Capacitor in AC circuit

[Julie.]

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I got this graph from the formula. Apparently it will be fine that way, but there is no way to verify it. My idea was about a smooth decrease in resistance with frequency. To the layman's eye, this looks like a high frequency "breakthrough". A formula is one thing, a practice of measurement, and a certain logic is another. As I say, I'm not well versed in electricity. It was just a question.

The formula is Xc = 1/(2. pi. f. c)

Since in your example 2, pi, and c are constant your result should have the shape ~1/x (actually 1/f) which is exactly what you are getting.

This is a constant/smooth change with frequency - double the frequency - half the reactance - the curve does illustrate this.

I am not sure what curve you are expecting - use a graphing calculator, or online graphing or just plot a simple 1/x on paper.

Take x = 0.1 (y=10), x = 0.5 (y=2), x = 1 (y=1), x = 2 (y=0.5), x = 10 (y=0.1)

And you will get a curve of the same form you initially presented.

I also think that the given waveforms of the current shape will not be correct. If we charged the capacitor with a triangular voltage, the current graph would be flat. After a certain period, the capacitor was charged again with the same amount of current until it was fully charged. So the course would not have to be rising and falling, but in some way jumpy.
In the case of a sine wave, the waveform of one wave would initially be flat or nearly flat and then gradually decrease more rapidly. In the case of a sinusoidal waveform, I would say that the current will not have a sinusoidal shape.

But I see no other answer here.

If you apply a triangular voltage to a capacitor, the current would be a square wave. (The maths is probably a bit too much for you but in a capacitor i = C dv/dt - since the change in voltage is a constant rate - say 0 to 10V in 0 to 1 sec then dv/dt = 10/1 - a constant - when the voltage falls from 10V to 0 in 1 sec, then dv/dt = -10/1 - hence a square wave)

You can do exactly the same with a sine wave, if you apply the voltage at zero and look at the current (i=C dv/dt) as the voltage follows the sine wave in growth, you will find the current follows a cosine curve - this is exactly the same as a sine wave with a 90 deg phase shift.

This is always the case for a steady state situation, a sine wave voltage applied to a capacitor will produce a sine wave current (with 90 deg phase shift), a sine wave current through a capacitor will produce a sine wave voltage across the capacitor (with a 90 deg phase shift) - end of.

You can confirm it mathematically, by experiment (oscilloscope), or actually intuitively if you understand it properly.


You can get a quite different picture on initial start if the point at which the voltage is applied is not at zero, again use i= C dv/dt, however remember that when you apply the (say 6V) voltage at say t=0 it actually goes from 0 to 6V very quickly, then follows the sine wave from 6V to 10V and onwards, so initially it goes from zero current to extremely high almost instantly, after which it decays as an exponential decay, whilst at the same time, the normal sine waveform is superimposed on it.

This is why the initial transient is often ignored, it depends on the point of energisation and is over usually very quickly at which point you get the steady state - sine wave creates sine wave with phase shift.

 

[marconi]

View attachment 106252
I got this graph from the formula. Apparently it will be fine that way, but there is no way to verify it. My idea was about a smooth decrease in resistance with frequency. To the layman's eye, this looks like a high frequency "breakthrough". A formula is one thing, a practice of measurement, and a certain logic is another. As I say, I'm not well versed in electricity. It was just a question.

This graph gives a major distortion of the visual relationship between the independent variable of frequency along the horizontal axis and the dependent variable of reactance along the vertical y axis because it uses markedly different numerical/distance scaling along each axis.

If you use the same numerical/distance scale along x and y axis then the relationship between reactance and frequency will be seen without scaling visual distortion.