Would you help me understand the behaviour of capacitor with AC? - Page 1
Forum section: electronics / beginners

author: Michael George
author_rank: Contributor
timestamp: June 20, 2016, 02:59:38 am
content: Hello,I know the equations and phasor diagram of this circuit and I can use math to calculate phase shift, gain, impedance... and so on.Although I draw phasor diagram and do algebra correctly, I don't understand the physical meaning of the circuit and I can not understand the following waveform:Please, explain the waveform to me in a very simple and basic way.Why is the current very high at the beginning of charging? and why is it very low at the beginning of discharging? there are many things in this wave form that does not make sense for me.Thank you very much,

author: helius
author_rank: Super Contributor
timestamp: June 20, 2016, 04:09:15 am
content: There is no "beginning" of charging at t=0. The waveform is infinite in both directions.

author: rstofer
author_rank: Super Contributor
timestamp: June 20, 2016, 04:19:06 am
content: As the sine wave crosses through zero, the rate of change of voltage is 1 unit/time interval. The sine is never steeper than this.But we also know that capacitor current is IC= C dv/dt - the capacitance times the rate of change of voltage with respect to time. We just said the rate of change of voltage was a maximum when the sine wave moved through zero (steepest slope) so that's the point where the capacitor current is maximum.Look at 90 degrees: The rate of change of voltage is zero (peak of curve) and the capacitor current is zero.

author: cvriv
author_rank: Frequent Contributor
timestamp: June 20, 2016, 04:35:24 am
content: Current keads voltage by 90deg in a capacitive circuit. ICE is a good way to remember that. That's basically what your looking at with that waveform. It's a bit confusing to visualize. When a capacitor is fully charged it blocks current, meaning its resistance is infinitely high, which is the equivalent of an open. In a simple capacitive circuit, if you check the voltage across the open, you get the voltage across the power supply or maximum circuit voltage. Now, in an AC circuit the capacitor is fully charged when the CHANGE in current is the greatest. The change in current. The greatest change in current being at 0 rad and pi rad. Now, knowing thatlook at the waveforms. You see that the voltage wave is peaked when the current waveform is either at 0 rad or pi rad. Look at the current waveform at 3 pi/2. Assume that is the start point, 0 rad. You see as the current waveform reaches its peak(which signifies the least amount of changein current) the voltage waveform begins. The voktage waveform begins there(90deg lag) because at that point the current change is beginning to increase. When the current change is peaked, the voltage is peaked. Remember the peak change is current is at 0 and pi rad. I hope this helps. Forgive me for poor writing of this, i wrote this out on my phone.

author: rs20
author_rank: Super Contributor
timestamp: June 20, 2016, 05:08:25 am
content: As mentioned by rstofer, the current through the capacitor isproportionalto therate of changeof the voltage across it. When the voltage is at its highest, it has reached a peak, and so it's barely changing at all. Hence, the current is small. When the voltage is at zero, that is the point of most rapid rate of change (i.e., the graph is steepest), so the current is highest.Quote from: cvriv on June 20, 2016, 04:35:24 amWrong, wrong, wrong. Take a big, fully-charged capacitor and short its leads out. The massive spark that results somewhat calls into question the claim that the capacitor has an "infinite resistance" when fully charged.An ideal capacitor has zero series resistance regardless of bias voltage. Resistance has no bearing at all in the diagram/graph that the OP presented.

author: Neganur
author_rank: Unknown
timestamp: June 20, 2016, 05:10:29 am
content: Quote from: cvriv on June 20, 2016, 04:35:24 amThe current is constant at those points (at 0 and pi) so there is no change in current. Is this a language thing or did you just think backwards?

author: cvriv
author_rank: Frequent Contributor
timestamp: June 20, 2016, 10:13:58 am
content: Quote from: Neganur on June 20, 2016, 05:10:29 amI was taught the rate of change is greatest at those points.

