If the AC EMF is 0 at t=0, then how does current pass through a capacitor?

[Zam0la]

Hi, I have a problem understanding the ac capacitive circuit, where Vc = Vm * sin(ωt), and, current leads the voltage, so at t = 0, v = 0 which means no emf, no pushing energy for the electrons, I think, then how current is max? what is the energy source that supplied the electrons with the kinetic energy to move especially the capacitor was still uncharged.

 

[berkeman]

Welcome to the PF.

You should define more of the circuit so it's easier to talk about it. Are you maybe referring to the voltage and current waveforms in an LC "tank" (resonant) circuit?

https://www.electrical4u.com/images/march16/1468320756.PNG

 

[Zam0la]

Thanks and I'm sorry for the confusion, I mean this, the "pure" ac capacitive circuit.
https://www.electronicshub.org/wp-content/uploads/2015/09/AC-Applied-Across-a-Pure-Capacitor.jpg

at the very beginning voltage is zero but we still have a maximum current, what does make the electrons flow though voltage is zero at that moment?

 

[berkeman]

Thanks and I'm sorry for the confusion, I mean this, the "pure" ac capacitive circuit.
View attachment 223894
at the very beginning voltage is zero but we still have a maximum current, what does make the electrons flow though voltage is zero at that moment?

There is no such thing. Be sure to include the output resistance (or impedance) of the signal source in the model. Can you show how that affects things?

EDIT -- Deprecated my first sentence...

 

[Zam0la]

We have studied this case in college, where the prof assumed that the equivalent resistance is zero, and thus we have only capacitive reactance, and I don't how that would affect thing.

 

[berkeman]

Yeah, maybe I over-stated that -- I've deprecated what I said now. Since you are talking about a shift in the current waveform only, maybe it's okay to use an ideal voltage source for that.

But it makes it harder to see physically what is going on. At least when you have a resonant LC tank circuit that is given an initial capacitor charge and let go, you can see how the energy flows back and forth between the L and C with the changing voltages and currents.

which means no emf, no pushing energy for the electrons, I think, then how current is max? what is the energy source that supplied the electrons with the kinetic energy to move especially the capacitor was still uncharged.

So given that circuit, the answer is that the voltage source is supplying the energy to drive the current waveform. The current is maximum at Vc=0 because that is the point where Vc is changing the fastest.

 

[Borek]

Thinks about it this way: t=0 and ωt=2πn for n>0 are not equivalent.

(In other words: at t=0 there is no current, but there is current every time Vc goes through zero during later cycles).

 

[jim hardy]

We have to train our mind to think in new ways.

You have forgot that AC circuit analysis is taught using sine waves.
Sine waves are a mathematical oddity in that they don't change shape when you differentiate them. d(sinωt) = ωcos(ωt) , both have same shape but are shifted ninety degrees from one another.
So really they're a special case and we forget about that when we're just learning classic AC circuit analysis.
Observe sinewaves have zero slope at their peak , and maximum slope at their zero crossing.
That's counterintuitive eh ?

Next , recall what teacher said, current through a capacitor is derivative of voltage across it. i = c X dv/dt .
Now look at your sine voltage wave - when is its slope the steepest (ie derivative at maximum) ? Why, at the zero crossing of course.
Math says for the special case of sinewave, current is maximum at the same instant voltage is zero.
And current is zero at instant voltage is max.

It's all in the slopes.

Try drawing it for a triangle shaped wave. It'll help cement that derivative relationship in your mind.

old jim.

 

[NascentOxygen]

Hi, I have a problem understanding the ac capacitive circuit, where Vc = Vm * sin(ωt), and, current leads the voltage, so at t = 0, v = 0 which means no emf, no pushing energy for the electrons, I think, then how current is max? what is the energy source that supplied the electrons with the kinetic energy to move especially the capacitor was still uncharged.

I think you are picturing a switch in the circuit, and declaring that time t=0 will be the moment that switch is closed? That is okay here, except you can't take a current reading off the graph right at the moment the switch is being closed, viz., t=0. In general, though, steady-state AC graphs have their t=0 axis placed arbitrarily at a point well away from switch-on transient behaviour. In the ideal circuit depicted, the transient is over in an instant while current jumps from 0 to its peak value right at t=0 if that's when you want to close a switch, but in real RLC circuits this transient can extend over some cycles. So the graphs here are correct for t>0 because for time just a little after 0 (call it t=0⁺) the voltage is also > 0 and the capacitor carries the charge appropriate to the voltage.

A physical switch is often not shown in RLC circuits depicting phase principles because it complicates things by suggesting wrongly that t=0 is going to be the moment the switch is closed. If there is no switch, don't put one in. Rather, just assume the circuit has been in operation for some time and the positioning of the t=0 axis is being drawn somewhere purely as a convenient reference point, and the sinewaves are continuous to the left of the vertical axis as well as to the right.