Question 157: AC fundamentals

Question

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 :

Accepted Answer

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

Answer

$$\frac{1}{\sqrt{2}} A$$

Answer

$$2 \sqrt{2} A$$

Answer

$$\sqrt{2} A$$

Answer

$$2 A$$

AC fundamentals

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 :

Choose the correct answer: