I guess I'm calling out "The Original Guru" but any difference in battery life is going to come from efficiencies of the conversion circuit. The input voltage (the batteries) isn't changing and to get 9W out you need at least 9W coming out of the battery, regardless of output voltage. This means at 5.0V, 9W on a 3.7 volt VV device, your battery will be dumping at least 2.45A.
I don't understand this. Anyone?
I think I got the main idea, and it brings up some questions I had anyway.
Adami is wondering the same basic thing I was. I think of watts as heat, when used for our vaping purposes. (A slight amount of light as well too, since a coil does glow a little bit. In an incandescent light bulb the combination of heat and light is just more readily observable).
So I think he's basically saying, and IMO pretty correctly, that a PV with a nominal 3.7V battery putting out 9 watts, regardless of the actual voltage setting or coil resistance [well, he used 5 volts at whatever resistance that would happen to me, but that doesn't really affect the essential point] will have its battery flowing out that 9 watts at 3.7 x 3.7 / x = 9 watts at 2.45 amps.
His question [and now mine] is how exactly does this apparent magic occur, that a battery is "less taxed" if its providing x watts to a coil at lower amps as opposed to putting out the same x watts at higher amps?
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OK, I'm going to leave that as I just typed it, flaws and all. Let's ditch the 3.7 volt device putting out 5 volts and go to an example that I would have used if I had asked the question I was going to ask before adami asked his.
PV with 8.4 volts available to it, regulated to 7.4 volts, asuuming no voltage drop for simplicity.
Powering a 4 ohm single coil at 6 volts its providing 9 watts to the coil, at 1.5 amps.
Powering a 2 ohm single coil at 4.24 volts its providing 9 watts to the coil at 2.12 amps.
Since it would seem to be a free physics lunch, why is it easier for the battery to provide what appears would be the same amount of "heat", 9 watts, to the coil just because the amperage is lower?
And I kindly ask for a gentle reply, no matter who gives it. I would like some credit for hereby admitting that I'm unclear on something. (I know, OMG!)
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Additional details relating to this same question, different application:
I have a 25HP 3-phase air compressor. At 230 volts, it draws 63 amps. At 460 volts, it would draw half of that, 31.5 amps, and smaller wire could be used. Why could smaller wire be used when it seems that the electric motor would be consuming the same overall amount of energy (total watts) to do its job?