simploerob Posted March 31, 2009 Share Posted March 31, 2009 what is the amps of a 1000w parcan and a 1200w strand cantana 18-32 thanks simon Link to comment Share on other sites More sharing options...
mark_s Posted March 31, 2009 Share Posted March 31, 2009 P = IV, where P is power (watts), I is current (amps), and V is potential difference (volts).Rearrange to get I = P/V I = 1000/240I = 4.2ish (for your parcan) Repeat the process for the 1.2kW fresnel <_< Link to comment Share on other sites More sharing options...
Robin D Posted March 31, 2009 Share Posted March 31, 2009 A = W / V when I went to school. In rough terms at 240v your examples are 4A and 5A. Slightly concerned that you are having to ask this question if using high power gear! Link to comment Share on other sites More sharing options...
Tom Posted March 31, 2009 Share Posted March 31, 2009 Amps = Watts / Volts Link to comment Share on other sites More sharing options...
simploerob Posted March 31, 2009 Author Share Posted March 31, 2009 thankfully im the sound enginer and can get hold of lampie to check with him this was quicker !!! although I realy should have remembered the formula ..... Link to comment Share on other sites More sharing options...
MarkPAman Posted March 31, 2009 Share Posted March 31, 2009 thankfully im the sound enginer So how much current for a 1000W power amp? ;) Link to comment Share on other sites More sharing options...
paradoxism Posted March 31, 2009 Share Posted March 31, 2009 A = W / V when I went to school. In rough terms at 240v your examples are 4A and 5A. Slightly concerned that you are having to ask this question if using high power gear!Area = Energy / Voltage? Time to go back to school! And while we're at it - the answer to the question is as Mark_S said - 4.2A for the Par64 and 5A for the 1.2k - rounding down to 4A isn't necessary and in fact may end up overloading a circuit somewhere if the conditions are wrong - always round up if you round at all. Is there really any need for a sarcastic comment every time someone asks for help with something that to us seems obvious but to them is simply a piece of knowledge that they don't have? The idea of this forum is surely to help each other out - not berate them for asking for help. If they were asking which of the three thicknesses of fuse wire on their cardboard blister pack they should use to connect the blue and red phases together in the 63A mains board then yes - show concern! But asking for help with the current load of a lantern that uses less power than a typical steam iron (presumably to avoid overloading a channel) is hardly worthy of such a great degree of concern! Not an attack on you particularly Robin - just indicative of a lot of the replies on the forum! Link to comment Share on other sites More sharing options...
jonhole Posted March 31, 2009 Share Posted March 31, 2009 When I started doing lighting at high school I wouldn't have known this.I just new that as long as I only put one lantern per plug (two per channel) I would be fine.Yes, it was a very simple rig, with only 18 channels, and the most powerful lanterns we had were the 1K codas, but using this 'rule' meant we wouldn't reach 10A per channel, and students are a lot less likely to make mistakes and overload the dimmers. Link to comment Share on other sites More sharing options...
Robin D Posted March 31, 2009 Share Posted March 31, 2009 A = W / V when I went to school. In rough terms at 240v your examples are 4A and 5A. Slightly concerned that you are having to ask this question if using high power gear!Area = Energy / Voltage? Time to go back to school! And while we're at it - the answer to the question is as Mark_S said - 4.2A for the Par64 and 5A for the 1.2k - rounding down to 4A isn't necessary and in fact may end up overloading a circuit somewhere if the conditions are wrong - always round up if you round at all. Is there really any need for a sarcastic comment every time someone asks for help with something that to us seems obvious but to them is simply a piece of knowledge that they don't have? The idea of this forum is surely to help each other out - not berate them for asking for help. If they were asking which of the three thicknesses of fuse wire on their cardboard blister pack they should use to connect the blue and red phases together in the 63A mains board then yes - show concern! But asking for help with the current load of a lantern that uses less power than a typical steam iron (presumably to avoid overloading a channel) is hardly worthy of such a great degree of concern! Not an attack on you particularly Robin - just indicative of a lot of the replies on the forum! OK. I take the point about the use of incorrect symbols, but was actually trying to be helpful by responding in a way that the OP would understand if he had never learned the formula. Mark_S put it much better. The fact that the word 'amps' was used rather than 'current' rung a tiny alarm bell. Hence I said I was 'Slightly concerned' and did not state I was suffering a 'great degree of concern'. Nor did I intend to berate the OP for asking the question. It certainly was not sarcasm on my part, and if it read that way I unconditionally apologise to you and the OP. OT I know, and I believe you apropos not attacking me personally, but are you obviously sensitive to other posters who raise similar concerns which is a perfectly valid view for you to hold. However, although a much less active member here than many, I've been around more years than I like to remember and all too often have come up against those who really have so little understanding that they are a danger to themselves and others. Reading between the lines, there would appear to be several BR regulars who have had similar experiences. Surely as responsible people we have a duty to at least pose the question and give the poster a chance to either show that they do have some level of understanding as in the case here, or perhaps to reveal a situation where the advice really ought to be to get someone experienced to help? Nuff said. ;) best wishes ...... Robin Link to comment Share on other sites More sharing options...
adam2 Posted March 31, 2009 Share Posted March 31, 2009 The simple calculations given above will tell you the amps required by any incandescent lantern or other light.In the case of flourescent lamps, discharge sources, LEDs or other non incandescent lights, the position is more complex and current cant be directly determined from the wattage. In such cases either consult the manufactures data, or use a plug in monitor such as the one sold by maplin. Link to comment Share on other sites More sharing options...
Guest lightnix Posted March 31, 2009 Share Posted March 31, 2009 ...I = 1000/240I = 4.2ish......at 240v your examples are...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 :D Not that it stops more coming out of your sockets at certain times of day. This means that 1kW actually requires about 4.35A - a small, but significant difference ;) Design for 4.5A per kW and you will at least build in a smidge of overhead - although it's always preferable to have a tad more than a smidge, IME :D P=IVI=P/VV=P/I:D *May not be true. Link to comment Share on other sites More sharing options...
simploerob Posted March 31, 2009 Author Share Posted March 31, 2009 thanks for the help I am dyslexic and could not remember the formula to do it!! I thought it was better to check here than have a show come crashing down around my head!!!! thanks for the concerns simon Link to comment Share on other sites More sharing options...
J Pearce Posted March 31, 2009 Share Posted March 31, 2009 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. Link to comment Share on other sites More sharing options...
Guest lightnix Posted March 31, 2009 Share Posted March 31, 2009 ...So the P=IV doesn't hold true.Well don't just stop there - tell us what does hold true ;) Quick - before we all blow ourselves and the client up! :D Link to comment Share on other sites More sharing options...
J Pearce Posted March 31, 2009 Share Posted March 31, 2009 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! Link to comment Share on other sites More sharing options...
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