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how to calculate this?


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If I have 8 units of 7 x10 LED Lights...

- Per Unit PAR CAN is 7x10w = 70w

- 8 Units PAR CAN - 70w x 8 = 560w

Malaysia 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

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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.

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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.

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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. 

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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).

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