Ah, my bad, you are correct - I got confused with power factor concerns but indeed for pure sine AC the RMS will make AC power identical to DC power in both load and wires. Thanks for clarifying and setting me straight!
Cor van de Water Chief Scientist Proxim Wireless office +1 408 383 7626 Skype: cor_van_de_water XoIP +31 87 784 1130 private: cvandewater.info http://www.proxim.com This email message (including any attachments) contains confidential and proprietary information of Proxim Wireless Corporation. If you received this message in error, please delete it and notify the sender. Any unauthorized use, disclosure, distribution, or copying of any part of this message is prohibited. -----Original Message----- From: EV [mailto:[email protected]] On Behalf Of Roger Stockton via EV Sent: Wednesday, June 08, 2016 2:08 PM To: Electric Vehicle Discussion List Subject: Re: [EVDL] Off-grid solar house and electric car charging Cor van de Water wrote: > Roger, > You used the wrong definition. The Wikipedia quote is about the power > delivered to the *load* but the discussion was about the power losses in > the *wire* feeding the load. No; just as the power dissipated in a resistive load remains the same, the power dissipated in the wire feeding the load remains the same. To see that this must be true, consider the case of a source connected to a non-zero load by wires with zero resistance: the dissipation in the load is the same for 1A DC or 1Arms. Now, consider a zero-ohm load connected to the source by wires with non-zero resistance: the dissipation in the wires is the same for 1A DC or 1Arms. The principle of superposition allows us to see that this situation holds for any combination of load and wire resistances: 1A DC and 1Arms will dissipate the same energy in the wires to the load, and they also dissipate the same energy in the load, though of course the amounts dissipated in the wire or load depends upon their respective resistance. > To use a simple and somewhat excessive example as illustration: > Case A: DC power to an (arbitrary) load: 1 Amp continuous, 1 Ohm line > resistance so 1 Volt drop, meaning 1V * 1A = 1Watt of loss in the line. > > Case B: power provided with 10% duty cycle so 10A during 10% of the > period and 0 during 90% of the period. Average current still 1A and same > power delivered to the load as in case A. > However, the line load is still 1 Ohm so the 10A current causes a 10V > drop and thus 100 Watt power is lost in the line during the 10% that the > current is flowing, the average power loss in the line is therefor: 100W > * 10% = 10W > so the average power loss in the line is 10 times as high as in case A > due to the peak current being 10 times as high, even though it is only > flowing 1/10th of the time. No. You are confusing average with RMS; the two are not necessarily the same. The RMS value of your 10% duty 10A peak square wave is 3.16Arms, and this will indeed have higher I2R loss than a 1A DC current. For your 10A peak square wave to have an RMS value of 1A, it needs a duty cycle of 1%: RMS = peak * sqrt(duty cycle); average = peak * duty cycle. Cheers, Roger. _______________________________________________ UNSUBSCRIBE: http://www.evdl.org/help/index.html#usub http://lists.evdl.org/listinfo.cgi/ev-evdl.org Read EVAngel's EV News at http://evdl.org/evln/ Please discuss EV drag racing at NEDRA (http://groups.yahoo.com/group/NEDRA) _______________________________________________ UNSUBSCRIBE: http://www.evdl.org/help/index.html#usub http://lists.evdl.org/listinfo.cgi/ev-evdl.org Read EVAngel's EV News at http://evdl.org/evln/ Please discuss EV drag racing at NEDRA (http://groups.yahoo.com/group/NEDRA)
