An inductor of inductance 2 is connected in series with a resistance… | JEE Main 2023 (29 Jan, Shift 2)

Q55JEE Main 2023 · 29 JAN · SHIFT 2Easy

Resonance, Bandwidth & Q Factor

An inductor of inductance 2 $\mu\text{H}$ is connected in series with a resistance, a variable capacitor and an AC source of frequency 7 kHz. The value of capacitance for which maximum current is drawn into the circuit $\frac{1}{x}\text{ F}$ , where the value of $x$ is ___________.

(Take $\pi = \frac{22}{7}$ )

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

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Answer: 3872

Maximum current at resonance: $C = \dfrac{1}{4\pi^2 f^2 L}$ , so $x = 4\pi^2 f^2 L$
$x = 4 \left(\dfrac{22}{7}\right)^2 (7\times10^3)^2 (2\times10^{-6}) = 4 \times \dfrac{484}{49} \times 49 \times 2$
$x = 4 \times 484 \times 2 = 3872$
So the answer is 3872.

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