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Alternating Current question

2023 · 25 Jan · Shift 1 · Q61
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  5. /2023 · 25 Jan · Shift 1 · Q61

Alternating Current question

2023 · 25 Jan · Shift 1 · Q61

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes xxx times its initial resonant frequency ω0\omega_0ω0​. The value of xxx is :
  1. A
    1/4
  2. B
    1/16
  3. C
    4
  4. D
    16
View written solutionFree

Correct answer: A

  1. Resonant frequency of an LC oscillator

    The angular resonant frequency is

    ω=1LC.\omega = \frac{1}{\sqrt{LC}}.ω=LC​1​.

    Initially,

    ω0=1LC.\omega_0 = \frac{1}{\sqrt{LC}}.ω0​=LC​1​.
  2. New values of inductance and capacitance

    Given:

    L′=2L,C′=8C.L' = 2L, \qquad C' = 8C.L′=2L,C′=8C.
  3. New resonant frequency

    ω′=1L′C′=1(2L)(8C).\omega' = \frac{1}{\sqrt{L'C'}} = \frac{1}{\sqrt{(2L)(8C)}}.ω′=L′C′​1​=(2L)(8C)​1​. ω′=116LC=14LC.\omega' = \frac{1}{\sqrt{16LC}} = \frac{1}{4\sqrt{LC}}.ω′=16LC​1​=4LC​1​.

    Since

    ω0=1LC,\omega_0 = \frac{1}{\sqrt{LC}},ω0​=LC​1​,

    therefore,

    ω′=14 ω0.\omega' = \frac{1}{4}\,\omega_0.ω′=41​ω0​.
  4. Find xxx

    We are given

    ω′=x ω0.\omega' = x\,\omega_0.ω′=xω0​.

    Comparing,

    x=14.x = \frac{1}{4}.x=41​.
  5. Option check

    • A: 14\frac{1}{4}41​ ✅
    • B: 116\frac{1}{16}161​ ❌
    • C: 444 ❌
    • D: 161616 ❌

Final Answer: x=14x = \frac{1}{4}x=41​

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