QJEEM23Ph1ab9: JEE Main 2023 Physics - Class 12 JEEM23Ph1ab9 Solution: A capacitor of capacitance is connected to an alternating…

For a capacitor connected to an AC source, the maximum current $I_\text{max}$ can be calculated using the formula:

$I_\text{max} = E_\text{max} \cdot \omega C$

where $E_\text{max}$ is the maximum voltage, $\omega$ is the angular frequency, and $C$ is the capacitance.

Given the emf equation: $E = 36 \sin(120\pi t) \, \text{V}$ , we can determine that $E_\text{max} = 36\, \text{V}$ and $\omega = 120\pi \, \text{rad/s}$ .

The capacitance is given as $150.0\, \mu\text{F} = 150.0 \times 10^{-6}\, \text{F}$ .

Now, we can calculate the maximum current:

$I_\text{max} = 36 \cdot (120\pi) \cdot (150.0 \times 10^{-6})$

$I_\text{max} \approx 2\, \text{A}$

Thus, the correct answer is $2\, \text{A}$ .