Physical Meaning of Leading/Lagging Voltage/Current

[scothoward]

Hey, I'm trying to grasp a more physical meaning of what exactly is meant by a leading/lagging voltage/current.

In terms of capacitors and inductors, I understand mathematically that the differential i-v relationship causes one waveform to lead and the other to lag, but in a physical circuit sense, what does this mean? Would it correspond to some sort of time-delay?

Thanks

 

[ice109]

Hey, I'm trying to grasp a more physical meaning of what exactly is meant by a leading/lagging voltage/current.

In terms of capacitors and inductors, I understand mathematically that the differential i-v relationship causes one waveform to lead and the other to lag, but in a physical circuit sense, what does this mean? Would it correspond to some sort of time-delay?

Thanks

of course, it corresponds to whatever fraction of $$2 \pi$$ times the period it is.

 

[nrqed]

Hey, I'm trying to grasp a more physical meaning of what exactly is meant by a leading/lagging voltage/current.

In terms of capacitors and inductors, I understand mathematically that the differential i-v relationship causes one waveform to lead and the other to lag, but in a physical circuit sense, what does this mean? Would it correspond to some sort of time-delay?

Thanks

Yes. It just means that if you were to plot as a function of time the voltage and the current (which both vary sinusoidally), they won't reach their peak at the same instant. In a purely resistive circuit (an ideal resistor connected to an ideal AC power supply) there is no lag which simply means that whenever the current is maximum in the circuit, the voltage across the resistor is maximum and so on. So the two are in phase. But fi you have a capacitor, say, the voltage across the capacitor is not maximum at the same time as the current is maximum in the circuit.

 

[berkeman]

I don't understand ice109's comment, but just to add a little to nrqued's comments...

Think of the current in the inductor -- it is caused by the voltage across the inductance, but once you apply an instantaneous voltage across the inductor, it takes a while for the current to build. If you are using sinusoidal voltage excitation of the inductor, the time it takes the current to build and change direction causes the current waveform to lag the voltage sinusoid by 90 degrees.

And for a capacitor, if you switch on an instantaneous current source, it will take a while for the voltage across the capacitor to build up. So with a sinusoidal current excitation, the voltage across the capacitor lags the current waveform by 90 degrees.

It's best to understand what the source is (voltage or current), and why the dependent waveform (current or voltage) lags behind the driving source waveform. The concept of "leading" is misleading, IMO, and definitely non-physical. So I prefer to just think in the physical terms of what causes the lagging behaviors.

 

[ice109]

berkeman if you draw a phasor diagram with phasors for the voltage across a cap and a resistor the two will have a phi in between them. for an ideal case the voltage across the cap will be $$\frac{\pi}{2}$$ behind the resistor, if $$2 \pi$$ is the entire period then $$\frac{ \frac{\pi}{2}}{2 \pi} = \frac{1}{4}$$ of the period , however long it maybe 1/60th or 1/50th of a second or w/e. I am sure you already knew this but i was just clarifying

 

[scothoward]

I don't understand ice109's comment, but just to add a little to nrqued's comments...

Think of the current in the inductor -- it is caused by the voltage across the inductance, but once you apply an instantaneous voltage across the inductor, it takes a while for the current to build. If you are using sinusoidal voltage excitation of the inductor, the time it takes the current to build and change direction causes the current waveform to lag the voltage sinusoid by 90 degrees.

And for a capacitor, if you switch on an instantaneous current source, it will take a while for the voltage across the capacitor to build up. So with a sinusoidal current excitation, the voltage across the capacitor lags the current waveform by 90 degrees.

It's best to understand what the source is (voltage or current), and why the dependent waveform (current or voltage) lags behind the driving source waveform. The concept of "leading" is misleading, IMO, and definitely non-physical. So I prefer to just think in the physical terms of what causes the lagging behaviors.

Thanks for the explanation! So just to confirm my thoughts...If I did have a capacitor connected to a sinusoidal voltage source, the instant it is turned on, the charge moving from one plate to the other (current) is at a maximum (corresponds with the lead). As the capacitor voltage in increasing in correspondance with the source, the amount of displacement charge is steadily decreasing. Then when the voltage reaches its peak, the capicitor is fully energized, and as a result, charge is no longer being displaced, so current is zero.

