Given a device that uses electricity from the wall socket (220-240V AC@50Hz or 110V AC@60HZ), and it has its current draw specific (i.e 1A), how can I calculate the power (in watts) that it consumes?
I know that for DC current, the calculation is very simple (Current multiplied by voltage) but I understood that for AC the calculation is rather more complicated.
Also, its quite unreasonable that my charger, that uses 1.5A AC, would consume 345W and only output 19V * 4.74A = 90.06W (255W turning into heat) and it won't get extremely hot. Also, it is rated above 80% efficiency, which doesn't meet the previous calculation.
How can I calculate an electric device's power usage given its current and AC voltage?
Best Answer
That would be unreasonable but that doesn't happen. The 1.5 A of current is at the lowest AC supply voltage that your charger can tolerate and still work. From what you are saying that sounds like 110V RMS. So now your input power will be 165 watts but that's still sounding a bit high so maybe your charger can operate at the "generally accepted" "world-standard" of 85V to 265V. At 85 volts RMS and 1.5 amps, the input power is 127.5 watts and this is a power efficiency of: -
Power efficiency = 100% x output power / inputted power = 71%.
If it is rated at above 80% efficiency this may be at the 230V range with efficiency dropping to 71% at the lower end of the range. Or you may have measured the value incorrectly.
Whether DC or AC, instantaneous power is instantaneous voltage x instantanous current. You then find that if you average the instantanous power over time, it becomes real power of which your can be billed for. Here are various scenarios: -
If you want to do it accurately you have to use a wattmeter and this gives you true average power - at the heart of any wattmeter is a device that multiplies the waveforms of voltage and current.
These days power supplies like yours are usually what is known as "power factor corrected" but if yours is an older type supply, the current and voltage waveforms may both look very different and not be truly in-phase. That makes simple power estimations problematic.