amps per light

[themadhippy]

If were going to be picky then the above equations only hold true on DC circuits,if its AC then youll have to factor in the power factor,and R dont exist either its Z ;)

[J Pearce]

Quite true.

Though for pure resistive loads (like halogen lamps) Z approximates to R (unless you dim your halogen lamps...).

[adam2]

As others post, the nominal or declared voltage in the UK is now 230 volts, not 240 volts and has been for some years.

 

However for all practical purposes it is in order to use 240 volts for calculations for the following reasons.

1) the actuall measured voltage is still nearer to 240 than 230 most of the time in most places

 

2) The lamps in the lanterns are designed for a 240 volt supply, not for 230 volts

 

3) If a lamp designed for 240 volts is used on a 230 volt supply, it will draw LESS current. (not more as is commonly supposed)

 

One should avoid over-complicating what should be simple matter. After all the UK supply voltage can be up to 253 volts (230+10%) does anyone allow for the slight extra current that a lamp would then use ? no of course not!

Actual lamp wattages can vary by 1 or 2% from the stated nominal wattage, does anyone allow for the fact that a "1KW" par might in fact be a 1,020 watt lamp ? no of course not !

 

In the real world, for incandescent lamps, dividing the wattage by 240 will be fine.

 

In the (very unlikely) event that lamps actually designed for 230 volts are used then one should allow for this. In practice 230 volt lamps are seldom used in the UK since they would have a very short life on our supply which in practice is still 240+ volts.

[Oovis]

P=I^2R where R is the resistance of the lamp.

 

P=IV only holds true for constant power devices. If lamps did this we'd never be able to dim them!

But in the world outside the lab it's easier to work with the rated power and voltage printed on a lamp than to try and find out the impedance of its filament when it's hot (there's no point measuring the resistance of a cold filament as it'll be wildly different to operating values).

 

Coat-Hat-Door ;)

[Ynot]

Except it wont for a lamp.

A lamp is essentially a resistor.

 

Therefore, a 230V lamp run at 240V will overconsume (and run with a shorter life), while a 240V lamp run at 230V lamp will underconsume.

 

So the P=IV doesn't hold true.

 

Er, WHAT???

 

Of course P=IV holds true, as will the rest of Ohm's law - in it's most basic format anyhow.

Whilst it is true that dimmer outputs are a slightly more complex waveform due to the SCR switching or whatever, the sums you need to be concerned with should really be applied to the maximum potential draw on the supply. So you're looking at the dimmer being at 100% as your optimum.

 

So - once your 1000W/230v rated lamp (ignoring tolerances for the moment) running at full on a 230V fed dimmer will consume 4.347A. This equates to a HOT resistance/impedance across that lamp of 52.9R. And yes, it WILL be different to the cold res.

 

Any variation in any element of the formulae can therefore be calculated, should you wish, to work out the effect on any other element.

 

The variations, however, can be dismissed on the whole for most changes. One to be avoided, however, is the use of (eg) 230V lamps on 240V supplies - as has already been said, they would likely burn out in a shorter time than a properly spec'd lamp.

[JMackenzie]

thankfully im the sound enginer

 

So how much current for a 1000W power amp? ;)

 

This a bit of a difficult question. What is the output voltage to the speakers?

Or, is it measured by its VA consumption at the wall socket, when running at full output. I am surprised no-one has taken the bait to answer this question.

[Guest lightnix]

P=I^2R where R is the resistance of the lamp.

 

P=IV only holds true for constant power devices. If lamps did this we'd never be able to dim them!

So why didn't you say so the first time?

 

I think I'd better stop offering advice in here, before I get someone killed.

 

So often these days, it seems that all the things I thought I knew and relied upon to get me safely through the working day were, in fact, just complete bullsh1t

 

;)

[adam2]

thankfully im the sound enginer

 

So how much current for a 1000W power amp? :unsure:

 

 

 

There is not a simple answer to that one !

 

1000 watts at 240 volts is clearly about 4.2 amps, however remember that amplifiers are nothing like 100% efficient, therefore after allowing for the losses in the amplifier, one might expect the input current to be at least twice 4.2 amps and perhaps more.

