this post was submitted on 15 Mar 2025
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[–] uneatable@lemmy.world 14 points 21 hours ago (3 children)

Ok. I was acountless on lemmy for a long time, your comment made me finally register. Thanks!

So, yeah, with double the voltage you get 4x the power. But you you put 4 times the power at 50% of the time, you get only 2x the power. And the other half of the time, you get 0 power. On the average you get the same power output.

[–] ch00f@lemmy.world 23 points 21 hours ago (1 children)

You double counted there.

You said 4x power 50% of the time and then said “the other half of the time.”

So you’re calculating 50% of 50% which is 25% duty cycle.

[–] uneatable@lemmy.world 16 points 20 hours ago (2 children)

Oh no, I didn't. Should I draw a graph? Pop out some equations?

Let's say P is the nominal power. When I said "The other half" I meant when the solder iron is not plugged. So:

50% of the time at 4xP 50% of the time at 0...

Oh shizzzz, you're right!

[–] Riprif@lemmy.world 2 points 14 hours ago

I imagine this is more or less what it felt like to be in the room at the time. A whole group of people discussing electrical theory and optimal soldering techniques and meanwhile the one guy standing there holding the actual device notices the power cord is a little loose and pushed it in another 1/8" without mentioning it because everyone is so involved in their nerdy conversation.

[–] Juvyn00b@lemmy.world 2 points 19 hours ago

Oh man I was going through it in my head too...

[–] brucethemoose@lemmy.world 7 points 18 hours ago (1 children)
[–] uneatable@lemmy.world 5 points 17 hours ago
[–] brucethemoose@lemmy.world 1 points 19 hours ago* (last edited 18 hours ago)

I had to think about it too, lol. This is an equation for DC/instantaneous power, and if you want to get into AC math, this is more like a square wave. Averaging the power out over time doesn’t necessarily work with the equation, as you figured out, as it doesn’t when you try to measure AC (sinusoidal) power by average voltage or whatever.