# Calculate Energy consumption by the adapter from its rating

1. Oct 6, 2014

### kverma

I want to calculate the total energy consumed by my lan switch in one month.
It has a AC adapter with rating-->
Input = 100-240V ~ 0.6A
Output = 5V - 1500 mA
Note:First I will calculate in joules then convert it in kWh.
At first, i was considering calculating using the input rating so the formula goes,
E = I2Rt
E = IVt .....as V=IR
I = 0.6A
V=240V(230-240V in India)
t = 60 x 60 x 24 x 30 (for a month )
Therefore,
E = 240 x 0.6 x 60 x 60 x 24 x 30
E = 124416000 J
E = 103.68 kWh
And this is not possible as it is too much.
It will cost me around Rs.777.6 and by total bill is Rs.3000.
How is it possible?
Should I take the output rating or i am correct?

2. Oct 6, 2014

### f95toli

The rating on the adapter tells you the maximum current it can supply (0.6A); but the switch is probably not drawing anywhere near that much. If you want to know how much power is used by the switch you have to look in the specifications for the switch itself, not the adapter.

3. Oct 6, 2014

### Staff: Mentor

Welcome to PF!

First, your procedure was a bit more complicated than necessary: you had Watts and wanted [kilo]Watt hours, so there was no need to multiply by 60x60 to get Joules and then divide by 60x60 to convert back to watts. Somehow there is an error in that intermediate step (12 MJ is wrong: you didn't multiply by 30), but the 103.68 kW is "correct".

Now, what does it mean? Is it high?

Yes, it is too high. A few reasons:
1. The device probably runs at a fraction of peak load. Figure half.
2. Inductive loads (the power supply) are subject to power factor, which makes the wattage calculation too high. Again, figure half.
3. The first part of the power supply circuitry is probably a bit like a transformer, so higher voltage means lower amperage to satisfy conservation of energy. So you should reduce the result by another 100/240.

So that leaves you with about 10.8 kWh.

Compare that with the output: 5.4 kWh.

Since I could see the efficiency being 50% or less, those numbers match-up pretty well. 10.8 kWh would be my conservative estimate.