author: cvriv
author_rank: Frequent Contributor
timestamp: June 20, 2016, 10:19:31 am
content: I didnt say anything about shorting the capacitor. If you hook a capacitor in series with a dc power source, when the capacitor is fully charged, current stops flowing. The current stops flowing because capacitors reactance becomes infinitely high. If i measure the voltage across the capacitor it should be the same as the voltage across the power supply.I was taught that the rate of change is greatest at 0 rad and pi rad and that the capacitor is charged and that a peak voltage can be measured across the capacitor.

author: Rerouter
author_rank: Super Contributor
timestamp: June 20, 2016, 10:27:28 am
content: The way i learned to think of it is, capacitors store voltage, so in your given example while the voltage is at a peak, the voltage change is tiny, so the current change is tiny, but as the waveform gets closer to the 0V reference, the voltage is changing at its fastest rate, so the current is at its largest, And because its storing voltage, it makes the voltage lag the current,The vice versa is true for inductors, my crude analogy is they store current (in a magnetic field), so they make the current lag the voltage, and with them, the largest rate of change in current is at the 0 crossing, resulting in the largest voltage,

author: rsjsouza
author_rank: Super Contributor
timestamp: June 20, 2016, 11:06:48 am
content: rstofer's explanation is good. One thing I used to do to remember the relationship between voltage and current in capacitors is they love when voltage is steady. Any change and they will try to sustain it at its previous level by increasing the current in an attempt to maintain their previous charge.Similarly to inductors: they love when current is steady. Any change and they will try to sustain it by increasing the voltage and maintain their previous charge.

author: Neganur
author_rank: Unknown
timestamp: June 20, 2016, 11:11:10 am
content: Quote from: cvriv on June 20, 2016, 10:13:58 amBut look at the current curve at those points...is the rate of change _really_ big there? You have local extrema there, the di/dt is zero

author: Gyro
author_rank: Super Contributor
timestamp: June 20, 2016, 11:28:20 am
content: I'm not sure if this is too simplistic an explanation, but it might help you get a 'feel' for it.Imagine that the capacitor is a large heavy mass (capacitance) on a smooth, preferably frictionless, surface. Now imagine that you try to slide it across the table. The faster you try to accelerate it, the more effort (current) you need to apply. The same when you try to stop it moving.Now imagine that you are trying to move it backwards and forwards, following the profile (voltage) of a sine wave. 'Feel' how much effort (current again) you are applying at different points in the cycle as you accelerate and decelerate it.Edit: Actually that isn't a perfect example, it implies that absolute position on the table is voltage, which it isn't. It's only relative position that is relevant.Edit edit: Note that in order to make the mass follow the sine path you need to start applying effort in the opposite direction before you reach the peak position on the table... this represents the phase lead of the current versus the voltage.

author: cvriv
author_rank: Frequent Contributor
timestamp: June 20, 2016, 12:21:46 pm
content: Quote from: Neganur on June 20, 2016, 11:11:10 amOh I see what your saying. My head is elsewhere. That graph at 0 rad, the current change is zero, therefore the voltage peak is 0v. I was see another graph in my head where the change in current starts at 0 rad. The graph in my head starts where 3pi/2 is on this graph. My bad:)

author: helius
author_rank: Super Contributor
timestamp: June 20, 2016, 05:53:12 pm
content: Quote from: Neganur on June 20, 2016, 05:10:29 amIt may be. A value can only be "constant" iff its first and second derivatives are zero.

author: luxfx
author_rank: Contributor
timestamp: June 20, 2016, 06:09:26 pm
content: I found a video yesterday while trying to understand impedance, which helped immensely. It uses a nearly identical circuit to what you've got and might help your understanding as well.

