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 :

<p>For a capacitor connected to an AC source, the maximum current $I_\text{max}$ can be calculated using the formula:</p> <p>$I_\text{max} = E_\text{max} \cdot \omega C$</p> <p>where $E_\text{max}$ is the maximum voltage, $\omega$ is the angular frequency, and $C$ is the capacitance.</p> <p>Given the emf equation: $E = 36 \sin(120\pi t) \, \text{V}$, we can determine that $E_\text{max} = 36\, \text{V}$ and $\omega = 120\pi \, \text{rad/s}$.</p> <p>The capacitance is given as $150.0\, \mu\text{F} = 150.0 \times 10^{-6}\, \text{F}$.</p> <p>Now, we can calculate the maximum current:</p> <p>$I_\text{max} = 36 \cdot (120\pi) \cdot (150.0 \times 10^{-6})$</p> <p>$I_\text{max} \approx 2\, \text{A}$</p> <p>Thus, the correct answer is $2\, \text{A}$.</p>