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I've gone through school and done all the work but looking back at it i still don't quite get it.

The rule is current in a capacitor leads and in an inductor lags. The inductor makes perfect sense but the capacitor doesn't.

We live in a causal world. Nothing from the future can affect the present. That said the inductor lags but then why does the capacitor not lag even more?

Is this only a mathematical construct?

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    \$\begingroup\$ For an inductor, voltage causes the current to change. For a capacitor, current causes the voltage to change. So it makes sense for voltage to come first ("lead") in an inductor and current to lead in a capacitor. \$\endgroup\$ – The Photon Jun 21 '16 at 16:51
  • \$\begingroup\$ Why are you assuming the present is between the leading and the lagging? They can just as well both be in the future or both be in the past. There is nothing magic about the present so far as temporal relationships are concerned. \$\endgroup\$ – David Schwartz Jun 21 '16 at 22:12
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To use terms like lead or lag is to imply that your are refering to the AC sinusoidal analysis of capacitors and inductors and that means the steady state AC situation in which currents can lead voltages or voltages can lead currents. Lead or lag does not imply transient analysis.

Go back to the basic formula for a capacitor Q = CV. Then differentiate both sides to get dQ/dt = C dv/dt and of course dQ/dt = current so: -

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

If voltage rises at a certain rate the current will be constant - nothing to do with leading or lagging here until you apply a sinewave and the differential of a sinewave voltage is a cosine wave hence current leads voltage by 90 degrees BUT we're taling steady state AC analysis and that's when the terms leading and lagging apply - they don't make sense when talking about transient analysis.

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The voltage of a capacitor can't change instantaneously it needs a certain time. A good way to visualize this behavior is by charging the capacitor with a current source. First we have the current then the voltage builds up. The voltage lags the current (or the current leads the voltage).

For an inductor the current needs some time to build up. A voltage is applied and a current starts to flow. The current lags, the voltage leads.

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This is a mathematical construct for when we are dealing with sinusoidal waveforms and to say that the current in a capacitor leads the voltage is just an equivalent way of saying that the voltage lags the current.

For a capacitor \$ i = C \cdot \dfrac{dv}{dt} \Rightarrow v = \dfrac{1}{C} \int i \ dt\$

For an inductor \$ v = L \cdot \dfrac{di}{dt} \Rightarrow i = \dfrac{1}{L} \int v \ dt\$

Now if we differentiate \$ \sin \$ we get \$ \cos \$ which appears to lead and if we integrate \$ \sin \$ we get \$ (- \cos) \$ which appears to lag.

If we want to know the current in any component we can use \$ i = \dfrac{v}{Z} \$

For a capacitor \$ Z = \dfrac{1}{j \cdot \omega \cdot C}\$ thus \$ i = \dfrac{v}{\dfrac{1}{j \cdot \omega \cdot C}} \Rightarrow \dfrac{i}{v} = j \cdot \omega \cdot C\$ and we can see that the current leads the voltage.

For an inductor \$ Z = j \cdot \omega \cdot L\$ thus \$ i = \dfrac{v}{j \cdot \omega \cdot L} \Rightarrow \dfrac{i}{v} = -j \cdot \dfrac{1}{\omega \cdot L} \$ and we can see that the current lags the voltage, or the voltage leads the current.

But it must be remembered that this is a steady state response after the system has had time to settle.

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It may seem capacitors are clairvoyant since the current appears to lead the voltage. However, what they're really doing is making the current follow the derivative of the voltage.

When the voltage is a sine, then the current can be said to lead the voltage. That's only because the derivative of a sinusoid is another sinusoid, so the signals appear the same with one leading the other by ¼ cycle. The "leading" part is just one way to look at this special case, although it is a common and useful special case.

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Put another way;

In the steady state, a leading phase of 90 is actually a lagging phase of 270. The lead of one cycle is the lag from the previous cycle.

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In a PERIODIC environment "lead" is essentially indistinguishable from a very long "lag".

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Sorry, I won't give any equations because I have not dabbled enough with the maths of passive components in AC. I can somewhat understand the first order and a lot less in second derivative equations.

But what we have to understand is what is causal and not. While this might seem like an un-resolvable philosophical question, this one is not. The equation and description is not causal. It is a relationship. Which means as long as the voltage a given value at a given time, then the current should exist and we have this certain value.

You can turn the question into a causal one. You can ask, "OK, I have a reliable voltage source (which can supply any amount of stream of charges to maintain the voltage across it) that is AC, what's the current?". Which is one side of the question. One can also ask, "If I have a current source that is AC, what would the voltage across the capacitor be?".

By the way, a current source charging a capacitor is hardly ever given consideration, but if we were to implement a setup like that. You can have a high enough voltage source (to provide the peak current) then have a transistor (BJT or MOSFET) provide the current. Incidentally, you can also say "the voltage drop across the transistor vary to create the voltage across it and the rest across the capacitor", as the equation tells us. You can see it from different points of view.

Equations only give us relationship and because we can plug-in numbers, value, but hardly ever the context.

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I'm gonna get a lot of fire for posting this simple idea, but if it wasn't for this, I wouldn't have resolved a system that I was solving in a purely mathematical sense a while back.

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