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We say that in capacitive circuit the voltage and current are out of phase. Current is 90 (degrees) ahead of voltage. What is the physical explanation for this effect? How can current flow through a capacitive circuit, when voltage is zero i.e when voltage has a phase angle of 0 and current has a phase angle of 90?

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If you want to gain an intuitive understanding of how this can be true, let's consider first an inductor, because this makes a better physical analogy. In an AC circuit with an inductive load, voltage leads current by 90 degrees. It's the opposite of a capacitive load.

Why? Well, an inductor is like a flywheel that gives inertia to current. The proper name for voltage is electromotive force. That is, it's a force that causes electricity to move. When electricity moves, we call it a current.

Imagine a flywheel. The angular inertia (size and weight) of the flywheel is the value of the inductor. The voltage is a force you apply to the flywheel. The current is the speed the flywheel is spinning. Now, say you apply a force to this flywheel. It does not begin spinning instantly. Rather, the force you apply accelerates it. Now, you apply force in the other direction. It does not immediately reverse direction. First it must slow, and eventually it will turn the other way. But by the time it has done this, you have moved on and have changed your direction of force yet again.

If the force you apply is sinusoidal, and there is no friction (resistance) in the spinning of the flywheel, then the speed of the flywheel will be 90 degrees out of phase with the force that's being applied to it.

Now, go develop a good mental model of a capacitor, and consider the same sort of thing. It should make more sense, just with current and voltage reversed, or the phase shift in the other direction.

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    \$\begingroup\$ These analogies don't describe at all why the phase shift of both inductor and capacitor is exactly 90 degrees. Neither flywheel, nor waterpump doesn't any sense to that 90 degree rule. \$\endgroup\$ – Al Kepp Jan 21 '13 at 13:58
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    \$\begingroup\$ @AlKepp true, but that wasn't the point. The point was to provide a basis for an intuitive understanding. If you want a mathematical explanation, see Oli's answer. \$\endgroup\$ – Phil Frost Jan 21 '13 at 14:13
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The formula for current through a capacitor is:

I = C * (dV / dt)

The small d stands for a tiny change, known as delta(δ)
This means the faster the voltage change, the higher the current through the capacitor. The capacitor acts as a differentiator.
Now if we connect a sine wave voltage across a capacitor, the calculation for the current is the derivative of this voltage.

From calculus, we know that the derivative of sin(ωt) is ω cos(ωt):

Diff Sin

If we plot these values:

Capacitor I-V plot

You can see that when the voltage is changing fastest (at it's zero crossing), the current is at the maximum, and when the voltage is not changing (at the peak of the sine wave) the current is zero. We can see the 90° phase shift clearly.
This also explains why a capacitor blocks DC but passes AC.

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    \$\begingroup\$ Worth pointing out that the "d" in differential equations is a lazy lower-case delta (\$\delta\$). Properly speaking, upper-case delta (\$\Delta\$) is only appropriate when we are considering the change in something over a non-zero amount of (usually) time. \$\endgroup\$ – Phil Frost Jan 21 '13 at 14:37
  • \$\begingroup\$ You're right, I accidentally capitalised it - fixed. \$\endgroup\$ – Oli Glaser Jan 21 '13 at 17:11
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Think of a tank of water where you either pump water in or out so that the tank level follows a sine over time. Now think about what the water current going into the tank looks like as a function of time. When the tank level is at either peak, it is not changing so there is no current into the tank. When the tank level is in the middle (tank level sine is 0) is when the maximum water is being pumped in or out, depending on whether the tank level is on the way up or down.

If you think about this more, you realize that the current being pumped in is directly proportional to how fast the tank level is going up. In mathematical terms, the current is the derivative of the level. It shouldn't be hard to see now that the current is also a sine and is leading the tank level by 90°.

A capacitor is pretty much the same thing, except now the tank level is the voltage and the water current is now the electrical current.

Added in response to comments:

Yes, I know this is not a great analogy of how a capacitor works. The flexible membrane is a better analogy for that. But, the question wasn't about what a capacitor is but why voltage and current were 90° out of phase with each other. I thought the tank analogy made it easier to visualize that.

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    \$\begingroup\$ Good analogy, just don't fall into the mental trap of thinking that capacitors are like tanks, and electricity can flow in to them without an equal amount flowing out. \$\endgroup\$ – Phil Frost Jan 21 '13 at 13:20
  • \$\begingroup\$ @PhilFrost: Indeed; a better view of a capacitor would be as a spring-loaded piston (where the amount of water flowing in once side must equal the amount flowing out the other); the same general analogy regarding flow versus charge level still holds, though. \$\endgroup\$ – supercat Jan 21 '13 at 21:36
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    \$\begingroup\$ It's funny how everyone has their favourite analogy - mine is of a pipe with a flexible rubber membrane stretched across inside it. Same kind of idea as Supercat's I think. I like Olin's too though, quite interesting way to think of it - I'll give it a +1. The trouble with analogies is none of them are perfect. For someone just starting out though, as long as they realise this fact, they are good to help visualise things. \$\endgroup\$ – Oli Glaser Jan 21 '13 at 23:20
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In an inductor, voltage leads current, because in an inductor, there is drag on the current flow. You could call it inertia, but basically it is the electromagnetic field the inductor produces as it energizes. This field gives current "momentum" because when the supply voltage changes, the magnetic field which has already established itself will attempt to maintain the same current flow, slowing down the "response time" of the current. The field also resists initial power up, due to the same "inertia". Imagine a guy with a steel ball chained to his leg - he is the voltage and the ball is the current he's dragging around with him. Once he can get the ball rolling, it does not want to stop.

