For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is 2.5 nF. If resistance of 200 Ω and 100 mH inductor is being used in the given circuit. The frequency of ac source is _________ × 10³ Hz (given π² = 10) Numerical answer type. Work it out, then open the answer.

Answer: 10. Maximum current at resonance: f = (1)/(2π√(LC)) = (1)/(√(4π² LC)) 4π² LC = 40 × (0.1)(2.5×10⁻⁹) = 10⁻⁸ f = 1√10⁻⁸ = 10⁴ Hz = 10×10³ Hz So the answer is 10.

For a given series LCR circuit it is found that maximum current is… | JEE Main 2024 (6 Apr, Shift 2)

Q40JEE Main 2024 · 6 APR · SHIFT 2Easy

Resonance, Bandwidth & Q Factor

For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is $2.5\text{ nF}$ . If resistance of $200\ \Omega$ and $100\text{ mH}$ inductor is being used in the given circuit. The frequency of ac source is _________ $\times 10^{3}\text{ Hz}$ (given $\pi^2 = 10$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 10

Maximum current at resonance: $f = \dfrac{1}{2\pi\sqrt{LC}} = \dfrac{1}{\sqrt{4\pi^2 LC}}$
$4\pi^2 LC = 40 \times (0.1)(2.5\times10^{-9}) = 10^{-8}$
$f = \dfrac{1}{\sqrt{10^{-8}}} = 10^{4}\text{ Hz} = 10\times10^{3}\text{ Hz}$
So the answer is 10.

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