In other words, this situation would correspond to current having a cosine waveform and the voltage (lagging) having a sine waveform. Is my intuition serving me right?

Thanks again

 

[waht]

Also note that power dissipated by an inductor or a capacitor is imaginary. That means instead of dissipating power as heat like a resistor, capacitor or an inductor is absorbing power (because of lead or lag).

 

[ice109]

Also note that power dissipated by an inductor or a capacitor is imaginary. That means instead of dissipating power as heat like a resistor, capacitor or an inductor is absorbing power (because of lead or lag).

lol @ borrowing power from the power company

 

[berkeman]

Thanks for the explanation! So just to confirm my thoughts...If I did have a capacitor connected to a sinusoidal voltage source, the instant it is turned on, the charge moving from one plate to the other (current) is at a maximum (corresponds with the lead). As the capacitor voltage in increasing in correspondance with the source, the amount of displacement charge is steadily decreasing. Then when the voltage reaches its peak, the capicitor is fully energized, and as a result, charge is no longer being displaced, so current is zero.

In other words, this situation would correspond to current having a cosine waveform and the voltage (lagging) having a sine waveform. Is my intuition serving me right?

Thanks again

You're on the right track, but be careful about what you turn on "instantaneously". It would take infinite current to charge up the capacitor to the peak of the sine voltage waveform at t=0. If you are driving the capacitor with a sine wave voltage source, then the output impedance of the voltage source comes into play, and it looks like the RC circuit that ice109 was describing, where the differential voltage across the capacitor lags the differential voltage drop across the output impedance of the signal source.

The current through the capacitor (displacement current flowing in the external circuit to move charge onto and off of the capacitor plates) is maximum for sine wave excitation 90 degrees before the voltage maximum. The voltage waveform lags the current waveform by 90 degrees.

 

[Keith rigby]

I apologise if I'm not allowed to ask further questions as I am new to this site and haven't figured out the protocols.
I too have been struggling to rationalise the AC theory which I've been using in a lifetime of engineering calculations with the actual physical process.(I tried to type phenonemum here but can't spell it)
I can get a grip of the capacitor charging with the current at minimum when the voltage reaches the max by imagining the flow of electrons into the capacitor meeting a more and more crowded and repulsive environment,but I can't imagine a similar explanation of the magnetic processes.Most expositions explain the capacitive case but gloss over the inductive one. Can anyone help without maths,I can do the diff eq bit but still can't envisage what the pesky electrons,magnetic dipoles etc are up to.

Old Engineer.

 

[alvaros]

scothoward:

In terms of capacitors and inductors, I understand mathematically that the differential i-v relationship causes one waveform to lead and the other to lag, but in a physical circuit sense, what does this mean?

I want to point out that phisical magnitudes never lead. This would mean that future can be predicted.

In the case of sinusoidal waves in a capacitor V is 90º delayed respect to I. If you want I is 270º delayed.
In an inductor I is 90º delayed respect to V.

 

[alvaros]

Keith rigby:

but I can't imagine a similar explanation of the magnetic processes.Most expositions explain the capacitive case but gloss over the inductive one. Can anyone help without maths,I can do the diff eq bit but still can't envisage what the pesky electrons,magnetic dipoles etc are up to.

In an inductor at constant current the magnetic field is constant, so there is no induced voltage. I you change I , magnetic field changes and there will be a voltage:

V = L dI/dt

A change in B produces a E field ( Maxwell equations ):

E = dB/dt ( may be some terms/signs lost ) Why ? Nobody knows.

 

[Keith rigby]

insomnia cure and Maxwell

You just couldn't resist the di/dt bit could you?
Perhaps Maxwell knew,but it was so obvious to him that he didn't think it worth mentioning.
Anyway if you can't get to sleep,just try to imagine what physically happens as an inductor is energised at various points on the voltage waveform.Zonk,its morning already.

Hopefully your 'nobody knows' will goad someone into at least an attempt to explain.

By the way this in my case I hope will lead me to the supplementary answer as to why big transformers go 'boing'and audibly take many hertz to settle down when switched on,and yes I know the transient formulae but that still doesn't help my physical visualisation.

