# Why does my Kill-a-Watt read Power of only half the Voltage times Current?

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.)

• Your math is confused. Vrms * Irms = 120V * 0.16A = 20VA. And 10 != 2 * 20. (Which is lucky, because a power factor of 2.0 is impossible) – Ben Voigt May 16 '16 at 5:04
• @BenVoigt: Fixed the typo – Mehrdad May 16 '16 at 5:20

## 2 Answers

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].

• Kill-a-Watt is supposedly to display input power, not output power. Due to efficiency, the output current will be less than 2 A, possibly by as much as half. The difference ends up as heat in the charger. – Ben Voigt May 16 '16 at 5:03
• The portion of your answer concerning phase is correct, but runs into the problem in the presentation of your question where you claim that the power factor is 2.0 -- There's no phase leading to a power factor > 1.0, because power factor is cos(Vphase-Iphase) – Ben Voigt May 16 '16 at 5:08
• @BenVoigt: Hm, looking at it again it seems correct. "Possibly by as much as half?" No way, I don't believe that... that'd be nuts. I don't think the difference is measurable at all. Is there anything actually wrong with the answer? Looking at it again I think the only mistake was in the question, but the answer got downvoted instead of the question which is confusing... – Mehrdad May 16 '16 at 5:19
• It's not the Kill-a-Watt that's burning power (plus, it should show its own output not its input). Consider the Kill-a-Watt reading to be a good measure of the power into the AC input of your USB charger. Then the USB charger loses some of that power to heat, and the remainder goes out the USB port. Some of that is lost in the USB cable, and the rest goes into your device being charged. – Ben Voigt May 16 '16 at 5:21
• The "2A" rating on the charger is the peak supported, but the actual amount being drawn depends on the device. The voltage is (mostly) pushed by the charger, the current is (mostly) pulled by the device, not pushed. – Ben Voigt May 16 '16 at 5:27

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).