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NEET PHYSICSALTERNATING CURRENTMedium

Question

A series LCR circuit containing 5.0 H inductor, 80 \mu F capacitor and 40 \Omega resistor is connected to 230 V variable frequency AC source. The angular frequencies of the source at which power transferred to the circuit is half the power at the resonant angular frequency are likely to be:

A

46 rad/s and 54 rad/s

B

42 rad/s and 58 rad/s

C

25 rad/s and 75 rad/s

D

50 rad/s and 25 rad/s

Step-by-Step Solution

To find the angular frequencies where the power is half of the maximum power (resonant power), we use the resonant frequency (ω0\omega_0) and the half-power bandwidth (Δω\Delta \omega).

  1. Resonant angular frequency (ω0\omega_0): ω0=1LC=15.0×80×10−6=1400×10−6=12×10−2=50\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{5.0 \times 80 \times 10^{-6}}} = \frac{1}{\sqrt{400 \times 10^{-6}}} = \frac{1}{2 \times 10^{-2}} = 50 rad/s.
  2. Half-bandwidth (Δω\Delta \omega): The frequencies at which power is halved occur at ω=ω0±Δω\omega = \omega_0 \pm \Delta \omega, where Δω=R2L\Delta \omega = \frac{R}{2L}. Substituting the values: Δω=402×5.0=4\Delta \omega = \frac{40}{2 \times 5.0} = 4 rad/s.
  3. Calculation: The required angular frequencies are ω1=50−4=46\omega_1 = 50 - 4 = 46 rad/s and ω2=50+4=54\omega_2 = 50 + 4 = 54 rad/s.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ALTERNATING CURRENT. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSALTERNATING CURRENTseriescircuitcontaininginductorcapacitor

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