Question:

A series LCR circuit is connected to a 45 sin(πœ”π‘‘) Volt source. The resonant angular frequency of the circuit is \(10^5\) rad sβˆ’1 and current amplitude at resonance is I0

When the angular frequency of the source is πœ” =\(8 Γ—\) \(10^4\) rad sβˆ’1, the current amplitude in the circuit is 0.05 I0. 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) I0 in mA (1)  44.4
(Q) The quality factor of the circuit(2) 18 
(R) The bandwidth of the circuit in rad sβˆ’1(3) 400
(S) The peak power dissipated at resonance in Watt(4) 2250 
 (4) 2250

Updated On: May 23, 2024
  • P β†’2, Qβ†’ 3, R β†’5, S β†’1 

  • P β†’3, Qβ†’ 1, R β†’4, S β†’2 

  • P β†’4, Qβ†’ 5, R β†’3, S β†’1 

  • P β†’4, Qβ†’ 2, R β†’1, S β†’5 

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The Correct Option is B

Solution and Explanation

Correct option is(B): P β†’ 3, Q β†’ 1, R β†’ 4, S β†’ 2.

As given 

\(0.05l_0=\frac{45}{\sqrt{R^2+(0.8X_{LO}-\frac{5}{4}X_{C0}})^2}\)

Where XLO=XCO are at resonant frequencies

on solving, \(R β‰… \frac{450Ξ©}{4}β‡’l0β‰…400 \,mA\)

Quality factor\( Q= \frac{1}{R}\sqrt{\frac{L}{C}}β‰…44.44\)

\(Q=\frac{Ο‰_0}{β–³Ο‰}β‡’ β‰…2250\,rad/s\)

peak power = \(45Γ—\frac{400}{1000}\,W\) 

=18.

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Concepts Used:

LCR Circuit

An LCR circuit, also known as a resonant circuit, or an RLC circuit, is an electrical circuit consist of an inductor (L), capacitor (C) and resistor (R) connected in series or parallel.

Series LCR circuit

When a constant voltage source is connected across a resistor a current is induced in it. This current has a unique direction and flows from the negative to positive terminal. Magnitude of current remains constant.

Alternating current is the current if the direction of current through this resistor changes periodically. An AC generator or AC dynamo can be used as AC voltage source.