 

In practice though the current will almost certainly be far less for a number of reasons

 

1) the amplifier rating is probably in "peak music power watts" or some other marketing term, not true RMS watts

2) the rating is probably in chinese watts, achievable only briefly under artificial conditions, not in normal use.

3) very few amplifiers are ever worked at full power for more than a second or two.

 

I have seen several amplifiers with a stated total output of 8,000 watts, worked from a single 16 amp supply.

The 16 amp MCB never tripped in some years, despite the fact that at 50% efficiency, the input current would have been about 70 amps !

 

Likewise I have seen "500watt" car audio systems with a recomended fuse size of 30 amps, thats only an INPUT power of 360 to 400 watts, making an output of 500 a bit unlikely!

 

In practice, the only answer is an informed guess, or testing, allowing a generous margin in either case.

[discopete]

thankfully im the sound enginer

 

So how much current for a 1000W power amp? :P

 

There is not a simple answer to that one !....

 

 

Agreed :unsure: It all gets a lot more complicated when you start talking about inductance instead of resistance. The noise boys will usually think of number and treble it for good measure. You certainly don't to have power problems, under voltage and the resultant square wave output.

 

I learnt a long time ago, DC kills speakers. Any part of the signal path clipping is a bad thing, not just the final foh amp.

[DangerousMark]

Ermmm.... oookay....

 

Yer resistance vs the voltage determines your current...

then the current and voltage determines the power... right?

 

Or was it the other way round? :)

(sorry, long lunch :** laughs out loud **:)

 

Either way - P=IV holds true in the case of a known steady voltage and current, it's how you can determine how much power a thing will draw if you're only given the PD & current, or what current it requires if you're only given the rated power and PD.

There's nothing in it that guarantees that any of the variables is a constant, however, or that they don't have other more complex relationships to each other behind the scenes. Only that this one holds true as well as the other ones.

 

The variation in the power supply is part due to demand, part to design. As was the reduction from 240 to 230V - it was actually meant as an energy saving measure; drop the supply voltage by ~4%, and the overall power demand will fall by ~8% if you assume a perfect system with no constant-power devices connected or no human/thermostatic/etc regulation. Most people won't notice the small change in their lightbulb/TV brightness, and will simply turn the volume control up slightly higher on their radio again without noticing a difference. The fridge motors, microwave elements, electric fires etc will end up running just slightly longer at a lower speed, using the same energy on the way. So maybe you'll only get 3-4% overall reduction, but when you're having to haul/pump many thousands of tons of coal, gas, uranium and what-have-you per day to run the power stations it can be an important difference.

 

BTW the supply voltage as far as I knew is 230 +/- a good 10% or thereabouts, so it's closer to 210-250V, and I've seen close to both ends of the scale when metering mains lines for fun and profit (well, salary and demos to schoolkids), though part of it may have been down to the old workplace's knackered wiring giving typically 205-210V at the wall B-) (luckily almost all the things we plugged in were "100-250v"). It'll vary throughout the day along with the frequency (something like 1.5% for that, 48.5 - 51.5Hz) which is a function of the voltage and load somehow. It's all very complex and related to demand, and the need to ensure that there are near-as-dammit 4.3200000x10^6 cycles over the course of a day by tweaking the number of stations running and at what capacity (and therefore the load, voltage and hertz), and to try and keep the overall voltage on the lower side if possible to conserve power. There's a page out there somewhere that'll even show you the current Hz live in a java applet thing. They're thinking of putting sensors in future intermittent-demand devices (e.g. fridges, permanent-power AC adaptors, even E7 immersion & storage heaters) so that they'll only turn on when the frequency is high showing low load, in order to better balance the network, and flexibly reduce their demand (let the fridge temperature go up another 0.1'C, the hot water down by 1'C or the attached cordless phone's battery discharge slightly) if a high load is detected. Presumably a frequency guage is easier and cheaper to implement than a voltmeter?

 

Anyway ...... what was my point ..... yeah. Figure out your impedance as well as just current/power, and work out what it'll need at the extreme ends of the scale just to be sure, if it looks borderline. Shouldn't be more than +/- 20% though.

