Why is there a phase difference in RC circuit?

Question

In a capacitor circuit,the there is 90 degree phase shift between current and voltage in capacitor(when supplied sinusoidalvoltage and it varies from 0 to 90 degrees in a RC circuit.Can somebody tell me why exactly there is a phase difference between 0 to 90 degrees in a RC circuit while a purely capacitive circuit has a phase difference of 90 degrees

Answer 1

This phenomenon can be explained by taking a look at the electrical impedance, which is caused by the R and C components. The electrical impedance $Z$ is composed as \begin{align} Z = R + i X \end{align} where $X$ is the reactance. This is for capacitors $X_C = -\frac{1}{\omega C}$ and for inductors $X_L = \omega L$. We can make a sketch of this in a complex plane, which looks like

Now you can see, that it depends on the ratio of $R$ and $X$ what the phase difference $\phi$ between the real electrical current ($R$ - direction) and the voltage ($Z$ - direction). The actual dependence being \begin{align} \varphi= \arctan{\left(\frac{X}{R}\right)} \, . \end{align} (in the electrical engineers notation with $j$ as the imaginary unit) From this you can also see that if you have nearly no electrical resistance $R$ you end up with a $90°$ phase difference.

I hope this answers your questions.

Answer 2

Let us apply a sinus formed voltage $u$ over a capacitor $C$: $$u = U_0 \sin \omega t.$$

The charge $q$ in the capacitor is given by $$q = C u = CU_0 \sin \omega t.$$

The current $i$ through the capacitor is then given by $$i = \frac{dq}{dt} = C U_0 \omega \cos \omega t = I_0 \cos \omega t,$$ where $I_0 = C U_0 \omega.$

Now you should be able to see the phase shift since $\sin$ and $\cos$ have a phase shift of $90^\circ.$