JEE Main 2023 — Alternating Current Question with Solution

The maximum current in an LC circuit can be found using the following formula related to simple harmonic motion:

$I_{\text{max}}=\omega Q_{\text{max}}$,

where:

• $I_{\text{max}}$ is the maximum current,
• $\omega$ is the angular frequency, and
• $Q_{\text{max}}$ is the maximum charge on the capacitor.

The angular frequency $\omega$ for an LC circuit is given by:

$\omega =\frac{1}{\sqrt{LC}}$,

where:

• $L$ is the inductance, and
• $C$ is the capacitance.

Given that $L=75\text{mH}=75\times 10^{-3}\text{H}$, $C=1.2\mu \text{F}=1.2\times 10^{-6}\text{F}$,

and $Q_{\text{max}}=2.7\mu \text{C}=2.7\times 10^{-6}\text{C}$,

we can substitute these values into the formulas to find $I_{\text{max}}$:

$\omega =\frac{1}{\sqrt{(75\times 10^{-3})(1.2\times 10^{-6})}}=3333.33\text{rad/s}$,

$I_{\text{max}}=\omega Q_{\text{max}}=3333.33\times 2.7\times 10^{-6}=0.009\text{A}$.

Therefore, the maximum current in the circuit is $0.009\text{A}$, or equivalently, $9\text{mA}$.