrm{s}^{-1}$ and current amplitude at resonance is $I_{0}$. When the angular frequency of the source is $\omega=8 \times 10^{4} \mathrm{rad} \mathrm{s}^{-1}$, the current amplitude in the circuit is $0.05 I_{0}$. If $L=50 \mathrm{mH}$, 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 $\mathrm{rad} \mathrm{s}^{-1}$ | (3) | 400 | | (S) | The peak power dissipated at resonance in Watt | (4) | 2250 | | | | (5) | 500 |

(B) $P \rightarrow 3, Q \rightarrow 1, R \rightarrow 4, S \rightarrow 2$

(A) $P \rightarrow 2, Q \rightarrow 3, R \rightarrow 5, S \rightarrow 1$

(C) $P \rightarrow 4, Q \rightarrow 5, R \rightarrow 3, S \rightarrow 1$

(D) $P \rightarrow 4, Q \rightarrow 2, R \rightarrow 1, S \rightarrow 5$

Alternating Current — JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 Question with Solution

A series LCR circuit is connected to a

45 sin (ω t)

Volt source. The resonant angular frequency of the circuit is

1 0^{5} rad s^{- 1}

and current amplitude at resonance is

I_{0}

. When the angular frequency of the source is

ω = 8 \times 1 0^{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

Answer: (B)

Given:

V_{0} = 45

ω_{0} = 1 0^{5}

rad/s,

L = 50

mH =

5 \times 1 0^{- 2}

H, at

ω = 8 \times 1 0^{4}

rad/s,

I = 0.05 I_{0}

.

At resonance:

I_{0} = \frac{V _{0}}{R}

ω = 8 \times 1 0^{4}

I = \frac{V _{0}}{Z}

where

Z = R^{2} + (X_{L} - X_{C})^{2}

Since

I = 0.05 I_{0} = \frac{I _{0}}{20}

, we have

Z = 20 R

X_{L} = ω L = 8 \times 1 0^{4} \times 5 \times 1 0^{- 2} = 4000

C = \frac{1}{L ω _{0}^{2}} = \frac{1}{5 \times 1 0 ^{- 2} \times 1 0 ^{10}} = 2 \times 1 0^{- 9}

X_{C} = \frac{1}{ω C} = \frac{1}{8 \times 1 0 ^{4} \times 2 \times 1 0 ^{- 9}} = 6250

Z^{2} = R^{2} + (6250 - 4000)^{2} = R^{2} + 225 0^{2}

400 R^{2} = R^{2} + 225 0^{2}

R = \frac{2250}{399} \approx 112.5

Ω

(P)

I_{0} = \frac{45}{112.5} = 0.4

A = 400 mA → 3

(Q)

Q = \frac{ω _{0} L}{R} = \frac{1 0 ^{5} \times 5 \times 1 0 ^{- 2}}{112.5} \approx 44.4

→ 1

(R) Bandwidth

= \frac{ω _{0}}{Q} = \frac{1 0 ^{5}}{44.4} \approx 2250

rad/s → 4

(S) Peak power

= \frac{V _{0}^{2}}{2 R} = \frac{4 5 ^{2}}{2 \times 112.5} \approx 9

W... Actually

I_{0}^{2} R = 0. 4^{2} \times 112.5 = 18

W → 2

Answer: (B)

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

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Single correct option