Members stunningbabe Posted April 23, 2013 Members Share Posted April 23, 2013 If I have 8 units of 7 x10 LED Lights...- Per Unit PAR CAN is 7x10w = 70w- 8 Units PAR CAN - 70w x 8 = 560wMalaysia uses 230/240v... So 560w divide by 230 =2.43ampere ?If this formula correct to calculate how many amperes my 8 Par Cans will draw?I have posted this topic in Lighting forum too. I wanna see if I can get the accurate answer from both sides :-D Link to comment Share on other sites More sharing options...
Members Unalaska Posted April 23, 2013 Members Share Posted April 23, 2013 Read the manual on this one, there's also power supply draw too. So little draw to me says put it on a circuit with not much already going on (like back line). Link to comment Share on other sites More sharing options...
Members Bugzie Posted April 23, 2013 Members Share Posted April 23, 2013 Google "Ohm's law" Link to comment Share on other sites More sharing options...
Members BillESC Posted April 23, 2013 Members Share Posted April 23, 2013 Looks about right to me. Link to comment Share on other sites More sharing options...
Members Audiopile Posted April 23, 2013 Members Share Posted April 23, 2013 stunningbabe wrote: So 560w divide by 230 =2.43ampere ? Is this formula correct to calculate how many amperes my 8 Par Cans will draw? I'll suggest: Yes, the formula (computation) is correct. The result (2.43A) is as correct as the specifications of the lighting fixtures. I believe the appropriate term for the result of your computation would be "par value".If it matters: you could verify the actual current draw with an induction amp meter (read and follow instructions of the meter). Chances are it wouldn't matter if the actual draw was 10% or even 20% more-or-less than what the computation suggests. Link to comment Share on other sites More sharing options...
Members Okie4Cards Posted April 23, 2013 Members Share Posted April 23, 2013 You got it. Link to comment Share on other sites More sharing options...
Members stunningbabe Posted April 23, 2013 Author Members Share Posted April 23, 2013 Thank you to all the fine gentlemen here. Link to comment Share on other sites More sharing options...
Members agedhorse Posted April 23, 2013 Members Share Posted April 23, 2013 Power amp is not as easy since there are multiple conditions involved. Generally it's probably close to what is listed in the specs for 1/8 power consumption at whatever impedance load you are using Link to comment Share on other sites More sharing options...
Members Tomm Williams Posted April 23, 2013 Members Share Posted April 23, 2013 That's a good tip, I'll keep that one on file Link to comment Share on other sites More sharing options...
Members stunningbabe Posted April 24, 2013 Author Members Share Posted April 24, 2013 Thx guys. Audiopile...yes you are right. I am scared to trip the breaker. Link to comment Share on other sites More sharing options...
Members agedhorse Posted April 24, 2013 Members Share Posted April 24, 2013 What kind of amp? Class AB? G/H? D? Is this total power of per channel? How is this power rated? DB RMS? Peak? Behringer? Link to comment Share on other sites More sharing options...
Members stunningbabe Posted April 24, 2013 Author Members Share Posted April 24, 2013 What kind of amp? Class AB? G/H? D? Is this total power of per channel? How is this power rated? DB RMS? Peak? Behringer? WOW! so many factors involved? OMG! Link to comment Share on other sites More sharing options...
Members agedhorse Posted April 24, 2013 Members Share Posted April 24, 2013 You guys are all way off base in your estimates. For a class d amp, 1000 watts rms (really) per channel is about 85% efficient so 2000 watts x 1.15 = 2300 watts x some factor between 0.125 and 0.250 gives average audio power at 85% efficiency and divide this by 230 v to get average current at 230v at 8 ohms There will be a different average power consumption at different load impedances based on the amplifiers rated load capacity. If the amp is rated at peak power, the power delivered and consumed will be derated by 50% because of the math difference. Link to comment Share on other sites More sharing options...
Members stunningbabe Posted April 24, 2013 Author Members Share Posted April 24, 2013 Once again...agedhorse has managed to answer my question in depth. Now...if only I can understand what he said I feel so stupid sometimes as I cant understand some of the things he said...due to my poor English Link to comment Share on other sites More sharing options...
Members Okie4Cards Posted April 24, 2013 Members Share Posted April 24, 2013 At 230V, the QSC PL340 is rated at 7A current draw with typical music program material with occasional clipping. This is straight from their spec. I have specified AC requirements for many projects. I use this spec to determine my AC requirements. agedhorse wrote: You guys are all way off base in your estimates. For a class d amp, 1000 watts rms (really) per channel is about 85% efficient so 2000 watts x 1.15 = 2300 watts x some factor between 0.125 and 0.250 gives average audio power at 85% efficiency and divide this by 230 v to get average current at 230v at 8 ohms There will be a different average power consumption at different load impedances based on the amplifiers rated load capacity. If the amp is rated at peak power, the power delivered and consumed will be derated by 50% because of the math difference. Please show me how this is inaccurate. Link to comment Share on other sites More sharing options...
Members Gregidon Posted April 24, 2013 Members Share Posted April 24, 2013 agedhorse wrote: You guys are all way off base in your estimates. For a class d amp, 1000 watts rms (really) per channel is about 85% efficient so 2000 watts x 1.15 = 2300 watts x some factor between 0.125 and 0.250 gives average audio power at 85% efficiency and divide this by 230 v to get average current at 230v at 8 ohms There will be a different average power consumption at different load impedances based on the amplifiers rated load capacity. If the amp is rated at peak power, the power delivered and consumed will be derated by 50% because of the math difference. Not sure I agree with "you guys are all way off base". My estimate comes in pretty darn close to yours (though you explained the specifics a little more clearly). Link to comment Share on other sites More sharing options...
Members StratGuy22 Posted April 27, 2013 Members Share Posted April 27, 2013 How many of what do you want to plug into what type of circuit at the same time. Link to comment Share on other sites More sharing options...
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