author: IanB
author_rank: Super Contributor
timestamp: June 20, 2016, 06:15:45 pm
content: Quote from: Michael George on June 20, 2016, 02:59:38 amOK, you asked about the physical meaning, so let's try to do that.Imagine that we start at some initial state where the supply voltage is at zero and the capacitor is discharged with a voltage also zero. (Someone above said there is no beginning of charge at t=0, that the wave is infinite in both directions--but in the real world everything has a beginning, and we can set up our physical experiment as described.)Now, what happens as time advances and the supply voltage starts to increase? If there is no resistance in the system (ideal assumption), then the capacitor voltage must always be the same as the supply voltage. So what must happen for the capacitor voltage to increase? It must get charged up, so a large current must flow from the supply into the capacitor to charge it.Here we have a first observation. When the supply voltage is near zero (and changing), there is a large current flowing in the circuit.Now let's follow the supply voltage as it increases. When the voltage starts to reach the top of the cycle its rate of increase slows down. So less current is flowing in the circuit (the capacitor doesn't have to work so hard to keep up.) What happens when the supply voltage is at the maximum? At this point its rate of change slows to zero. We know the capacitor voltage is equal to the supply voltage, and the supply voltage is not changing, so the capacitor has nothing to do. No current needs to flow in the circuit, so the current becomes zero.Does this help with physical understanding?

author: Electric flower
author_rank: Regular Contributor
timestamp: June 20, 2016, 06:49:18 pm
content: I have always looked at this behaviour in a intuitive way.If you only look at voltage source (+ or - terminal of battery, or voltage at generator) there is excess of charged particles on one side, and at the same time less of them on the other side.If you only look at one side of voltage source inside of the material (Copper) there are lots of charged particles (particles that give voltage potential) that are repelling each other (electrostatics, same charges repell) and when you connect capacitor, those particles inside conductor are still repelling each other inside conductor and when you connect capacitor those particles have a path to flow toward capacitors plates.At first moment you connect capacitor, there are those reppeling particles in conductor (voltage source) and that reppeling gives them potential energy that we more often call voltage potential, but at that moment on capacitor plates there are no reppeling or attracting particles so capacitor has voltage potential = 0So when you connect capacitor to voltage source you have "bigger" voltage difference and that is why current is higher at start, as capacitor is charing its plate get electrons than start repelling other electrons and start getting voltage potential, but that voltage potential is getting closer and closer to sources voltage so there is less voltage and current gets smaller.When capacitors current is zero, capacitor is charged to its max value (max. voltage of source) and at that moment those two thing have same voltage potential and thete is no current flow. But then sine voltage source starts to drop, and then capacitor starts responding to this change, at start there is small difference between him and source so current is quite small, as the voltage on source drops further this difference becomes bigger and current increasesThis is how stuff works inside of partly dyslexic head.This analogy i used works in real world, but it does not work on college exams.(explained by 19 year old electronic geek who fucked up his college education)This analogy you are now learning at college ignores facts that capacitors also has resistive and inductive properties.Please ask me anything.

author: rstofer
author_rank: Super Contributor
timestamp: June 20, 2016, 06:51:37 pm
content: Another important observation: At no time during the positive half cycle is the rate of change of voltage higher than it is at t=0. This is important! It tells us that at no time during that interval is the current any higher than it is at t=0 and that's why it starts to immediately decline. It's hard to see on a sine wave but the slope is continually decreasing after t=0 so the rate is continually decreasing until it peaks out at pi/2.

author: rs20
author_rank: Super Contributor
timestamp: June 21, 2016, 12:46:49 am
content: Quote from: cvriv on June 20, 2016, 10:19:31 amWrong, wrong, wrong. Formula for a capacitor's reactance is 1/(2 pi f C). It is a function offrequency, not charge, voltage, nor time. You need to stop assuming that everything is a resistor and that any time when no current is flowing, it must be due to an infinite reactance, impedance or reactance. Sometimes, there's no current because there's just no voltage differential.

author: cvriv
author_rank: Frequent Contributor
timestamp: June 21, 2016, 06:21:27 pm
content: DC has a frequency of zero. Whats 2PI x 0 x C? Its 0. Whats 1/0? Undefined or infinite. No matter what the capacitance is, the cap's reactance will eventually become infinitely high, blocking current. Thats what I was taught at least.

author: rs20
author_rank: Super Contributor
timestamp: June 21, 2016, 09:38:56 pm
content: Quote from: cvriv on June 21, 2016, 06:21:27 pmOP's question has a frequency of 1, and is completely free of any DC component.