Our friend who states that the current can't really lead the voltage is quite right as this would transpose cause and effect,but the word 'lead' has a special meaning in this case,referring to the relative phase of the waveforms after the initial transient state.
Anyway I am off to bed and visualizing if the capacitive transient and steady state really supports this is another guaranteed insomnia cure.

 

[seang]

Another similar topic: how many times have you seen someone say 'device x supplies so many VAR of reactive power?'

Is this really correct? In terms of reactive power, there is no sending or recieving, is there? It is simply established?

Even my professors say, 'A sync generator can provide real power, but absorb reactive power.' Does this make sense?

If so, can you both supply and absorb positive reactive and negative reactive power? Are there 4 possible combinations?

 

[cabraham]

scothoward:


I want to point out that phisical magnitudes never lead. This would mean that future can be predicted.

In the case of sinusoidal waves in a capacitor V is 90º delayed respect to I. If you want I is 270º delayed.
In an inductor I is 90º delayed respect to V.

I agree about V being 90 deg delayed wrt I in a capacitor, but I being delayed 270 deg wrt V makes no sense. If the cap is excited with a step function of current, the voltage takes a finite time to "catch up". In a cap, a change in I always precedes a change in V. I is NOT delayed 270 deg wrt V. That would imply that a change in V precedes a change in I which NEVER happens in a capacitor.

Also, it is worth noting the following. In a capacitor, just because I leads V does not mean that I causes V. In an inductor, V leads I but V does not cause I.

Claude

 

[cabraham]

I agree about V being 90 deg delayed wrt I in a capacitor, but I being delayed 270 deg wrt V makes no sense. If the cap is excited with a step function of current, the voltage takes a finite time to "catch up". In a cap, a change in I always precedes a change in V. I is NOT delayed 270 deg wrt V. That would imply that a change in V precedes a change in I which NEVER happens in a capacitor.

Also, it is worth noting the following. In a capacitor, just because I leads V does not mean that I causes V. In an inductor, V leads I but V does not cause I.

Claude

I said "step function of current", but I meant to say "pulse of current". A steady step function of current into a cap would result in voltage ramping indefinitely towards infinity.

Sorry. BR.

Claude

 

[nitsnits.sher]

significance of phasor

sir i would like to ask that why do we use this stuff called "phasor" in our operation...
the reason i got to make calculation easier,or to show 90 degree out of phase quantity in some equation...reply soon...
thanks

 

[sophiecentaur]

Also note that power dissipated by an inductor or a capacitor is imaginary. That means instead of dissipating power as heat like a resistor, capacitor or an inductor is absorbing power (because of lead or lag).

Power is not "dissipated" in a pure inductor. Energy is merely stored in the magnetic field around it and can all be reclaimed, in principle.

 

[Okefenokee]

You can solve problems for AC by using real numbers but it requires complex analysis in the time domain. That's complex as in complicated, not complex as in a complex number. Differential equations will be involved.

Phasor diagrams are just one way to represent complex numbers. Complex numbers can be used to transform differential equations in the time domain into the familiar Ohm's law equivalent in the frequency domain. That way, we can use algebra to solve problems easily for AC. It's also a good way to conceptually understand how a system responds to frequencies

For example:

The time domain relationship for a capacitor is this:

i = C(dv/dt)

In the frequency domain, it becomes this:

$$V = \frac{1}{j C \times 2 \pi f} \times I$$
(j is the imaginary number)

That looks a lot like Ohm's law doesn't it?

Define:

$$V = R_C \times I$$
-with-
$$R_C = \frac{1}{j C \times 2 \pi f}$$

For completeness, here is the formula for an inductor and resistor:

$$V = R_L \times I$$
-with-
$$R_L = j L \times 2 \pi f$$


$$V = R_R \times I$$
-with-
$$R_R = R$$




If we pick a certain frequency, we can treat the capacitor just like a resistor with a complex resistance. That makes circuit analysis easier.

Also, this tells us exactly how capacitors respond to frequency. Notice that the reactance is a fraction (1/f). That means that it becomes smaller at higher frequencies. A capacitor effectively becomes a short circuit when the frequency is high enough. It also becomes infinite (open) at very low frequencies. Inductors do the opposite and become an open circuit at high frequencies and a short circuit at low frequencies. The resistor is still just a plain old resistor at any frequency.

