Timeline for Why does my Kill-a-Watt read Power of only half the Voltage times Current?
Current License: CC BY-SA 3.0
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| May 16, 2016 at 5:31 | history | edited | user541686 | CC BY-SA 3.0 |
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| May 16, 2016 at 5:30 | comment | added | user541686 | @BenVoigt: Ahh... that's where the problem is. Even though I knew that part, I assumed since the charger came with the phone and the phone was low on battery, the phone would draw maximum power. Didn't realize that probably wasn't the case, but that makes sense. Thanks! | |
| May 16, 2016 at 5:27 | comment | added | Ben Voigt | 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. | |
| May 16, 2016 at 5:26 | comment | added | Ben Voigt | For your "don't think the difference is measurable", check out some data from someone who measured As a rule of thumb you can expect about 70% efficiency from small AC/DC converters. | |
| May 16, 2016 at 5:26 | comment | added | user541686 | @BenVoigt: Sorry, yes, I got that and then somehow I momentarily forgot that you were talking about the adapter. But again, I don't think that's right... in fact the adapter actually says its USB output is supposed to be 5 V and 2 A, so I think any loss due to it is negligible here... | |
| May 16, 2016 at 5:21 | comment | added | Ben Voigt | 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. | |
| May 16, 2016 at 5:19 | comment | added | user541686 | @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... | |
| May 16, 2016 at 5:17 | history | undeleted | user541686 | ||
| May 16, 2016 at 5:11 | history | deleted | user541686 | via Vote | |
| May 16, 2016 at 5:08 | comment | added | Ben Voigt |
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)
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| May 16, 2016 at 5:03 | comment | added | Ben Voigt | 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. | |
| May 16, 2016 at 4:40 | history | answered | user541686 | CC BY-SA 3.0 |