Question

in a capacitor in an AC circuit the voltage leads​

Answer 1

Answer:

We say that current leads the voltage across a capacitor by 90∘. In the graph VC(t)=Vmaxsinωt and so the current is I(t)=ωCVmaxcosωt with a peak current Imax=ωCVmax.

Answer 2

Answer:

When an alternating voltage is put across a capacitor, the current is 90 degrees behind the voltage. This signifies that the current exceeds the voltage by a quarter of a cycle.

Explanation:

The derivation for this is:

Considering phase$=\frac{\pi }{2}$

If,$V=V_{o} sin{\omega}t$

Since,

$q=CVq=CV_{o} sin{\omega}t$

So, now the current is,

$I= \frac{dq}{dt}$

$I=\frac{d(CV_{0} sin{\omega}t)}{dt}$

$I={\omega}CV_{0} cos{\omega}t\\I={\omega}CV_{0} sin({\omega}t+\frac{\pi}{2} )$

Therefore, the current is ${\omega}CV_{0}sin({\omega}t+ \frac{\pi}{2} )$