 

[stewartcs]

Power is not "dissipated" in a pure inductor. Energy is merely stored in the magnetic field around it and can all be reclaimed, in principle.

I don't believe he was saying it was dissipated...i.e. he said it was imaginary.

CS

 

[sophiecentaur]

You can solve problems for AC by using real numbers but it requires complex analysis in the time domain. That's complex as in complicated, not complex as in a complex number. Differential equations will be involved.

.

The ruse of going into and coming out of the world of Complex numbers is a very smart one but can confuse people unnecessarily. There is nothing 'inherently' imaginary about what is happening - it's just that, if you can do that particular bit of Maths, things come out very nicely. It's quite possible to start with
V = V₀ cos(ωt+θ) and apply that varying voltage to a network containing 'reactive' components, which may differentiate or integrate the signal and get a result from it. In fact, with a calculator - or better still, a spreadsheet like Excel - you can produce the shape of the resulting signal, numerically, without ever going into Complex stuff.
The clever thing about the Complex approach is that it replaces a sinewave with a spiral / helix - along the time axis. When projected into one plane, the helix produces a sinewave. It gives you a Voltage vector which is turning at the same frequency as the signal - if you view this helix along its (time) axis. Other signals at the same frequency are going around at the same rate.
If you let your 'graph paper' rotate at the signal frequency you 'freeze the rotation' and you just see stationary lines (the voltage vectors) stationary and at different angles (the phases). These are the Phasors. A signal at a different frequency will appear as a phasor which rotates at the 'difference frequency'.

 

[sophiecentaur]

I don't believe he was saying it was dissipated...i.e. he said it was imaginary.

CS

It was the fact that he actually used the word "dissipated" that I made my comment. Perhaps it was a bit picky but dissipation has a particular meaning and could confuse if used inappropriately.

 

[Okefenokee]

The ruse of going into and coming out of the world of Complex numbers is a very smart one but can confuse people unnecessarily. There is nothing 'inherently' imaginary about what is happening - it's just that, if you can do that particular bit of Maths, things come out very nicely. It's quite possible to start with
V = V₀ cos(ωt+θ) and apply that varying voltage to a network containing 'reactive' components, which may differentiate or integrate the signal and get a result from it. In fact, with a calculator - or better still, a spreadsheet like Excel - you can produce the shape of the resulting signal, numerically, without ever going into Complex stuff.

In the end you'll be doing math that has the exact same properties as complex numbers even if you call it something else, like a phasor diagram.

The imaginary number isn't arbitrary. It's not just a tradition. It the natural choice for transforming time->frequency. Imagine trying to do a Fourier transform without ever bringing in the imaginary number. You'd have trigonometric identities coming out of your ears.

 

[sophiecentaur]

I agree that Complex Maths is the best way to deal with it but people can get carried away with the 'significance' of it and feel the need for some physical interpretation. It is only a mathematical 'trick' and introduces ideas which, although very productive, are an added complication first time through. It is not a trivial idea to introduce an imaginary space and then throw it away at the end of a procedure. After all, you can only see 'real' voltages on an oscilloscope.
Of course, if you're already happy with Complex operations then there's no problem. I got the impression that, for some posts on this thread, it was representing just another conceptual problem. It should be comforting that you don't actually have to do it that way.

 

[john72]

Hi, someone correct me if I'm totally wrong but the way I understand it is; Think of water flowing in a pipe! Voltage = Pressure, Current = Volume flow rate, Resistance = Resistance.
If water flowing along a pipe and meets a blockage (resistance), you get a buildup of pressure (voltage) and also the flow rate decreases (current).
Instead of a blockage in the water pipe we have it connected to a wider pipe (less resistance), Voltage/Pressure goes down and Current/Flow rate goes up.
For AC think of the water pump, as it cycles (at say 60 times a sec 60Hz) 0 psi to 120 psi you have a build up of pressure (Voltage) then water flows (Current). as the water flows the pressure will drop. So the current lags the voltage.
A capacitor holds a charge then relases it, if its on a 120v RMS AC circuit, at the top of the waveform about 170v the current is zero, there is no current flowing. If we put a load on the circuit, the cap stores then replaces, this causes a phase shift in the waveform.

