JEE Main 2023 — Alternating Current Question with Solution

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 $$2\text{A}$$.