Capacitor in AC circuit

[tomlib]

I have a question about a school circuit called a capacitor in an AC circuit. The capacitor is connected directly to the AC source and the phase shift is measured, which should be 90° voltage to current. My question is whether this phase shift will be 90° in every case or this also depends on the frequency of the source and the capacitance of the capacitor.
Can it happen that the capacitor is charged to its maximum capacity even if the sine wave of the source is not yet at its peak? Or, on the contrary, that the frequency will be so great that the electrolyte will not respond quickly enough to the supplied current? I know that a capacitor has a different resistance in the circuit for different frequencies. Is it also related to phase shift?

Please don't answer me with a formula

 

[Julie.]

A pure capacitor will have a phase difference of 90 deg between voltage and current on ac sine wave.

Irrespective of the capacitance, voltage, current, frequency, dielectric etc.

In practice a real capacitor will have resistance between the plates (parallel resistance), it will have resistance between the plates and the connections (series resistance) and similarly it will have inductance.

So in reality you could have a capacitor which has ~100% capacitance at one frequency, ~100% inductance at the other end of the frequency range, and actually pure resistance somewhere in the middle!


Most of the time though, the resistance and inductance is too small to make a significant difference - the capacitor can be regarded as a capacitor and nothing more.

The actual time of connection on the ac sine wave does make a difference, but only during that particular part of the wave, after that it settles down to the situation described above. This aspect is pretty advanced maths and is known as a transient response.

 

[Lucien Nunes]

At AC utility power frequencies (50 - 60Hz) most capacitors behave quite well as true capacitors and the phase shift will be very close to 90°. The resistive effects @Julie mentions do exist but they are small. There is another deviation from true capacitor behaviour caused by loss in the dielectric, which dissipates energy each time the electrostatic field across it changes direction. This makes the capacitor appear to have a resistor in parallel, although it is not actually a resistor. All of these effects reduce the phase shift slightly.

At radio frequencies the picture is different. Dielectric loss and series inductance can be quite serious and special construction techniques are sometimes needed to prevent the effects interfering with the use of the capacitor. You won't find any wound foil capacitors in the radio end of a mobile phone.

 

[tomlib]

So the capacitive reactance is based on what? What will be the reactance graph of the capacitor as the frequency changes?

 

[Lucien Nunes]

Xc = 1/2.π.f.C
So the reactance is inversely proportional to frequency, a standard 1/x graph.
You can think of it intuitively as the capacitor letting through a certain amount of charge at each cycle of the AC waveform. At a given voltage, the charge passed through per unit time is proportional to the frequency. So the voltage required to pass a given current is inversely proportional to frequency.

 

[tomlib]

The curve doesn't seem right to me, maybe I have a mistake somewhere.

 

[DPG]

The curve doesn't seem right to me, maybe I have a mistake somewhere.

Reactance of a capacitor is definitely inversely proportional to frequency. Show us your results/graphs.

 

[tomlib]

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.

 

[tomlib]

Now I would say that this curve is strange in that at different values of capacitance in the order of 10pF - 1μF, which is actually the main range of capacitances used for AC capacitors, I noticed the main bend of this curve always before 1kHz. I'd like to see some actual readings and graphs.
Possibly on behavior and connection in series with resistors.

 

[tomlib]

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.

 

[brianmoooore]

Are you suggesting that the equation Xc = 1/2.π.f.C, posted by Lucien is incorrect? It's an equation that I was taught more than 50 years ago. It was correct then, and still is today.
Don't think it has a name attributing it to any particular physicist, like Ohms Law or similar, but definitely correct.

 

[Lucien Nunes]

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.

It most certainly is sinusoidal.

Q=CV where C is constant.
I = dQ/dt
= C.dv/dt.
Everybody with me so far? What does sin(x) differentiate to? And what shape is that?

Please get yourself an oscilloscope, a function generator and a capacitor and satisfy yourself that this is true. I did when I was about 8 and the scope was bigger than me. It's still true, although I've grown and the scope has shrunk since then.

 

[pc1966]

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.

If you plot with a log scale you will find it is a straight line with slope -1

 

[pc1966]

As @Julie. and @Lucien Nunes have said real capacitors are never exactly 90 deg phase shift, though the sort used for power circuits at power frequencies are pretty close.

The concept of reactive impedance though is one which lends itself to easier calculation, but fundamentally what a capacitor is doing is passing a current that is proportional to the rate of change of voltage. In mathematical terms it is the time-derivative so:
I = C * dV/dt
A capacitor stores electrical charge, and the flow of charge is current. So if you want to change a capacitor by a certain voltage then a proportional amount of charge has to be moved around. The faster that happens the higher the current, as current is the rate of charge flowing.

 

[littlespark]

Is it a bird… is it a plane?

No

It’s this thread flying over my head too early in the morning

 

[Julie.]

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.

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.