Question

What is the phase difference between voltage and current in an AC
circuit containing pure capacitor?​

Answer 1

Given:

An AC circuit with pure capacitor.

To find:

Phase difference between voltage and current?

Calculation:

$Q = CV$

$\implies Q = C \{V_{0} \sin( \omega t) \}$

$\implies i = \dfrac{dQ}{dt}$

$\implies i = \dfrac{d \: C \{V_{0} \sin( \omega t) \}}{dt}$

$\implies i = \: C V_{0} \dfrac{d \{ \sin( \omega t) \}}{dt}$

$\implies i = \: \omega C V_{0} \cos( \omega t)$

$\implies i = \: \omega C V_{0} \sin( \omega t + \dfrac{\pi}{2} )$

$\implies i = \: \dfrac{ V_{0} }{ (\frac{1}{\omega C} )} \sin( \omega t + \dfrac{\pi}{2} )$

$\implies i = \: \dfrac{ V_{0} }{ \chi_{C}} \sin( \omega t + \dfrac{\pi}{2} )$

So, current leads voltage by 90°.