1
$\begingroup$

I have a question about charging a capacitator in an alternating current. If the process starts the current charges the capacitator, therefore the voltage on the capacitator will be increased. Now the current negates itself, the capacitator gets discharged.

But the overall Charge should be:

$$Q_{max}=\int_{0}^{T/2} I_0 \cdot sin(2t/T)\,dt$$ $$Q_{min}=\int_{0}^{T} I_0 \cdot sin(2t/T)\,dt = 0$$

So how the charge $Q$ can be smaller as 0? How does the charge of a capacitator behave in an alternating current, so that the voltage can be negated?

EDIT:

Basically, I cannot imagine how exactly the phaseshift ($\alpha \neq \pi/2$) comes. If the voltage of the voltage-source is high, basically many electrons should be pushed into the capacitator, this decreases the current. If the voltage is high, there should be current pushing out of the capacitator. This does not explain the phase shift.

I have seen this equation $i = C\frac{dU_{current}}{dt}$ (which seems obvious to me). This would mean that if $i$ is $0$, $U$ would have to be maximal. This is not the case in every (capacitator) circuit, is it?

$$i = C\frac{dU_{current}}{dt} \implies Q = \int_{U_0}^{U_{T}} CdU_{current} \implies Q_{min} = 0 \implies U_{min,capacitator} = 0$$

But this conclusion is a contradiction to obvious observations.

$\endgroup$
  • $\begingroup$ electronics-tutorials.ws/capacitor/cap_8.html $\endgroup$ – Jasper Apr 14 '19 at 7:02
  • $\begingroup$ @TVSuchty It should be $I=I_0sin(2\pi t/T)$ $\endgroup$ – Karthik Apr 14 '19 at 7:21
  • $\begingroup$ @KV18 You are right, though it should not change something... $\endgroup$ – TVSuchty Apr 14 '19 at 9:55
  • $\begingroup$ @Jasper This is exactly the point, I do not understand how exactly the delay comes and what exactly causes it... $\endgroup$ – TVSuchty Apr 14 '19 at 10:00
  • $\begingroup$ I shall edit my answer. Meanwhile could you please correct the typos in your question? $\endgroup$ – Karthik Apr 14 '19 at 10:14
0
$\begingroup$

First of all, the driving voltage is going to be a sinusoidal one as you are picking an AC source.

$$V=V_0sin(\omega t)$$

The current as a function of time is given by: $$I=I_0cos(\omega t)$$

This is because: $Q=CV$ $\implies$ $I=\frac{CdV}{dt}$ (by taking the derivative of the LHS and RHS with respect to time.

Now you asked why there is a delay which I am assuming to be the phase shift between the voltage and the current - which is $\pi /2$ in this case. This is why the voltage was assumed to be a sinusoidal one (because the capacitor plates were not charged and hence no potential difference could exist between two neutral plates). and the current turned out to be a cosine wave, much to your surprise.

But this makes sense of course because the current through the circuit will decrease when the voltage increases throughout the charging process in the capacitor. But what you should realize is that the AC source can only help in generating $Q$ amount of charge in the capacitor (after charging one cycle). So this means there would be increased repulsion with time when the electrons are moving across the wire. So, the rate of charging would reduce although charge would build up anyway with time.

Rate of charging is literally the current $I$ in this scenario.

EDIT: As to the scenarios where there is a phase shit other than $\pi /2$, you can use: $tan \phi =\frac{X_L-X_C}{R}$.

In other words, you get such phase shifts when you add different components (like resistors and inductors) into the circuit in addition.

$\endgroup$
  • $\begingroup$ Thank you for your help, it would be great if you might comment to my edit additionally and explain where I am thinking wrongly there :) $\endgroup$ – TVSuchty Apr 14 '19 at 10:24
  • $\begingroup$ I have explained the "delay" part. I'll add the rest now. $\endgroup$ – Karthik Apr 14 '19 at 10:25
  • $\begingroup$ I understand the phase shift in general if it is exactly $\pi/2$, but otherwise, I am wondering... In my mind, this contradicts the equations (given above) and the intuition... $\endgroup$ – TVSuchty Apr 14 '19 at 10:27
  • $\begingroup$ Have you understood why $Q>0$ ? It is because $I$ is a cosine function between $0 \to T/2$. $\endgroup$ – Karthik Apr 14 '19 at 10:29
  • 1
    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – TVSuchty Apr 14 '19 at 10:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.