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So far I've established the following:

Current is the movement of charge over time, measured in coulombs/second.

Charge is the electron in an atom.

Voltage is the potential difference between two points and the energy per unit of charge.

Still, I don't understand why this happens:

Voltage versus current in capacitor

How is it possible that at time t=0 the current is present in an RC circuit without the potential difference? What caused the charge to flow in the first place?

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    \$\begingroup\$ How is it possible that at t=0 current is present without voltage? Well, remember that what is plotted is the voltage across the capacitor, not the voltage across the resistor. In fact, there is voltage across the resistor! For a resistor, current can only be present if voltage is simultaneously across the resistor; for a capacitor, this isn't always true. You can have current without voltage, positive current with positive voltage, or even positive current with negative voltage (depending, of course, on what the capacitor is connected to). \$\endgroup\$ – Zulu May 5 '15 at 17:47
  • \$\begingroup\$ So in the beginning the voltage is basically everywhere in the circuit except for the capacitor ? that seems too sketchy for my liking \$\endgroup\$ – Shady Programmer May 5 '15 at 22:38
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    \$\begingroup\$ is that so hard to believe? Imagine, if you will, that you initially have 0V on the capacitor, 0V on the voltage source, and 0V on the resistor. Suddenly the voltage source pops up to 1V, and proceeds to oscillate as a cosine. For a moment, right at the start, there was (and must have been) 0V across the capacitor, because its voltage couldn't change instantaneously (doing so would require infinite current). Therefore, for that moment, there was 1V across the resistor. So, yes, for that moment, there's voltage everywhere except the capacitor. \$\endgroup\$ – Zulu May 5 '15 at 22:49
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The picture in your question assumes that the voltage waveform started some time earlier and that the transient of it beginning is no longer affecting things.

Basically Q=CV and this translates to I = C dv/dt and, if you applied a sinewave the differential of that sinewave voltage gives rise to the cosine wave of current but, of course at t=0 things are a little different; For a start you can't suddenly start a sinewave from rest - that would imply infinite bandwidth. Given this fact, there is a small finite time which the current rapidly ramps up to the starting value in your picture. From thereon it pretty much follows the equation given above.


EDIT section, mechanical analogy

A mechancial analogy could be regarded as a flywheel i.e. a rotating mass. The force applied to the end of the flywheel will accelerate the speed at which the flywheel rotates but when the flywheel (lossless assumed) is at constant speed, no force is needed. You can imagine the flywheel speed like voltage; the flywheel has charged up to speed n and there is no longer any force needed to keep it charged at that speed. Just like a capacitor, once charged to a constant voltage there is no current needed to keep a perfect capacitor at that voltage.

However, if you applied a constant force to decelerate the flywheel, the speed decelerates linearly and if the constant force is a true constant force, the flywheel speed will decelerate through n=0 and start rotating in the opposite direction after a little while. Force is -X and speed ramps down linearly. Ditto with the capacitor, if you take a constant current from the capacitor the voltage falls linearly and eventually becomes negative and charges up to a negative voltage.

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    \$\begingroup\$ The current, as I explained in my answer, is C dv/dt. Slightly before or after t=0 the voltage will have values that imply a ramp and this ramp in voltage gets differentiated to a near-constant value that is the current. \$\endgroup\$ – Andy aka May 5 '15 at 13:38
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    \$\begingroup\$ @ShadyProgrammer the terms leading and lagging are just conveniences; one doesn't come before the other but it's convenient sometimes to think this way. There is no theory of leading and lagging. \$\endgroup\$ – Andy aka May 5 '15 at 14:17
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    \$\begingroup\$ @ShadyProgrammer, the instantaneous voltage across a capacitor is not dependent on the current through at that instant but, rather, on the history of the current through. Also, it is important to distinguish between AC analysis (sinusoidal steady state) and transient analysis. Only in AC analysis can we say the voltage lags the current by 90 degrees. \$\endgroup\$ – Alfred Centauri May 5 '15 at 14:18
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    \$\begingroup\$ @ShadyProgrammer, You asked "does a capacitor in some weird way track history of what is going on through it ?". Yes, but it's not weird. Just look at the integral form of the capacitor equation. \$V(t) = \frac{1}{C}\int_{-\infty}^{t}I(t') \mathrm{d}t'\$. \$\endgroup\$ – The Photon May 5 '15 at 15:59
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    \$\begingroup\$ @ShadyProgrammer, as an analogy, think of the pressure associated with a balloon filling with air. The pressure at any instant does not depend on the flow of air into (or out off) the balloon at that instant but, rather, the amount of air in the balloon at that instant. But the amount of air in the balloon at any instant depends on the history of the flow of air. Similarly, the voltage across a capacitor depends on the charge Q separated on the plates. But the separated charge Q depends on the history of the flow of charge (current). \$\endgroup\$ – Alfred Centauri May 5 '15 at 16:47
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To answer your question, let's start with a simple DC source e.g. a battery. Just when you turn on the circuit, the schematic appears like this:

