JEE MAIN - Physics (2023 - 6th April Evening Shift - No. 8)
A capacitor of capacitance $$150.0 ~\mu \mathrm{F}$$ is connected to an alternating source of emf given by $$\mathrm{E}=36 \sin (120 \pi \mathrm{t}) \mathrm{V}$$. The maximum value of current in the circuit is approximately equal to :
$$\frac{1}{\sqrt{2}} A$$
$$2 \sqrt{2} A$$
$$\sqrt{2} A$$
$$2 A$$
Explanation
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 $$2\, \text{A}$$.
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