Capacitors on the other hand work by loading up one side of a dielectric medium with electrons. This time we can imagine the same guy only plowing snow with a snow shovel. The snow (current) is leading by 90 degrees out of phase because the applied voltage is directly proportional to how much excess electrons (current) are stacked up one side of the capacitor. As the snow shovel gets full, there comes a point where we can't push any more - voltage between the capacitor and supply is zero, however measuring across the cap terminals will equal supply voltage. The electrons flowing is the catalyst that changes the voltage as it passes through the capacitor, thus current leads phase.

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The phase shift idea can be understood and explained intuitively by means of the water analogy. Imagine you fill (sinusoidally) a vessel with water and you picture graphically this process (choose the half of the maximum water height as a zero level - the ground).

Analogy. So, you first open and then close (sinusoidally) the supply faucet. But note no matter you close the faucet (in the second part of the process) the level of the water continues rising... it is strange that you close the faucet but the water still continue rising... Finally, the faucet completely closed (zero current), but the level of the water (the voltage) is maximum.

Now, at this point, you have to change the flow (current) direction to make the water level decrease. For this purpose, you open (and later close) another faucet at the bottom to draw the water (now you draw a current from the capacitor). But again, no matter you close the faucet the level of water continues falling... and it is strange again that you close the faucet but the water still 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 elements storing pressure-like quantities (water, air, sand, money, data...) named integrators is:

The sign of the output pressure-like quantitty 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.

Capacitor. Let's now explain this phenomenon fully electrically. Imagine we drive a capacitor by a sinusoidal current source ("current source" means that it produces and passes a sinusoidal current in spite of all). No matter what the voltage (drop) 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. The voltage across the capacitor does not impede the current (it impedes but the current source compensates it).

So, until the input current is positive (imagine the positive half-sine wave) it enters the capacitor and its voltage continously increases in spite of the current's magnitude (only the rate of change varies)... Imagine... the current rapidly increases -> slows down -> rapidly decreases... and finally becomes zero. At this moment there is a maximum voltage (drop) across the capacitor.

Thus, at the maximum voltage across the capacitor, there is no current through it... Now the current changes its direction and begins rapidly increasing again -> slows down -> rapidly decreases... and becomes zero again... and again and again and again...

So, in this arrangement, the phase shift is constant and exactly 90 degree because of the ideal input current source that compensates somehow the voltage drop (losses) across the capacitor.

RC circuit. Let's now consider the ubiquitous RC circuit. First, let's build it. Since it is incorrect to drive a capacitor directly by a voltage source, we have 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 a current; so, the resistor acts here as a voltage-to-current converter.

Imagine how the input voltage VIN changes in a sinusoidal manner. In the beginning, the voltage rapidly increases and a 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" a 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...

So, 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 the current direction change is when the input voltage crosses the zero (the situation is similar to the arrangement of a current-supplied capacitor).

The conclusion is that, in this arrangement, the phase shift varies from zero to 90 degree when the frequency varies from zero to infinity because of the imperfect input current source that cannot compensate the voltage drop (losses) across the capacitor.

These explanations are based on an old Wikipedia discussion.

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In an inductive circuit, the back emf produced is very high initially as the coil is de-energized and the change in voltage applied across it is max. This back emf opposes the current flow initially. Once the voltage applied across the Inductor becomes zero, the magnetic flux produced before induces a current called residual current which remains even after the has reached zero. Hence Inductive circuits produces a lag.

I don't have a proper explanation for current lead in a capacitive circuit this has helped me remember the leading concept: When the voltage applied across a capacitor is increased in one direction, it charges and when it is decreased, it discharges. Basically it stores charge when voltage is increasing. But when the capacitance is reached, it will not draw any current even if voltage is increasing. Similarly while discharging, capacitor gets discharged before voltage reaches zero then it cannot supply current any more. Hence capacitive circuit leads.

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4) At the beginning, we should know that the generating voltage produced from a rotating machine is Sinusoidal type, i.e. each cycle has 4 quarters. 1st quarter – drooping-rise, 2nd quarter – increasing-fall, 3rd quarter – reverse-drooping-rise, and 4th quarter – reverse-increasing-fall. In a capacitor, during AC 1st quadrant (drooping-rise), the charging happens and the back emf builds up from 0 to source-voltage with the gradual fill-up of charges. Here 2 things to note: First: As AC voltage is sinusoidal type, its marginal rise is drooping type (represented by Cos function). For example, the pattern of instantaneous voltage at contant time-interval would be say v1 = 20, v2=35, v3= 48, v4= 58, v5=66 and so on. Second: In a continuous charging process, while the source is some voltage say v3, the capacitor attains the previous source voltage (say v2) by that instant. As the instantaneous current happens due to the difference of instantaneous voltage (vs –vc) at any instant of time; so, with the progress of time, the voltage difference is drooping, the instantaneous current goes diminishing. At the instant of source-voltage maximum, the marginal difference is almost nil; hence the instantaneous current is zero. The capacitor becomes saturated. (Note: as the resistance is very small, the growth period is negligible as the time-constant τ = RC i.e. capacitor voltage attains the source voltage almost instantly. However, while vs = vmax Sin ωt, vc= vmax Sin (ωt - τ)) This back-emf is considered as if an equivalent resistance similar to the resistive circuit, called as the reactance. The instantaneous-reactance (xc) is a time-based parameter, varies from 0 to infinite unlike the resistance which is constant. For simplicity, the average reactance (Xc) is used in general application and is measured by Vmax at the end of charging divided by Imax at the start of charging (non-sense!). I have explained the process of charging. Similarly, the process of discharging, reverse charging, reverse-discharging may be visualised. This is analogous to hydraulic tank filling.

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