enter image description here Capacitor is like a hungry child and someone is serving him cookies on a plate. You are trying to measure his eating speed by monitoring his plate, which is a wrong plan because initially when the child is very hungry, you will see an empty plate. But as his stomach gets full, his eating speed will become zero and you will see a full plate. That's the case with the capacitor.

Initially, there will be a large current through the capacitor essentially making it equivalent to a short circuit. Assuming the wire to be of negligible resistance, you are essentially putting your probes together which will give you a zero voltage reading.

Now let the circuit sit for a while till the capacitor gets charged. Now the equivalent circuit looks somewhat like: enter image description here

It's open circuit now with zero current flow (ideally). Now you will be able to measure the true charging voltage (5V).

Now coming to your doubt, initially at t = 0, there was a potential source which made the electrons move. However, the current was moving so rapidly through the capacitor that you were unable to measure a potential drop across it.

At this point you might think, where did that potential go?

Well say you are using a 5V battery along with an ideal zero resistance capacitor. The potential drop will occur across the internal resistance of the battery, Giving you this scenario:

enter image description here

Well again you are putting the probes together at t = 0 and hence you will get zero voltage. You simply can't measure any voltage this way at t = 0.

So, how can anyone measure it:

There are two ways:

1) Impossible way - Split the battery into two components - an ideal battery and a resistor equivalent to internal resistance and put the probes across the resistor. This will give you battery potential at t= 0.

2) Possible way - Usually internal resistance is small. Take a bigger resistor and put it in series with the capacitor and measure the voltage across that resistor. At t= 0, this will give you almost the battery potential. Almost because some potential drop is across the internal resistor as well.

After a long time though, the current will diminish to zero and the circuit will essentially be open. In open circuits, there is no point of resistors and hence the circuit becomes equivalent to the initial charged circuit where you can measure all the battery potential across the capacitor.

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  • \$\begingroup\$ "Now coming to your doubt, initially at t = 0, there was a potential source which made the electrons move. However, the current was moving so rapidly through the capacitor that you were unable to measure a potential drop across it." --- From that I've just got this epiphany that initially there is no voltage on the capacitor because the amount of charge Leaving the capacitor on one plate is the same as the amount of charge arriving on the other plate and then the voltage starts building up when there is more and more charge on one plate in comparison to the other one. Am I right or am I right? \$\endgroup\$ – Shady Programmer May 6 '15 at 13:30
  • \$\begingroup\$ @Shady - I think you are right but this case holds for capacitor only, I guess. Considering a resistor, equal number of charges reach one end and leave the other end. Going by your logic, there shouldn't be a potential drop across a resistor. However this isn't true. We both know that. \$\endgroup\$ – Whiskeyjack May 6 '15 at 18:04
  • \$\begingroup\$ I've researched, I've thought about it long and hard and found an answer to that. Voltage is a broad term that describes couple of things. It's an electromotive force, potential difference, electrical potential energy per unit of charge. What we've done is we've talked about potential difference between capacitor plates measured in volts. Considering a resistor we're now talking about an electrical potential energy of EACH CHARGE unit that is "LOST" (converted to heat) also measured in volts. \$\endgroup\$ – Shady Programmer May 8 '15 at 20:24
  • \$\begingroup\$ It is because of this broad terminology that we can talk about an excess of charge on one plate of a capacitor in comparison to the other plate and call it "VOLTAGE over capacitor" and also talk about the transfer of energy (in joules) that each coulomb dissipates in a resistor (V=J/C) and call it "VOLTAGE over resistor". Now can a brother get an AMEN or are my findings flawed ? \$\endgroup\$ – Shady Programmer May 8 '15 at 20:54
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First, note that your waveform shows what happens in the sinusoidal steady state. This implies that the voltage and current have been stable sinusoids for all time. So there's no "in the first place" in your graph.

