MCQEasyJEE 2023LCR Circuits & Resonance

JEE Physics 2023 Question with Solution

In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes xx times its initial resonant frequency ω0\omega_0. The value of xx is:

  • A

    1/41/4

  • B

    1616

  • C

    1/161/16

  • D

    44

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: In an LC oscillator, inductance becomes 2L2L and capacitance becomes 8C8C.

Find: The factor xx such that the new resonant frequency is xω0x\omega_0.

The resonant frequency of an LC circuit is given by

ω0=1LC\omega_0 = \frac{1}{\sqrt{LC}}

When L2LL \to 2L and C8CC \to 8C, the new resonant frequency is

ω=12L8C=116LC=14ω0\omega = \frac{1}{\sqrt{2L \cdot 8C}} = \frac{1}{\sqrt{16LC}} = \frac{1}{4}\omega_0

Thus, the value of xx is 1/41/4. The correct option is A.

The resonant frequency of an LC oscillator decreases as the inductance and capacitance increase, since ω01/LC\omega_0 \propto 1/\sqrt{LC}.

Common mistakes

  • Using direct proportionality with LCLC instead of inverse square-root dependence is incorrect. The resonant frequency varies as ω01/LC\omega_0 \propto 1/\sqrt{LC}, so increasing LL and CC decreases the frequency. Always apply the square root in the denominator.

  • Multiplying the change factors incorrectly is a common error. Here, LL becomes 2L2L and CC becomes 8C8C, so the product becomes 16LC16LC, not 10LC10LC. First combine the scaling factors correctly, then take the square root.

  • Forgetting to compare the new frequency with the initial frequency leads to confusion about xx. The question asks for the multiplicative factor relative to ω0\omega_0, so rewrite the new result as a multiple of ω0\omega_0 before choosing the option.

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