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Kilowatt /hr conxumptio of: ff attached to a 500VA regulator
1 110 v 60 w TV..1 Dell laptap pc 110-240 V w/60 W adapter
1 DSL modem 220V w/adapter
thank ypu

2007-10-16 15:13:56 · 3 answers · asked by Ricardo S 1 in Science & Mathematics Engineering

3 answers

The Regulator itself does not consume power. To get the Total power consumption read the wattage or VA s printed on the name plates of each piece of equipment that is connected to the regulator. Add all those together and multiply this by the number of hours that you have the equipment turned on . This will give you the number of watt hours. Now divide the watt hours by 1000 and this will give you the number of kilowatt hours.Make sure you do not exceed 500 total watts connected to the 500 VA regulator. VAs multiplied by the power factor are the same as watts. The power factor is equal to the cosine of the phase angle between the voltage and current and is always equal to or less than one. I think the power company charges by the watt rather than the Volt Amp. In your case they will be almost equal anyway so it will be safe to count volt amps as watts. By the way you must be on a tight budget to worry about the cost of the power required to operate the devices you listed.

2007-10-16 17:01:12 · answer #1 · answered by Mr. Un-couth 7 · 0 0

Richard is pretty much right, This doesn't make any sense.

What is an FF?

I see that you are using 500 VA, but how many Watts (Unless you are a huge industrial building, you are getting charged for WATTS not Volt-Amps).

You need to find the Wattage for everything that you want to measure and calculate. Then, the amount of time that each of these are on.

1000W = 1kW
1kWh = 1kW * 1 hour

So, for every hour that you use 1kW, you get charged for 1 kWh.... hope that makes sense.

2007-10-16 16:26:53 · answer #2 · answered by solo 3 · 0 0

Ricardo S. -
Your question may make sense to you - but not to me. I don't seek any points for answering. I think you are careless - and insulting people out doing the mental exercise and trying to answer questions.

2007-10-16 16:18:15 · answer #3 · answered by Richard S 6 · 0 0

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