My Kill-a-Watt is displaying that its output is 120 V, 0.16 A, and 10 W.
I'm confused... I thought P = V × I from basic EE. Why do I see only half the expected value here?
(This is a USB charger, and I'm trying to calculate the charger output current indirectly.)
P = V × I only holds if I and V include phase information (i.e. are complex numbers), whereas the "V" and "I" displayed are actually the magnitudes of these variables.
The value we want to use is the P that is displayed, since it includes phase information.
So, in this case, the USB output current should be 10 W / 5 V × [efficiency].
cos(Vphase-Iphase)
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Commented
May 16, 2016 at 5:08
In AC circuits there's an important measurement that your Kill-a-Watt might not be giving you, the Power Factor. See with an AC circuit, peak current might not occur at the same time as peak voltage (this is true for pretty much anything not an ideal resistor). Resistive loads burn power but capacitive and inductive loads don't, they grab a bit of power then give it right back again, they 'hot potato' energy back and forth down the line. These "reactive" loads draw current but not (average) power, the fact that you're getting V*I > P means that your load is partially reactive as some of the energy being drawn is coming in then going right back out again (reactive power) and some is being used by the load to do some real work (called, funnily enough, real power). Real power is the instantaneous product of V and I as opposed to the average product of V and I (which is called the apparent power as it is the sum of real and reactive power and represents the, well... apparent power draw of the device). The power factor is a measure of the amount of Real to Reactive power and is the cosine of the phase angle between the AC current and the AC voltage (this only holds if the two are at the same frequency).
