Question

A small signal voltage $V\left( t \right) = {V_0}\sin \omega t$ is applied across an ideal capacitor C. Then:
A. current I(t) lags voltage V(t) by 90$^\circ $
B. over a full cycle the capacitor C does not consume any energy from the voltage source
C. current I(t) is in phase with voltage V(t)
D. current I(t) leads voltage V(t) by 180$^\circ $

Answer 1

Hint: First, we need to calculate expression for current from voltage and check the phase angle between the current and the voltage. Then we also need to calculate the power consumed by the capacitor from which we can calculate the work done by the capacitor.

Formula used:
The current in a capacitor is given as
$I\left( t \right) = C\dfrac{{dV\left( t \right)}}{{dt}}$
The power consumed by a capacitor is given as
$P = {V_{rms}}{I_{rms}}\cos \phi $

Complete step-by-step answer:
We are given a voltage signal which is applied to an ideal capacitor having capacitance C. The voltage signal is given as
$V\left( t \right) = {V_0}\sin \omega t$ …(i)
We know that in a capacitor, the current in the capacitor can be calculated by the following formula.
$I\left( t \right) = C\dfrac{{dV\left( t \right)}}{{dt}}$
Inserting the value of voltage in it, we get
$I\left( t \right) = C\dfrac{d}{{dt}}\left( {{V_0}\sin \omega t} \right) = C{V_0}\omega \cos \omega t$
This is the expression for the current which can be written as follows:
$I\left( t \right) = C{V_0}\omega \sin \left( {\omega t + 90^\circ } \right)$ …(ii)
Now when we compare equation (i) and (ii), we notice that the current is leading the voltage by a phase angle of 90$^\circ $. So, we can say that options A, C and D are wrong. Now let us check the accuracy of option B. For this, we need to calculate the power consumed by the capacitor which is given as
$P = {V_{rms}}{I_{rms}}\cos \phi $
Here $\phi $ is the phase angle between current and voltage which we have obtained to be 90$^\circ $. Therefore, we have
$P = {V_{rms}}{I_{rms}}\cos 90^\circ = 0$
Now the work done by the capacitor can be calculated as follows:
$W = P \times t = 0$
Hence, the capacitor does not take any energy from the source, so, the correct answer is option B.

So, the correct answer is “Option B”.

Note: It should be noted that there is a phase difference between the current and the voltage which results in the voltage lagging behind the current in the capacitor. This phase difference of 90$^\circ $ ensures that the power consumed and hence the work done by the capacitor for any duration is zero.