# Physical Meaning of Leading/Lagging Voltage/Current

1. Jun 20, 2007

### 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

2. Jun 20, 2007

### ice109

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

3. Jun 20, 2007

### nrqed

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.

Last edited: Jun 21, 2007
4. Jun 20, 2007

### Staff: Mentor

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.

5. Jun 20, 2007

### 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. im sure you already knew this but i was just clarifying

6. Jun 20, 2007

### scothoward

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

7. Jun 20, 2007

### 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).

8. Jun 20, 2007

### ice109

lol @ borrowing power from the power company

9. Jun 21, 2007

### Staff: Mentor

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.

10. Dec 1, 2007

### 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 cant 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 cant envisage what the pesky electrons,magnetic dipoles etc are up to.

Old Engineer.

11. Dec 2, 2007

### alvaros

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.

12. Dec 2, 2007

### alvaros

Keith rigby:
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.

13. Dec 2, 2007

### Keith rigby

insomnia cure and Maxwell

You just couldnt 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 cant 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 doesnt help my physical visualisation.

Our friend who states that the current cant 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 Im off to bed and visualizing if the capacitive transient and steady state really supports this is another guaranteed insomnia cure.

14. Dec 3, 2007

### seang

Another similiar 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?

15. Dec 3, 2007

### 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

16. Dec 3, 2007

### cabraham

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

17. Mar 19, 2010

### 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

18. Mar 19, 2010

### sophiecentaur

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.

19. Mar 19, 2010

### 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 alot 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.

20. Mar 19, 2010

### stewartcs

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

CS