A capacitor of capacitance 150.0 $$\mu$$F is connected to an alternating source of emf given by $$E = 36 \sin(120\pi t)$$ V. The maximum value of current in the circuit is approximately equal to:

A capacitor of $$C = 150$$ $$\mu$$F is connected to an alternating voltage source described by $$E = 36\sin(120\pi t)$$ V.

From this expression, the peak emf is $$E_0 = 36$$ V, the angular frequency is $$\omega = 120\pi$$ rad/s, and the capacitance is $$C = 150 \times 10^{-6}$$ F.

Next, the capacitive reactance is given by

$$X_C = \dfrac{1}{\omega C} = \dfrac{1}{120\pi \times 150 \times 10^{-6}} = \dfrac{1}{120 \times 150 \times \pi \times 10^{-6}}$$

$$= \dfrac{1}{18000\pi \times 10^{-6}} = \dfrac{1}{0.018\pi} = \dfrac{1000}{18\pi} = \dfrac{500}{9\pi}$$

Having found $$X_C,$$ the maximum current is

$$I_0 = \dfrac{E_0}{X_C} = \dfrac{36}{\dfrac{500}{9\pi}} = \dfrac{36 \times 9\pi}{500} = \dfrac{324\pi}{500}$$

Using $$\pi \approx 3.14$$, one obtains $$I_0 \approx \dfrac{324 \times 3.14}{500} = \dfrac{1017.36}{500} \approx 2.03$$ A.

Thus the current is approximately 2 A, giving the final result $$2$$ A.