The reason there's a current at t = 0 because the voltage is changing at t = 0. To get the voltage to start rising, you need to be pumping charge onto the plates of the capacitor. I think you're trying to apply DC thinking to an AC circuit. The voltage might be zero at t = 0, but its first derivative is not. That derivative has physical significance! It's what really matters to the capacitor.

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I asked myself such questions in the late 70's when I was studying the subject of Theoretical Electrical Engineering... where they unsuccessfully tried to “explain” me this phenomenon through strict definitions. I remember what I could not imagine was why as the current went down, the voltage on the capacitor kept rising. Many years later, in an interesting conversation with my former students and followers… and by the help of the hydraulic analogy, I finally managed to figure out what was really going on...

After the main question why the voltage across a capacitor lags the current through it, another logical question arises, "And why is this lag exactly 90 deg in the single capacitor and less than 90 deg in the RC circuit"? Here are possible intuitive explanations (such as I would like to hear years ago).

1. Single capacitor. I have come to the conclusion that textbooks fail to explain the phase shift between the current and voltage since they consider the case of a voltage-supplied capacitor. But this arrangement (an AC voltage source directly drives a capacitor) is fundamentally incorrect (like the case where a voltage source directly drives a diode)... although it is still used to make a voltage-to-current differentiator. But what is more important to us is this arrangement is not suitable for an intuitive explanation of what is happening.

The dual arrangement - current-supplied capacitor, can help us easily explain why voltage lags the current with exactly 90 deg. In this arrangement, an AC current source drives the capacitor that now acts as a current-to-voltage integrator. "Current source" means that it produces and passes sinusoidal current through the capacitor in spite of all. No matter what the voltage across the capacitor is - zero (empty capacitor), positive (charged capacitor) or even negative (reverse charged capacitor), our current source will pass the desired current with desired direction through the capacitor. So the voltage across capacitor does not impede the current (it tries... but the current source compensates it by increasing its internal voltage).

Until the input current is positive (imagine the positive half-sine wave) it charges the capacitor and its positive voltage continuously increases in spite of the current's magnitude. The strangest here is that even when the current decreases to zero the voltage continues to increase to maximum (my amazement in the past). Then the current changes its direction and during the negative half-sine wave it charges the capacitor with an opposite polarity... and the magnitude of its negative voltage continuously increases in spite of the decreasing current's magnitude. So, in this arrangement, the phase shift is constant and exactly 90 deg because of the ideal input current source that compensates the voltage drop across the capacitor.

Current-driven capacitor

Hydraulic analogy. The popular "water vessel analogy" ("electrical current - water flow" and "voltage - water level") can help us fully understand in an intuitive way the phase shift idea.

First half wave (0 - 180 deg): Imagine you fill a vessel with water and picture graphically this process. Choose the half of the maximum water height as a zero level (ground) and begin gradually, in a sinusoidal manner, opening (in the interval 0 - 90 deg) and then closing (90 - 180 deg) the supply faucet. Note that no matter you close the faucet (in the interval 90 - 180 deg), the water level will continue rising. It is strange that you close the faucet but the water continues rising. Finally, you have completely closed the faucet (zero current), but the level of the water will be maximum (maximum positive voltage).

Second half wave (180 - 360 deg): At this point, you have to change the flow (current) direction to make the water level decrease. For this purpose, you can begin gradually opening and then closing another faucet at the bottom to draw the water (i.e., you draw current from the capacitor). But again, no matter if you close the faucet the water level will continue falling. It is strange that you close the faucet but the water continues falling. Finally, you have completely closed the faucet (zero current), but the level of the water will be maximum negative (maximum negative voltage).

