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 $i=dq/dt$ is the instantaneous current in the circuit so,

$$\begin{array}{rl}i=& \frac{dq}{dt}=\frac{d}{dt}(C{V}_0\sin \omega t)\\ & =C{V}_0\omega \cos \omega t\\ & =\frac{V_0}{(1/\omega C)}\cos \omega t\\ & =i_0\cos \omega t=i_0\sin (\omega t+\frac{\pi }{2})\end{array}$$

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 $\underset{―}{\underset{―}{s}\text{, we see that in a perfect capacitor current leads the emf by a phase }}$ 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

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

Phase difference (between voltage and current):

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

Power :

$$P=0$$

Power factor :

$$\cos (\varphi )=0$$

Time difference:

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