Is a dimmer and incandescent light efficient?

[The Real Doctor]

"Some of us are asking questions about the proportionality of the police response"

Ian

[The Real Doctor]

I think you are getting a little confused here. Sketch the voltage waveform produced by a dimmer. Now sketch the current waveform you would expect if that voltage waveform went into a purely resistive load. Hint: I(t) = V(t)/R.

Ian

[The Real Doctor]

It's possible that the perpetual motion con artists go to that degree of trouble, but I don;t think they are bright enough in general. I suspect that they may well try several power meters before deciding which to demonstrate to their marks.

The electricity companies get very cross if one strays too far from a power factor of 1.0 on a domestic meter. My recollection is that they used to allow up to 0.9 lagging and no lead at all, but I am open to correction.

Ian

[Chris Charles]

Current can't flow until there is a circuit - switched on - and a voltage is applied. If the circuit is resistive then the current is in phase with the votage. It can't be any other way for a resistive load.

??? If the load is resistive - as the filament is - then no voltage = no current. see basic ohms law. I=E/R for a resistive circuit.

[<me9]

Yes, the voltage and current in the load will be in phase, but the delay in current flow will make the current in the /supply/ lag the voltage.

It will also introduce harmonics (and harmonic currents) into the supply.

[Derek Geldard]

Ho - Ho. (weedy pun).

DG

[Derek Geldard]

Well even that's not right, dI/dt cannot be infinite and I is at zero to start with, and also every time the "perfect switch" closes every half cycle.

That aside, if you were to carry out a fourier series analysis on the resultant I waveform you would find it has a strong component at the

50 Hz. fundamental (plus plenty of harmonics) and a constant delay term related to the phase angle at which the "perfect switch" closes.

It is this phase angle which governs the power transmitted by the dimmer, and also causes the current to lag the voltage.

I do restrict the discussion to cheap, simple phase angle devices such as lamp dimmers and drill speed contollers. If there are "true sinewave" dimmers (as there are "true sinewave" inverters.) I've not come across any, and probably couldn't afford them if I did.

DG

[Adam Aglionby]

Sine wave dimmers are available from a few manufacturers, heres one:

you have to ask the price.....

Adam

[Chris Charles]

Nonsense

You cannot have a power factor with a resistive load. By definition.

The current and voltage are always in phase.

[Derek Geldard]

Your definition does not encompass active components.

Clearly, nay obviously, they are not.

DG

[Chris Charles]

On Mon, 10 Mar 2008 19:48:38 +0000, Derek Geldard

Ok, if that is so what is producing the phase shift.

[Andy Hall]

Oh, he's not a trainspotter as well, is he?

[Derek Geldard]

It's implicit in there being a delay between the voltage impressed and the current drawn.

DG

[The Real Doctor]

Tiny, tiny second order effect.

I am not quite sure what you are getting at here (and not because I don't understand Fourier analysis, by the way - I've written a thesis on it). How do you think the current and voltage waveforms in a resistive load are related?

With all dues respect ... cobblers.

Ian

[The Real Doctor]

What voltage?

Ian

[The Natural Philosopher]

Its VERY fast..thats why you have an indiuctor in there to slow it down..

They exist all right. Fr motor control. Complex PWM plus a smoothing inductor to iron out the ripple in each phase..

[The Natural Philosopher]

It depends entirely on how you define power factor.

The concept becomes almost meanibngless when applied to non linear loads, which a dimmer in series with a resistor, is.

And in fact the current slightly leads the voltage in a normal bulb with no dimmer anyway as it heats and cools through the cycle..

You should have gone a little beyond O level Physics.

[The Natural Philosopher]

"If all the world ran on bicycles, real ale, beards, anoraks and electric trains; we could all sing folk songs and be a proper communitee'

[Andrew Gabriel]

Domestic meters neither know, nor care what the power factor is. They measure the true power (or more correctly, the energy used, the integral of power).

[The Real Doctor]

Not true. Unless you think that attaching a suitable reactive load in order to run the meter backwards means that your house is suddenly producing energy ...

But quite apart from that, it's not just a question of power. The infrastructure for supplying reactive amps costs just as much as the one for resistive amps, and if you hand them back unused no power meter will see 'em. Hence commercial users with large reactive amp requirements negotiate individual deals.

Ian