 

Discopete: What of the %$£)&!! modern phenomenon of heavily over-compressed (as in dynamic not mp3) recordings, which I'd hazard get close to flatlining in some cases of loud lowfrequency stuff? I often reduce the volume (anywhere from 98 down to 90%) of the worst offenders and run CoolEdit's "DC adjust" function on them just to reintroduce some significant movement to the peaks. No evidence it affects the sound or even the result at all, but it gives me a better sense of security, in not blowing speakers and in it compressing with slightly less additional clipping.

 

Paradoxism: 1) V=W/A... heh... I think they were shorthanding watts/amps in a rather nonscientific way, but good point. 2) The arrogance and snidey snobbery towards noobs and the uneducated on forums, it was ever thus. Even when the newbie turns out to be correct and/or actually have a good idea in the face of resistance from the veterans (been there, done that in a slightly oilier discipline, and probably been guilty of a lot of it in return). Not for nothing does exist the popular avatar that says "^ Not my real name ^ / > Things I Wouldn't Say To Your Face >"

[J Pearce]

Sorry my P=IV statement was misleading.

Of course it always holds true, but using it to calculate current from a lamp power rating and a voltage that is unrelated to the lamp in question will give false answers.

[mark_s]

This topic is ridiculous - so much necessary complexity and elaboration has been posted and argued about that the poor OP - who asked a very straightforward question - is probably scratching his head in confusion! The talk of P=IV not holding true and the need to use P=I^2R is completely irrelevant - how many people here actually measure (or know) the resistance of a hot lamp in order to calculate its power requirements? The value you get from P=IV is adequate in this context, and talking of theoretical complexities just clouds the issue - especially for the OP, who, with the greatest respect, probably isn't particularly comfortable with all these calculations and the theory behind them.

 

While I hate to be pedantic* I think it should be pointed out that the UK officially runs on 230V and has done for... Oooo, ages :)

 

While the nominal voltage may be 230V, I based my reply on the consensus that what comes out of the wall is still 240V - as discussed more than a few times on here over the past few years. A 240V lamp at 230V will underconsume as has been mentioned. Also, again as has been mentioned, lamps sold in the UK are designed for 240V (due to the supply being near enough 240V). As a 240V will underconsume at 230V, it makes sense to use the manufacturer's stated voltage and power figures in calculations, does it not? Any inaccuracies caused by a lower actual voltage will result in the calculations giving a slightly higher figure, so there's no danger of underestimating the power requirements for the rig.

 

The question that the OP asked was answered within the first few posts, and everything else since then seems to have added irrelevant ifs and buts.

 

Moderation: Couldn't have put it better myself. A simple question with an unnecessarily overcomplicated answer. Thread closed.

[andy_s]

erm... my brain aches after reading all that.

 

I'm not an electrical engineer, so for kosher info read all the above.

 

however in my touring days down in the west country I regularly wired my 24 ways of Tempus (2kW per channel) into the cooker switch in the kitchens of village halls dotted around dartmoor and exmoor. As far as I know, most of those halls are still there - I didn't blow any of them up. I worked on a basic rule of thumb of 4A per kW, so a 32A switch would allow me to rig 8kW worth of lights, or possibly more, so long as no state had more than 8kW's worth of lights. Generally speaking the only state likely to approach everything on at full would be the curtain call state, although one lighting design called for a "white-out" rather than a blackout at the end of the show. This was quite effective in the theatres we played, but had to be re-interpretted when we took that show into the village halls.

 

for those of you who like kit lists, we had a dozen 500W minuette fresnels, half a dozen 650W minuette profiles, 2 coda 3s, a couple of manky 1kW starlette Fs that needed rebuilding after every journey in the truck, and a couple of parcans. We later aquired six minuette PCs, which we also lamped at 500W.

 

Anyway, despite having that massive rig available to me, my rule of thumb of 4A per kW never let me down.

 

Should just add that wiring in to cooker switches with nothing more than a bit of common sense some rudimentry instructions from someone better qualified than me and a sense of one's own mortality is not likely to be acceptable as best practice in this day and age! :)