Physical Meaning of Leading/Lagging Voltage/Current

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Leading and lagging voltage and current in circuits relate to the phase difference between waveforms, particularly in capacitors and inductors. In a capacitor, the current reaches its peak 90 degrees before the voltage, indicating that the voltage lags the current. Conversely, in an inductor, the voltage peaks 90 degrees before the current, leading to the current lagging behind the voltage. This phase relationship corresponds to time delays in the response of the circuit elements to changes in voltage or current. Understanding these behaviors is crucial for analyzing AC circuits and their power dynamics, where capacitors and inductors do not dissipate power like resistors but instead store and release it.
  • #31
johnnybee said:
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!
 
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  • #32
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.
 
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  • #33
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
 
  • #34
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.
 
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  • #35
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.
 
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  • #36
Thanks for the additional info VK6kro...cool info and graph.

Thanks Jim for the water analogy...Sophie love those!

I've read all the info in this thread. Will all the info available...someone reading this thread should have no further questions on caps and inductors and the like.

If you do still have questions...keep asking. And I know 50% of you guys reading this still have questions!

Ask away! Until you fail...you cannot succeed.
 
  • #37
Here is where analogies can let you down. A capacitor is easy to model with a tank of water. Give us a water analogy for an inductor that makes easy sense without using the same Maths as you would need to use for the real thing.
 
  • #38
The flyback boost converter is analogous to the hydraulic ram patented in 1809

http://en.wikipedia.org/wiki/Hydraulic_ram

but one should really dig in and learn the electrical math.
1/2 MV^2 is for mass
1/2 LI^2 is for inductance
1/2 CV^2 is for capacitance
1/2 KX^2 is for a spring
1/2 Iω^2 is for a flywheel
Mother Nature is consistent.

if you can do one you can do them all.

old jim
 
  • #39
There is a common mnemonic for this: ELI the ICE man

ELI - Voltage(E) leads Current(I) in an Inductor(L). The reason why is quite simple, an inductor resists a change in current. By the way, "E" used to be a more common abbreviation than V, V being short for Voltage but E being short for Electromotive force.

ICE - Current(I) leads Voltage(E) in a Capacitor(C). And this is because a Capacitor resists a change in voltage.
 
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