1. Given data

• Capacitance: $C = 150.0\,\mu\text{F} = 150 \times 10^{-6}\,\text{F}$
• Applied emf: $E = 36\sin(120\pi t)\,\text{V}$

Comparing with the standard form $E = E_0 \sin(\omega t),$ we get:

• Peak voltage: $E_0 = 36\,\text{V}$
• Angular frequency: $\omega = 120\pi\,\text{rad/s}$

2. Formula for maximum current in a purely capacitive AC circuit

For a capacitor, $I_0 = \omega C E_0$

because capacitive reactance is $X_C = \frac{1}{\omega C},$ and hence $I_0 = \frac{E_0}{X_C} = E_0\omega C.$

3. Substitute the values

$I_0 = (120\pi)(150\times 10^{-6})(36)$

Now simplify:

$150\times 36 = 5400$

So, $I_0 = 120\pi \times 5400 \times 10^{-6}$

$5400 \times 10^{-6} = 5.4\times 10^{-3}$

Thus, $I_0 = 120\pi \times 5.4\times 10^{-3}$

$120 \times 5.4\times 10^{-3} = 0.648$

Therefore, $I_0 = 0.648\pi$

Using $\pi \approx 3.14$, $I_0 \approx 0.648 \times 3.14 \approx 2.04\,\text{A}$

4. Approximate answer

$I_0 \approx 2\,\text{A}$

5. Option check

• A: $\frac{1}{\sqrt{2}}\,\text{A} \approx 0.707\,\text{A}$
• B: $2\sqrt{2}\,\text{A} \approx 2.828\,\text{A}$
• C: $\sqrt{2}\,\text{A} \approx 1.414\,\text{A}$
• D: $2\,\text{A}$

So the correct option is D.