First look at my circuit. The voltage source has a value of 5V with a phase angle of zero, and the capacitor's impedance is 5Ω. So the current is obviously 1A with a phase angle of 90°.

Circuit Diagram

What is the physical reason behind this phase shift? I can prove mathematically that a capacitor can make a 90° leading phase shift. But I want to know the physical reason for it.

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    \$\begingroup\$ Ohms is not a unit of capacitance. \$\endgroup\$ – Olin Lathrop May 27 '15 at 17:02
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    \$\begingroup\$ @Olin Lathrop, I think the OP means 'of 5 ohm reactance'. \$\endgroup\$ – Chu May 27 '15 at 17:11
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    \$\begingroup\$ Try this analogy: The sun is at its most powerfull at 12:00, yet the hottest part of the day is later (~ 15:00). To when effect is cumulative, the effect lags the cause. \$\endgroup\$ – Wouter van Ooijen May 27 '15 at 17:28
  • \$\begingroup\$ I meant the reactance @OlinLathrop \$\endgroup\$ – Raihan Khalil May 27 '15 at 18:49
  • \$\begingroup\$ Nice analogy @WoutervanOoijen \$\endgroup\$ – Raihan Khalil May 27 '15 at 18:49

If, instead of a sine-wave, you consider a turning on the circuit for the first time, with a DC voltage source and a discharged capacitor.

Immediately after you turn on, the maximum current will be flowing, and the minimum voltage will be across the capacitor.

As you wait, the current will reduce as the capacitor charges up, but the voltage will increase.

As the voltage arrives at its maximum, the current will have reached minimum.

And that's basically it - that's a description of a pair of sine-waves (one voltage, one current), 90 degrees out of phase, with alternating mutually-exclusive minima and maxima.


I'll try a basic explanation.

Let the voltage source be a constant voltage, V. The charge on the capacitor is therefore constant (Q = CV). Now lets say the voltage changes. The charge on the capacitor must also change, therefore some current flows to add or remove charge. The amount of charge that moves is therefore proportional to the change in voltage.

Now lets represent voltage as a function of time, V(t). Then the amount of charge on the capacitor is Q(t) = CV(t). The derivative of the voltage is the rate of change of the voltage at some time t. Therefore the rate of change of charge is proportional to the rate of change of voltage. dQ(t)/dt = C dV(t)/dt. Current is the rate of change of charge - dQ(t)/dt. Therefore current is proportional to the rate of change (derivative) of the voltage.

The derivative of a sine wave is a cosine wave. Therefore the current for a sine wave voltage will be a cosine wave - 90 degrees out of phase.

  • \$\begingroup\$ The OP says he already knows how to express it mathematically. He wanted a physical reason for the phase relationship. \$\endgroup\$ – tcrosley May 27 '15 at 22:41
  • \$\begingroup\$ @tcrosley Well write up an answer then. \$\endgroup\$ – geometrikal May 27 '15 at 23:05
  • \$\begingroup\$ I would if I could. I'm a digital guy. \$\endgroup\$ – tcrosley May 27 '15 at 23:29
  • \$\begingroup\$ @tcrosley ^_^ same here actually :) \$\endgroup\$ – geometrikal May 27 '15 at 23:40

Well, after some 20 years of engineering digital circuits, I think I still know the answer.

As your voltage source moves past zero deg. it has 0 volts of output. However, the voltage is increasing quickly. So, the electric field strength in the dielectric of the cap is changing quickly, and as the field gets stronger, it pushes more electrons out of the positive side plate (due to increasing electric force on them created by the field). It's important here to realize that a cap is an open circuit essentially, just a specially shaped one. Therefore, current does not flow through a capacitor (ideal one here, we can talk about effect of leakage later if you like), but rather to or from one plate or the other. This causes an electric field to build in the dielectric which affects the free electrons on the other plate via electric force. To explain all that physics, we need to get into Gauss' law, etc. so I won't do that here.