 

[sophiecentaur]

@John72
If you really want to introduce a highly complicated notion like water flow into this topic then you are just making things hard for yourself. There are regular 'analogy' threads on this forum and I haven't read any good argument in favour of using the water analogy for electricity, which just brings in more imponderables like fluid dynamics, turbulence, static vs dynamic pressure blah blah blah. Why bother?

Electricity behaves, refreshingly, in a very straightforward way on many more occasions than does flowing water. Voltage is NOT pressure and Current is NOT volume flow rate. 'Resistance' is not defined in water flow because Resistance is the ratio of Volts to Current (which are strictly electrical quantities). Resistance is not 'something that impedes current flow'; it is just a ratio.

Steer clear of water flow if you want a deeper understanding of anything electrical.
There are, of course, analogues: the pressure in a water tank is proportional to the depth /volume of water (as in V=Q/C), for instance but there is no simple water analogue for an Inductor. Until you have really sussed out electricity you should avoid any of this or any 'predictions' will be suspect.

 

[johnnybee]

Please for give me if I am violating any forum rules by placing this statement here. I am interested in this very same question of the physical properties of current flow related ot voltage. Especially the leaking and lagging. I am more confused now after reading the forum than I was when in was just wondering about this subject. I am an electrician of 35 years and was taught that voltage is pressure and current is volume and power was the product of the two and is represented by work done. Also that resistance is a physical property of material. I always thought of water flow to compare it to. Now I am told here that that is all wrong. It seems to me that current can never lead voltage as voltage is required to cause the current to flow. The flow is current . The voltage pushes the and causes the current and the amount of current is determined by the resistance of the material which would be the load on the voltage (pressure). You can prove this by increasing the current flow by lowering the resistance and watch the voltage go down if you have a limited power supply. In a resistive only circuit this is rather easy to understand. As the component relationships change in capacitive and inductive circuits, I can see that the current flow and voltage (pressure ) are not peaking simoultaneously which causes them to be in a different phase of the sinusodial wave from each other at each moment in time. It seems (seems) to me that the 270 degree of lag appears top be a lead when it is graphed (superimposed) over each other. Of course this is only my conception of reality. In any case , electricity is some sort of physical movement of physical items (electrons) and should be able to be explained in a physical analogy. I am sure that many of you are smarter than I am about this and I am in the asking mode here. Please add to this subject if you know the answers.

 

[psparky]

Hey, I'm trying to grasp a more physical meaning of what exactly is meant by a leading/lagging voltage/current.

In terms of capacitors and inductors, I understand mathematically that the differential i-v relationship causes one waveform to lead and the other to lag, but in a physical circuit sense, what does this mean? Would it correspond to some sort of time-delay?

Thanks

I've spent a lot of time on this topic...so I'll throw my two cents in.

First of all, there are two time ways to look at the lag/lead.

There is the time domain...the sin wave of the voltage compared to the shifted sine wave of the current. The more they shift away from perfect phase (up to 90 degrees...the more power you are using to produce the same output simply because P=IV.

The other way is in vector or phasor form. The current is always compared to the Voltage. Set your voltage at zero degrees and the current is the one the goes out of phase in reactive circuits.

As Sophie is saying...watch the term "imaginary". Let's say I have a load of 50 KVA at 480 volts three phase. To find the current, 50 KVA =480*1.73*I.

The current you just found is very real! It is not imaginary! You must size your wire and breaker to carry that "imaginary" current!

I strongly encourage you to use the term reactive power over "imaginary power".

Ok...to the math. 1/jωc or JωL

Another way to write these...1/(ωC<90) or ωL<90...Cdv/dt or Ldi/dt will get you to the same place...

Since V=IR...you will have a phase shift in the current compared to the voltage. If the circuit is purely reactive...the phase shift will be exactly 90 degrees. So in a purely inductive circuit...the current will lag the voltage by 90 degrees. In almost all cases, there will be some real resistance...so the current lagging the voltage by 10 or 20 degrees is more realistic. The exact number of degrees is in the math.

Little about KVA, VARS and watts.

