Question

Derive the expression for the current flowing in an ideal capacitor

and its reactance when connected to an ac source of voltage

V=V0sinωt​

Answer 1

To derive:

The expression for the current flowing in an ideal capacitor and its reactance when connected to an ac source of voltage V=V_(0)sinωt

Derivation:

Let's consider that a capacitor C has been connected to an alternating current source.

So, the charge developed in the capacitor will be:

$\sf \therefore \: q = C \times V$

$\sf \implies \: q = C \times V_{0} \sin( \omega t)$

$\sf \implies \: q = C V_{0} \sin( \omega t)$

Now, current passing through the circuit :

$\sf \therefore \: i = \dfrac{dq}{dt}$

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

$\sf \implies \: i= CV_{0} \times \dfrac{d \bigg \{ \sin( \omega t) \bigg \} }{dt}$

$\sf \implies \: i= CV_{0} \omega \times \cos( \omega t)$

$\sf \implies \: i= CV_{0} \omega \times \sin( \omega t + \dfrac{\pi}{2} )$

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

If , we denote the reactance by X_(C):

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

Hence, reactance is :

$\boxed{ \bf \therefore \: X_{C} = \dfrac{1}{ \omega C} }$

Hope It Helps.