Each plate is a relatively large chunk of conductive metal, so lots of free electrons exist in it. Many many more than are involved in a reasonable level of current flow. So, the voltage difference between plates, generated by your source will push free electrons from the negative side of the source onto the plate it's connected to. This builds an electric field within the dielectric of the cap such that electrons are pushed by the electric force out of the opposite plate. The circuit carries them back to the positive leg of your source. As more and more charge is pushed into the negative plate, the field grows stronger and more electrons are pushed off the other plate.

However, since the rate of change of voltage is slowing as we reach max voltage (at 90 deg), our field strength is still increasing, but more slowly all the time. For that reason, fewer and fewer electrons are pushed off the positive plate per unit time (so current flow is getting smaller). At the point of max voltage, the rate of voltage change is zero, so there are zero more electrons being pushed off that positive plate. At that point the voltage begins to fall, and the field weakens. This allows some of the pushed out electrons from the positive plate to come back into it. As the voltage rate of change accelerates and the voltage itself falls back toward zero volts, the rate at which electrons return to the positive plate accelerates (current rises). When the voltage is at zero, it's changing at its max rate, so you have max current flow in the circuit (electrons are coming back to the plate as fast as they ever will for this circuit). The other half of the waveform (negative lobe of the voltage sinusoid) is the same, but switch the plates I'm calling negative and positive since voltage reverses at this point (current doesn't of course, it reversed at the 90 deg point, and will again at 270).

I suppose it could be written up more elegantly, but do you get my meaning here. Can you picture the effect of the field within the cap's dielectric and its relationship to to electrons flowing out of or into the plates? (positive and negative voltages are not really that, they just indicate that they are associated with current vectors of opposite direction)


Q = CV (a basic fundamental i.e. charge = capacitance x voltage)

Differentiate both sides keeping capacitance constant: -

\$\dfrac{dQ}{dt} = C \dfrac{dv}{dt}\$

\$\dfrac{dQ}{dt}\$ is current therefore I = C \$\dfrac{dv}{dt}\$

Now, if the voltage is a sine wave, the current will be a cosine (due to the differentiation of v w.r.t. t) i.e. it will be shifted forward by 90 degrees.

If you are after a deeper explanation then try the physics site.

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    \$\begingroup\$ Essentially the same answer as geometrikal's (I don't know who posted first). The OP says he already knows how to express it mathematically. He wanted a physical reason for the phase relationship. At least you suggested physics.stackexchange.com. But it seems someone on this site should be able to describe the physics of it also (Will Dean sorta does). \$\endgroup\$ – tcrosley May 27 '15 at 22:44

This might be best thought of in terms of what's called 'displacement current' if you want to get in to the physical behavior. How can there be current flow across the capacitor -- the plates aren't connected at all! Well, there's something in between the plates, a dielectric medium, like glass. Inside that glass there are positive charges of nuclei surrounded by negative charges of orbiting electrons. Because it is an insulator, the electrons are not free to flow through the material, but they aren't locked in place either -- they're in a cloud around each nucleus.

When there's an increase in voltage (but it has to be a change), then the electrons aren't centered on the atoms any more -- they move towards the positive plate of the capacitor (unlike charges attract). Because they're a negative charge, we see that as a current flow from positive plate to negative plate. But if the voltage stops changing, they stay in their new equilibrium position. It's only the CHANGE in voltage that causes some motion of the electron charges, because they are not able to leave the neighborhood of their atoms. Then, when you decrease the voltage, they move back towards being 'on center' and that's a negative current.

That's why current is proportional to the rate of change of voltage (dV/dt), and that's why the current is greatest when the voltage is changing fastest (which is when the voltage sinusoid is crossing zero), and drops to zero when the voltage has stopped changing (which is when the sinusoid of voltage has reached a maximium or minimum value) in a sinusoidal excitation.


ElectroBOOM explains it very simply in the following video: https://www.youtube.com/watch?v=Og9j5Wk0Di4 (6:15)

  • A capacitor resists against change of voltage across it.
  • An inductor resists against change of current through it.
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    \$\begingroup\$ Link only answers are not the best, maybe better to post relevant information so when the link goes down the answer remains. \$\endgroup\$ – Voltage Spike May 17 '18 at 17:02
  • \$\begingroup\$ "A capacitor resists against change of voltage across it." This explains very little, if anything. \$\endgroup\$ – Transistor May 17 '18 at 17:34

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