Q 12-07-071JEE MainJEE Main 2023 (29 Jan, Shift 2)Medium
An inductor of inductance $2\ \mu\text{H}$ is connected in series with a resistance, a variable capacitor and an AC source of frequency $7\ \text{kHz}$. The value of capacitance for which maximum current is drawn into the circuit is $\dfrac1x\ \text{F}$, where the value of $x$ is ______. (Take $\pi=\frac{22}{7}$)
Numerical value type. Enter your answer.
Answer: 3872
Maximum current at resonance: $\omega^2=\dfrac{1}{LC}$, so
$$C=\frac{1}{4\pi^2f^2L}=\frac{1}{4\times\frac{484}{49}\times49\times10^6\times2\times10^{-6}}=\frac{1}{4\times484\times2}=\frac{1}{3872}\ \text{F}$$
Solution by Sreeraj P, M.Sc Physics