Given the current and voltage of a consumer electronic device, how to calculate its power draw?

Question

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?

Answer 1

Also, its quite unreasonable that my charger, that uses 1.5A AC, would consume 345W and only output 19V * 4.74A = 90.06W

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%.

Also, it is rated above 80% efficiency, which doesn't meet the previous calculation.

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: -

How can I calculate an electric device's power usage given its current and AC voltage?

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.

Answer 2

Your DC formula is correct: P = V * I. The formula applies in AC too, provided the load is resistive. Things get a little more complicated if the 'power factor' is less than one as is the case with motors, for example. The formula then becomes P = V * I * PF.

Your laptop power supply should have a power factor close to one so we can stick with the basic formula for this.

• The input current quoted will be worst case and may be heavily influenced by the in-rush current on first powering up.
• On a switched mode power supply the maximum current will be drawn at lowest supply voltage. If you check the ratings you will probably see that it's rated for 80 - 240 V AC. As the voltage goes up less current will be required to provide the rated power. Conversely highest current will be drawn at low voltages.

If you can borrow a plug-in watt-meter (wall adaptor type) you can read the actual watts being drawn by your load and compare with your calculated values. Most will also give you the amps and power-factor too which can be quite instructive.

Answer 3

AC Power is calculated as RMS voltage times RMS current (provided the voltage is in phase with current which it typically is).

General formula is: \$ P_{AC} = V_{AC-rms} \cdot I_{AC-rms} \cdot PF \$

PF is Power Factor.

For example, my laptop charger has 1.2 A max current draw but it works with voltages from 100 V to 240 V (all rms).

Output power is 20V*4.5A = 90W. Assuming 80% power conversion efficiency, the power draw on the AC side is 113 W, which is close to the minimum voltage of 100 V times the maximum current draw of 1.2 A.

Charger cables/fuses are specified for maximum current. Lower voltage (as in the US) results in heavier current and higher voltage (in Europe) results in lower current but the components remain the same. The current is specified to design the power distribution panels and power strips. It's good to know how many chargers (worst case) can be connected to one, say, 110V 15 A outlet.

Answer 4

It is not just motors that have a low power factor. Low power factor is also caused by electronic devices that distort the supply current waveform. The current marked on a product is the maximum that could be drawn and may vary considerably with actual operating conditions. Because of those factors, any calculation of power used is likely to be higher than the actual power. Inexpensive plug-in wattmeters appear to be reasonably accurate. The best way to accurately determine power usage is to measure it.

Answer 5

The problem of higher current ratings required on many power supplies is related to the fact that current is absorbed from distribution net in narrow pulses around the peak voltage. That is due to the input filter capacitor that sinks current pulses of high peak value to keep the input voltage leveled. The new PFC corrected supplies should avoid this problem by distributing current over the whole cycle of mains voltage. This is different from reactive power factor of motors because power is absorbed in very short pulses while motors have only a delay between voltage and current. If you have a 100 W charger working with 90 W load you can see peaks of about 2 to 3 A at 220 V with a duration of 1 or 2 millisecond over about 8 to 10 milliseconds of half wave cycle of mains voltage. The problem is much greater with 110 V lines because current pulses are very high respect to medium value. To do a quick computation in case of dc power supplies you should start backwards by multiplying output current by output voltage increasing the product about 5 to 10 % to consider conversion losses. If you want to measure it, there are cheap digital power measurement instruments on the market at about 10 $ on chinese websites.