# Converting Instantaneous kW to kwH

I'm breaking my head over this. I've got 1200 samples of true RMS current using a BK Precision 5492B over a period of 10 minutes (this means a sample every 0.5 seconds). I'm using a power factor of .8 for this calculation.

P (apparent) = Irms x Vrms

P (real) = P (apparent) x Power Factor.

If I sum all of my 1200 samples of P (real) I get 0.8523W

Now, since I took samples every 0.5 seconds, that means in 1 hour there are 7200 samples, so if I divide that number by 7200 I get: 0.000118385kWh

Am I correct in my calculation?

I'm trying to figure out the cost of running this device for 1 hour.

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Yes, sir. The device was in stand-by mode, it's basically a very small PC, so I expect it to consume very little power in that mode. Alright, so that means I consume 0.000710311kWh in one hour. At $0.10 per kWh it would cost me$0.000071 USD to run it. Correct? – Gustavo Corona Dec 12 '12 at 2:18