So, the basic idea behind all kind of such storing elements (named integrators) is: The sign of the output pressure-like quantity (voltage, water level, air pressure, etc.) can be changed only by changing the direction of the input flow-like quantity (current, water flow, air flow, etc.); it cannot be changed by changing the magnitude of the flow-like quantity. At the final point, the current is zero but the voltage is maximum; this gives the 90 phase shift on the graph.

2. RC circuit (voltage-to-voltage integrator). We have already realized that it is incorrect to drive a capacitor directly by a voltage source; it is better to drive it by a current source. For this purpose, let's connect a resistor between the voltage source and the capacitor to convert the input voltage to current; so, the resistor acts as a voltage-to-current converter. Thus we have built a current source by the input voltage source and resistor. Let's now consider the circuit operation (I will do it electrically but the hydraulic analogy of communicating vessels is an impressive way to do it as well).

RC integrating circuit

Imagine how the input voltage VIN changes in a sinusoidal manner. In the beginning, the voltage rapidly increases and the current I = (VIN - VC)/R flows from the input source through the resistor and enters the capacitor; the output voltage begins increasing lazy. After some time, the input voltage approaches the sine peak and then begins decreasing. But until the input voltage is higher than the voltage across the capacitor the current continues flowing in the same direction. As above, it is strange that the input voltage decreases but the capacitor voltage continues increasing. Figuratively speaking, the two voltages "move" against each other... and finally meet. At this instant, the two voltages become equal; the current is zero and the capacitor voltage is maximum. The input voltage continues decreasing and becomes less than the capacitor voltage. The current changes its direction, begins flowing from the capacitor through the resistor and enters the input voltage source. It is very interesting that the capacitor acts as a voltage source that "pushes" current into the input voltage source acting as a load. Before the source was a source and the capacitor was a load; now, the source is a load and the capacitor is a source…

The moment where the two voltages become equal and the current changes its direction is the moment of the maximum output voltage. Note it depends on the rate of changing (the frequency) of the input voltage: as higher the frequency is, as low the maximum voltage across the capacitor is... as later the moment is... as bigger the phase shift between the two voltages is... At the maximum frequency, the voltage across the capacitor cannot move from the ground... and the moment of current direction change is when the input voltage crosses the zero (the situation is similar to the case of current-supplied capacitor).

So, in this arrangement, the phase shift varies from zero to 90 deg when the frequency varies from zero to infinity. This is because of the imperfect input current source that cannot neutralize the voltage drop across the capacitor.

Op-amp inverting integrator

If we want the phase shift between current and voltage in the RC circuit to be exactly 90 deg regardless of frequency (as in the case of a single capacitor), we should somehow compensate the voltage across the capacitor. This is done by the operational amplifier in the circuit of the op-amp inverting integrator. It makes its output voltage equal to the voltage drop across the capacitor and adds it in series. The result is zero voltage (the so-called virtual ground).

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    \$\begingroup\$ +1 for 4.5 years delay \$\endgroup\$ – muyustan Dec 1 at 15:45
  • \$\begingroup\$ Such fundamental ideas are eternal and it's never too late to find an explanation for them... \$\endgroup\$ – Circuit fantasist Dec 1 at 19:25
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    \$\begingroup\$ don't get me wrong please, I am just joking \$\endgroup\$ – muyustan Dec 1 at 20:20
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I think the main point here is that the notion of voltage lagging current by 90deg is a theoretical best case, and in practice the lag will be slightly less.

In reality the connecting leads have some resistance, so the point at which the capacitor's voltage is zero will occur slightly later in time than the point at which the AC generator's output is zero. Hence the PD driving the current.

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If you look back at the graph, the equation \$ I=C*{dV}/{dt} \$ does not equate a level to a level with a phase shift. It equates a level to a rate of change or a slope. It requires current to change the voltage, and that's exactly what's happening in the graph.

Instantaneous points are weird.


Now that we've got the math out of the way, I'll also mention that you will never get that graph in real life. Real capacitors also have some inductance, which will smooth out the sharp transition at the beginning, assuming \$V=I=0\$ to start.

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Capacitors needs current to develop voltage.

So first there should be current before the voltage.

Current leads voltage. (no pun intended)

Voltage lags current.

Just trying to visualize intuitively.

