Derive the expression for the current flowing in an ideal capacitorand

1.	Derive the expression for the current flowing in an ideal capacitorand its reactance when connected to an ac source of voltageV=V0sinωt
Answer» 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.

Derive the expression for the current flowing in an ideal capacitorand its reactance when connected to an ac source of voltageV=V0sinωt

Answer»

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

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.