Regulated power supply Calculation

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Paras Bhanot

Joined Jun 22, 2015
3
hello everyone ,

i am working on home automation project in which Micro-controller (atmega328p ) is used . This micro-Controller is powered by Regulated power supply . I want to calculate Total power consumption in KWhr so that i can calculate my monthly electricity bill if i run my circuit 24/7.

To simplify Calculations i replaced atmega32p with simple led. Then i measured input rms voltage and input rms current using multimeter on input side as shown in my attached circuit. i have got the following results ->

V(rms) = 220 v and I(rms) = 6.6 mA

So P(rms) =V(rms) X I(rms) = 220 X 6.6 mA = 1.45W

in KWhr = (1.45W x 24 hrs) /1000 = 0.03484 which is obviously wrong

Note : Cost of 1 unit = 8 bucks in my country

What is the correct way to calculate Total power consumption.
 

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hp1729

Joined Nov 23, 2015
2,304
hello everyone ,

i am working on home automation project in which Micro-controller (atmega328p ) is used . This micro-Controller is powered by Regulated power supply . I want to calculate Total power consumption in KWhr so that i can calculate my monthly electricity bill if i run my circuit 24/7.

To simplify Calculations i replaced atmega32p with simple led. Then i measured input rms voltage and input rms current using multimeter on input side as shown in my attached circuit. i have got the following results ->

V(rms) = 220 v and I(rms) = 6.6 mA

So P(rms) =V(rms) X I(rms) = 220 X 6.6 mA = 1.45W

in KWhr = (1.45W x 24 hrs) /1000 = 0.03484 which is obviously wrong

Note : Cost of 1 unit = 8 bucks in my country

What is the correct way to calculate Total power consumption.
34.8 watt-hours per day. Sounds right. 0.034 kilo-watt-hours per day.
What is your billing period? Monthly? Quarterly?
 

crutschow

Joined Mar 14, 2008
25,245
Isn't there a sine or cosine involved here?
For true power, yes.
They actually calculated apparent power since the power factor wasn't included, but that's likely close enough to the real power for the purposes of the calculation. It is a pessimistic value so the real power will always be less.
 
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