AC Voltage Applied To A Capacitor - Practice Questions & MCQ

AC voltage applied to a capacitor:

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

Suppose at any time $\mathrm{t}, \mathrm{q}$ 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, $i_0=\frac{V_0}{(1 / \omega C)}$ is the peak value of current.

Comparing equation of current  with V=V₀sinωt ,we see that in a perfect capacitor current leads the emf by a phase angle of π/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}
$