Richard Hammond said wiring to bulbs gets hotter when they're dimmed.

[Andrew Gabriel]

Capacitive and inductive loads generate a current phase shift, resulting in wattless power. This is relatively easily corrected by using the opposite inductive or capacitive wattless load to cancel out the phase shift.

However, modern loads such as power supplies used in electronic equipment, compact fluorescents, etc also generates a wattless load due to being non-linear, and drawing current mostly only at the voltage peaks of the mains waveform. However, there is no phase shift, so this can't be corrected by the simple addition of capacitors or inductors. It can be corrected by more complex design of the power supply so it looks more like a resistive load to the supply, and the EU now requires this integral power factor correction for most appliances > 25W, although older appliances won't have it.

You can see 3 different types of uncorrected low power factor at

play with the parameters to observe the change in power factor.

[Brian Morrison]

Just as long as the Chinese manufacturers (yes I know there are other ones) don't put the necessary stuff in to do it for CE testing and then make production batches without it like they have done for EMC filtering in SM PSUs.

[Sofa - Spud]

You miss the point - this is the BBC's take on the Discovery/Quest programmes - *Biggest highest etc etc* , it's prime time and has a popular BBC2 presenter in pace. Nothing new or spectacular - just filler.

What I'd like to know from the crossposting DiYers is how many Washing machines are damaged or destroyed by lost Bra Underwires each year? Earlier I took the washing machine apart for the Nth time to remove lost underwires!

[H. Neary]

And in the breakers and fuses feeding those conductors. I didn't see the demo that was the subject of the initial post, but methinks the supply cable was perhaps quite a bit thinner than the fuse feeding it.

HN

[Andy Wade]

That's not the case for tungsten filament lamps it seems.

Quick experiment (dangerous lash-ups-R-us):

- 100 W GLS bulb (from secret hoard)

- bog standard triac phase control dimmer (Hamilton)

- true RMS ammeter (Fluke 87)

- power and PF readout from Maplin 'energy monitor'

- supply voltage ~235 V

Result:

- as the dimmer is turned down from max. the RMS current falls monotonically. There is no current peak at less than full power due to the filament resistance falling with temperature.

Some readings:

Irms Power PF ------ ----- ---- Max. 412 mA 95 W 0.99 Mid. 324 mA 50 W 0.65 Min. 280 mA 33 W 0.51

No setting of the dimmer gave a current higher than 412 mA.

There's reasonable agreement between 235 V * Irms * PF and the measured power for each setting.

[The Natural Philosopher]

No. VA is voltamps. Although it has the dimension of power it says nothing about the actual power delivered, because the phase is undefined.

[The Natural Philosopher]

This is a misnomer. Power is not wattlesss.

VA has neither power not watts implicitly..

[Steve Terry]

Have you considered wearing lightweight sports bras instead?

Steve Terry

[doug]

Well, the `expert` was using an Agilent 100Mhz `scope to demonstrate the waveforms to the presenter ,no expense spared ! Aitch Aye

[The Natural Philosopher]

So waht? we are not talking RMS current we are talking power in the wire.

Assuming you even have a meter that reads true RMS current in a heavily clipped non sinusoidal wave.

What you dont do of course, was to measure how much power was actually being dissipated in the WIRES.

It must vary in relation to what is being dissipated in the bulb becuase the hot bulb takes several orders less current than the cold one, as we know that cold bulbs have big switch on surges, yet you are assuming that the proportion of power lost to the wire, is the same whether its n series with a hot bulb or a cod one?

Go back and rethink ohms law dude.

Hint. The voltage drop across the wire is not a constant between the full on and part on state,.

Hint: you wont see the current peaks when you are measuring RMS.

Hint: how sure are you that you are ersuin RMS anyay.

Hint: Consider a 50% square wave of 1A peak through a resistor of one ohm the power dissipated is 0.5W. (0.5 * 1^2 * 1)

The average current drawn is 0.5A So surely the power dissipated by that current is (0.5 * 0.5 * 1)= 0.25W.

Reconcile the difference.

[Truebrit]

Even the women. Truebrit.

[Andy Burns]

They didn't fit the bit which they persuaded to ignite the cottonwool/nailvarnish while it was at full brightness, only after it had been dimmed and then they wanted it to catch fire.

[Andy Wade]

You've really lost it now. The dissipation in the wires is just I^2 * R, where I is the RMS current and R is the resistance. This assumes that R is constant over the integration period (10 ms in this case), which is it is.

Yes I do. True-RMS reading multi-meters have been widely available for a long time now. You do have to watch the crest factor on very peaky waveforms, to ensure that clipping doesn't occur in the meter, but that's certainly not a concern in this case.

That was exactly the purpose of the experiment. The RMS current in the circuit and the resistance are all you need to know.

If you have a valid measurement of the RMS current in the wires, and their resistance is constant you don't need to know anything about the load. Oh and BTW the hot/cold resistance ratio for a GLS lamp is about

15:1 - hardly "several orders"

Pah, you need to go and re-sit electrical basics 101.

Agreed, so what.

So long as the meter sees them you're OK. See comments above re crest factor. Anyway the chopped dimmer waveform is an easy one to handle from this point of view.

Very sure. Very very sure. Better than 1% accuracy.

Agreed (assuming you mean a square wave between 0 and 1 A).

No, because you've used the mean current, not the RMS value. In this example the mean-squared current is 0.5 A, so the RMS current is 0.707 A and 0.707^2 * 1 ohm gives 0.5 W. QED.

Done. Trivial.

[Mark]

That was the OU, not Horizon IIRC.

[Ian Jackson]

A squarewave is the easiest example to get your head around (and without having to resort to 'hard sums'). One simplistic way of thinking about it is that (for a 50/50 squarewave), the power is 1W for 50% of the time, and zero for the other 50%. Over the cycle, the average is therefore 0.5W. The same logic can be applied to other mark-space ratios.

[therustyone]

Automatic Power factor corection is fairly common in large commercial buildings/estates to minimise corporate electricity costs. Nothing to do with overheating cables !

[Mark]

What people forget is this programme is for children, not adults. AFAIK it is aimed at getting young people interested in science and it achieves that aim IMHO.

The bit about the burning shed was crap though.

[Man at B&Q]

Which bit of "true RMS ammeter" are you struggling with?

MBQ

[Man at B&Q]

If it's AC with no DC offset then the power is 1W 100% of the time, regardless of the mark space ratio.

Square it, calculate the mean, take the square root. The RMS voltage of an AC square wave with no DC offset is just the peak voltage.

MBQ

[Man at B&Q]

Do they use switching or linear controllers? Bipolar or MOSFET? It could well be more to do with poor controller design than power factor.

MBQ