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  • \$\begingroup\$ Your answer is like saying "It happens this way because that's the way it works" \$\endgroup\$ – Shady Programmer May 8 '15 at 20:59
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To me, the answer to this question is very intuitive. Notwithstanding the math, it is really very simple if reduced to what happens with a capacitor in a DC circuit.

If you connect a battery to a capacitor, current must flow into the capacitor to charge it up. If the capacitor is not charged, then the voltage across the capacitor is zero before it is connected to the battery. The instant (and when I say instant, I mean an infinitely small point in time) the battery is connected to the capacitor the battery begins to charge the capacitor, but the capacitor does not charge up to the battery's voltage instantly. No matter the value of the capacitor, it takes some time for this to happen. It is very quick for a small value of capacitance, and it takes longer for a large value of capacitance, but no matter the size of the capacitor it takes some amount of time.

The current is initially large, but as the voltage charge across the capacitor approaches the battery voltage, the amount of current falls, until such time as the capacitor is fully charged. Thus, the voltage is behind (lagging) the current.

When the capacitor is charged to the battery's voltage, for a perfect capacitor, the current is zero; for a real-world capacitor in good working order, the current is extremely small.

Think about what would happen if you connect a 100,000 mfd capacitor across a 12 volt power source? If you do that, you better connect it through a resistor to limit the current to a safe value, or have a very large power capacity power source. When first connected, the capacitor would be almost a dead short. Current would be limited only by the value of the resistor. When the capacitor is charged to 12 volts, current will become almost zero for a good quality capacitor.

That is why large broadcast transmitters charge their oil filled rectifier power supply capacitors through a resistor of appropriate value, which is shorted out by a contactor once the capacitor is fully charged (typically after about 1 second after the power supply is turned on).

This is more complicated to visualize on an AC circuit, but it works exactly the same way. The math just becomes more complicated. But this is why a capacitor bank, shunted across an AC power line, can provide reactive power for voltage support when the line has inductive loads. Right after the sine wave just begins to move closer to zero, the capacitor voltage is still building up almost 90 degrees behind the power line's wave form and begins to discharge its energy to support the power line's voltage. Without such capacitor banks, our power system would be very inefficient.

Hope this helps your understanding.

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I like to think as follows:

Capacitors are basically two plates isolated by a dielectric. To have a voltage between the two plates you should charge them first. To charge them you must source current, so the voltage between the plates is like a response to the current you gave.

Inductors, on other side, behaves in respect of Lenz's Law. Voltage is correlated to electric field. So, when you apply voltage (electric field) on a winding you make a current flows on it, which leads us to think that the current on a inductor is like a response to the voltage applied.

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    \$\begingroup\$ Capacitors - okey seems fair enough and straight forward but why is there a delay in response to the current ? As soon as charge flows onto one plate of the capacitor it means that the potential between plates changes AS SOON as each particle arrives onto the plate so where does the lag come from ? \$\endgroup\$ – Shady Programmer May 6 '15 at 10:54
  • \$\begingroup\$ Like you, I don't know what really happens inside it. All I'm doing is guessing upon physics theory. I suppose the delay is due to the time the other plate takes to react. When you start charging a capacitor, firstly you charge one plate with V+. The other side, in response (due to the electric field) is charged with V-, but I suppose its not instantaneous, due to the dielectric properties. \$\endgroup\$ – Pedro Quadros May 6 '15 at 13:21
  • \$\begingroup\$ Now, supposing it really has a delay, let's see what's happen on AC. Firstly one side (plate 1) is charged with V+. The plate 2, at first instance, is neutral. After delay it is V- (and so we have the maximum voltage between them). When plate 1 is V- it again takes a time to plate 2 be V+ (maximum voltage on capacitor). Again, as I said, all I'm doing is supposition and hope, as you, to get it clarified by one which properly know about the subject. \$\endgroup\$ – Pedro Quadros May 6 '15 at 13:28
  • \$\begingroup\$ tf.uni-kiel.de/matwis/amat/elmat_en/kap_3/backbone/r3_7_2.html Like shown by the link, greater is the dielectric constant, greater is the refraction of the material. We also know that greater is the refraction index, lower it's speed of the wave on it. So, since the dielectric between the plates has a greater value of refraction than copper, we expect to have a delay when a wave crosses it. \$\endgroup\$ – Pedro Quadros May 6 '15 at 13:42

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