Have you read the previous 18 posts? If not, I would strongly suggest that those will provide your answer.

 

As we all know that when a capacitor is allowed to charge at the starting instant the maximum amount of current flows through it and the value of voltage at the very instant is very low
and according to the properties of phasors the y component of phasor gives the magnitude of the quantity thus current is perpendicular to the voltage.. Mathematically It can be explained as (Q = VC).
dQ=CdV

Current (I) = dQ/dt.

So I = CdV/dt.

If V is a sine curve, then I is the slope of the sine curve, which leads it by 90°.

 

Sir I need further explaination because this topic is realy ambiguous to me I want to clear it as far as possible. Would you like to explain it in a new way, by using a mathematical expression ??

 

This zombie thread won't stay dead!

 

Sir I need further explaination because this topic is realy ambiguous to me I want to clear it as far as possible. Would you like to explain it in a new way, by using a mathematical expression ??

What is it that is ambiguous? Given your previous post, you seem to understand that, from the differential equation that governs the relationship between the voltage and the current in a capacitor that if the voltage is a sine wave that the current is a sine wave that leads the voltage waveform by 90°.

What is it that you find ambiguous?

 

Why current leads voltage by 90 degree in capacitor?

This zombie thread won't stay dead!

That is because March 22nd leads April 1st by 10 days.

 

If you think of a resistor in series with a capacitor.
When say 10 volts if applied to the resistor to ground there is at the start 0 volts voltage across capacitor but maximum 10/ R current into the capacitor.
So it can be seen the current is at a maximum while the voltage is at a minimum.
After a period of time the voltage will be maximum across the capacitor and current will be zero.

So this means the voltage and current are out of phase with each other.