KVA is the power that must be delivered to a motor to make it run. Watts is the actual power put out at the shaft of the motor. Vars are the "imaginary" part that makes of the vertical part of the power triangle.

Since we hate wasting electricity in this day and age...what if we could add a vertical vector pointing straight up to our out of phase lagging current vector? Turns out we can. All we have to do is add a capacitor in parallel. The motor loves this because it doesn't change anything thru the motor! The voltage across the motor is the same...the current thru the motor is the same...and the power consumed by the motor is the same! Here's the difference. The power company sees the current vector and the capacitor current vector added together. With the addition of the right capacitor, the power company now sees a smaller current that is closer in phase with the voltage. It's magic! I know.

Good question...keep asking...there is much more to learn. Incidentally, I once asked my professor in sophomore year how exactly a capacitor behaves. He pointed at Cdv/d(t)=i(t) on the chalk board. That was the extent of his explanation...lol. Technically, he was right...that's how it behaves!

 

[psparky]

Throw in another two cents about resonance in RLC circuits:

When the resonant series RLC circuit reaches steady-state, the total
circuit impedance is resistive (current is a max). VL-Vc= jXLI-jXcI=0.
The voltage drop across either component could be gigantic but the voltage drop
across the two componets (L&C) in series is zero. What happens??
The capacitor's magnetic field energy oscillates with the electric
field energy of the inductor. Not from the source!

When the parallel RLC circuit reaches steady-state, the total circuit impedance
is resistive. The same thing happens except the capacitor and inductor currents
are 180 degress out of phase. Equal inductive and capacitive reactances in a
parallel circuit at resonance act as an open circuit. That's the only way it will
work since all components in a parallel circuit have the same voltage drop.

Make sense?

And you better believe this relates to power factor. If you are at resonance
frequency, your power factor is 1. Voltage and current are in perfect phase.

 

[johnnybee]

good info , I do not understand what the letters are in the formula : Cdv/d(t)=i(t) t is time constane ?? v must be voltage ? we always used E for votage. C is capacitance ; in farads, of micro f, or nano f ? and in the formula : 1/(ωC<90) or ωL<90 what is w ? and whaere is the equal sign ? and I understand reactive power but I am really having a difficult time with leading current . I see the graph makes it look like it is leading but in physical senses how can current come ahead of voltage when the voltage is what produces the current ? how does the term leading current work in physical bodies such as electrons ? Thanks ,

Johnnybee.

 

[psparky]

good info , I do not understand what the letters are in the formula : Cdv/d(t)=i(t) t is time constane ?? v must be voltage ? we always used E for votage. C is capacitance ; in farads, of micro f, or nano f ? and in the formula : 1/(ωC<90) or ωL<90 what is w ? and whaere is the equal sign ? and I understand reactive power but I am really having a difficult time with leading current . I see the graph makes it look like it is leading but in physical senses how can current come ahead of voltage when the voltage is what produces the current ? how does the term leading current work in physical bodies such as electrons ? Thanks ,

Johnnybee.

As far as the physical...you need to speak to someone in physics...or Sophie is pretty good too. I just understand the math more or less.

t is a way of saying "in respect to time".
Cdv/dt=it. You could say...the change in voltage over time multiplied by the capacitance...equals current over time. Or you could say...take the derivative of the voltage across the capacitor...multiply it by the capacitance to get the current over time.

ω= radians per second...or you could say frequency =2∏*ω
Capacitance is capactince. if you have 5 uF...you plug in the number .000005 for C

When i say ωL<90...that means a vector with the magnitude ωL at 90 degrees! JωL is identical! J=1<90! J*J=-1 or 1<180! (< in my explanation simply mean "angle"...not less than)

Let's say you have a voltage source of 170sin(377t) with a capacitor of .001 farads. Incidentally...ω=377 in that voltage source!
In this case, I take the derviative of the voltage source which is:
377*170cos(377t)...then multiply the magnitude times .001 for C.

Or...if you have a trianglar input wave...you use more the change in voltage over time...then multiply by the capacitance. You are basically just finding the slope of the line in either case.

How does the current lead the voltage in a capacitve circuit? Good question. Don't forget that the sin wave is cooking at 60 times per second. The current just leads the voltage at any given time. It can also be represented in the vector form...keep in mind that the vectors are also spinning at 60 times per second...or 60 Hz.

