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In a purely capacitive circuit it is $\pi/2$. But in a RC circuit $$V=v \sin \omega t(1-e^{-t/RC})$$ while $$I=i \sin\omega t (e^{-t/RC}).$$ For such a function how to calculate phase difference?

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  • $\begingroup$ Do you know about phasor diagrams? $\endgroup$ – Bob D Aug 22 at 20:27
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    $\begingroup$ This doesn't make much sense to me. In AC steady state, the voltage and current amplitudes are constant in time, and the phase difference between voltage and current is well defined. But these are not AC steady state solutions. There is obviously come kind of transient response involved and the current decays to zero! $\endgroup$ – Hal Hollis Aug 22 at 20:29
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Your solutions are not for steady state conditions but for transient conditions after some kind of turn on or turn off. I am not clear what conditions they apply to since they imply the current falls to zero as $t\to\infty$.

For steady state AC, you can easily find the solution using the impedance concept.

The impedance of a resistor and capacitor in series is given by

$$Z = R + 1/j\omega C$$

and the voltage-current relationship is given by

$$\tilde{v}=Z\tilde{i}$$

so that the phase difference between voltage and current is just $\angle{Z}$.

As you can see from the above, this will depend on the magnitudes of $R$ and $C$, and the operating frequency.

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