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NEET 2022
Exam
NEET
Year
2022
Subject
Physics
Chapter
Alternating Current

NEET 2022 · Physics · Alternating Current

A series LCR circuit with inductance 10H , capacitance 10 ,...

A series LCR circuit with inductance 10H10H, capacitance 10\muF10\muF, resistance 50Ω50Ω is connected to an AC source of voltage, V=200sinV = 200sin\left(100t\right)volt.IftheresonantfrequencyoftheLCRcircuitisvolt. If the resonant frequency of the LCR circuit is{ν}_{0} and the frequency of the AC source is νν, then

Options

  1. Aν0=ν=50πHz{ν}_{0} = ν = \dfrac{50}{\pi}Hz
  2. Bν0=50πHz,ν=50Hz{ν}_{0} = \dfrac{50}{\pi}Hz, ν = 50Hz
  3. Cν=100Hz;ν0=100πHzν = 100Hz; {ν}_{0} = \dfrac{100}{\pi}Hz
  4. Dν0=ν=50Hz{ν}_{0} = ν = 50Hz

Correct answer

A. ν0=ν=50πHz{ν}_{0} = ν = \dfrac{50}{\pi}Hz

Step-by-step solution

From the given equation V=200sin⁡100tV = 200\sin 100t, the angular frequency of the source is

ω=100rads−1\omega = 100rad{s}^{- 1}

⇒ν=ω2π=1002π=50πHz⇒ ν = \dfrac{\omega}{2\pi} = \dfrac{100}{2\pi} = \dfrac{50}{\pi}Hz

For resonance

ν0=12πLC=12π10×10×10−6{ν}_{0} = \dfrac{1}{2\pi\sqrt{LC}} = \dfrac{1}{2\pi\sqrt{10 \times 10 \times {10}^{- 6}}}

⇒ν0=1002π=50πHz⇒ {ν}_{0} = \dfrac{100}{2\pi} = \dfrac{50}{\pi}Hz

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