Keep asking questions...this stuff is not easy. It took me about 10 years to fully grasp it. So if you don't quite have a handle on it in 10 minutes...don't fret!

 

[psparky]

And before you ask about DC...yes, all the rules I just layed out apply the same way.

Capacitors open and inductors short in DC...right?

Ok...let's check the math.

Caps:
Cdv/dt= i(t). What's the change in voltage over time in DC? There is no change...there for the current is zero...or open ciruit. Check.

Reactance (resistance) of cap is 1/Jωc. ω=0 in DC. Plug it into the formula...and you get infinite resistance...or open circuit. Check.

Inductors:
Ldi/dt=v(t). What's the change in current in steady state. Zero. Therefore there is zero voltage drop...behaves like a short. Check.

Reactance (resistance) of an inductor is JωL. ω=0 in DC. Zero resistance...short...check.

Ok...now you are going to say what about transient. Think about it...when you first get a voltage source across a capacitor...there is a big change in voltage...so there is full current at t=0+...then it steady states into zero current in DC according to it's time constant.

Same for inductors...but opposite. Upon hooking up voltage source...there is a big change in current thru inductor...hence full voltage...inductor acts like open at t=0+...then steady states into a short at steady state in DC.

 

[jim hardy]

I always thought of water flow to compare it to. Now I am told here that that is all wrong. It seems to me that current can never lead voltage as voltage is required to cause the current to flow. The flow is current . The voltage pushes the and causes the current and the amount of current is determined by the resistance of the material which would be the load on the voltage (pressure).

Water analogy is not perfect but will get you along way in your understanding.

Your capacitor is a storage tank that can hold water.
Flow into it causes pressure to change gradually as water inventory builds, not immediately as with resistance.
That accumulation effect is cause of time delay.

Next recall that a sinewave is a special case in nature. We use it everywhere because it's so mathematically beautiful.
For a sinewave a time delay is a phase shift.
Now - since pressure is delayed wrt flow, volts is delayed wrt current and the question is reduced to "which sinewave comes first"

volts behind current means current leads.

Does that help?

Now I'm ducking for cover from the slings & arrows this post will invoke.

If the water analogy gets you past this hurdle it has served its purpose.
Next - inductance is inertia...

then on to phasors

 

[vk6kro]

The current in a capacitor is not directly caused by the supply voltage.

It is caused by the difference between the supply voltage and the capacitor voltage

Assume you have a capacitor across a power source giving a sine wave and insert a small resistor in series with one of the connections.

The resistor isn't necessary, but it makes the principle easier to explain.

The voltage from the power source will vary and the voltage across the capacitor will follow along behind this voltage.
The current through the resistor will depend on the difference between the two voltages. At times, the capacitor will actually be feeding current back into the supply if the capacitor voltage is greater than the input voltage.

It looks like this:
http://dl.dropbox.com/u/4222062/Capacitor%20current.PNG

The green trace is the supply voltage. The blue one is the capacitor voltage and the red one is the current in the resistor, but also the capacitor current.

Can you see that the red trace starts off at zero like the other traces, but within a half cycle, it leads the input voltage? This is because the current is greatest when the input voltage is varying most, not when it is at its maximum value. A sine wave varies most when it is at the zero crossing point.

 

[psparky]

Another little trickle on resonance.

At resonance we are obviously looking for ZERO reactance.

So in a LC circuit...or RLC circuit...there is one reactance vector pointing up...and one point down. When the magnitude of these differ...you will have reactive power.

So if we want one reactance vector to equal the other vector...we should simply add the two together to obtain zero...and to derive the resonance formula.

So let's say JωL+1/(JωC)=0

This gives JωL = -1/(JωC)

This should satisfy zero reactance.

Multiplying J*J gives -1...which gets rid of the negative sign and the J's.

Now, solve for ω...and you get

ω^2=1/(LC)...aka...the resonance formula.

If you have a parallel circuit (LC) with an AC source in parallel with a cap and inductor...if you are at resonance...and you graph the reactance...you will have an empty graph!

If it's a RLC circuit at resonance...if you graph the resistance and reactance, you will have a graph with one vector on the real plane...aka...sitting at zero degrees.