Why is there a phase difference in RC circuit? 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
 A: 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.
A: 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 $90^\circ.$
