AC Voltage Applied To A Capacitor MCQ - Practice Questions with Answers

AC voltage applied to a capacitor:

 The circuit containing alternating voltage source V=V₀sinωt is connected to a capacitor of capacitance C.

Suppose at any time t, q be the charge on the capacitor and i be the current in the circuit. Since there is no resistance

in the circuit, so the instantaneous potential drop $q / C$ across the capacitor must be equal to applied alternating voltage so,

$$
\frac{q}{C}=V_0 \sin \omega t
$$

Since $i=d q / d t$ is the instantaneous current in the circuit so,

$$
\begin{aligned}
i= & \frac{d q}{d t}=\frac{d}{d t}\left(C V_0 \sin \omega t\right) \\
& =C V_0 \omega \cos \omega t \\
& =\frac{V_0}{(1 / \omega C)} \cos \omega t \\
& =i_0 \cos \omega t=i_0 \sin \left(\omega t+\frac{\pi}{2}\right)
\end{aligned}
$$

Where,

$$
i_0=\frac{V_0}{(1 / \omega C)} \text { is the peak value of current. }
$$

Comparing equation of current with $\mathrm{V}=\mathrm{V}_0$ sin $\underline{\underline{s} \text {, we see that in a perfect capacitor current leads the emf by a phase }}$ angle of $\pi / 2$.

Again comparing peak value of current with ohm's law ,we find that quantity 1/ωC has the dimension of the resistance. 

Thus the quantity

$$
X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C} \text { is known as capacitive reactance. }
$$

Phase difference (between voltage and current):

$$
\phi=-\frac{\pi}{2}
$$

Power :

$$
P=0
$$

Power factor :

$$
\cos (\phi)=0
$$

Time difference:

$$
\mathrm{T} . \mathrm{D}=\frac{T}{4}
$$