Current is not controlled here, resistance (aka the soldering iron) and voltage are.
Power = Voltage ^ 2 / Resistance. Double the voltage, that quadruples the power. So you only want to plug in 25% of the time to get the equivalent power of 120V.
But it might not melt at double power? Maybe the extra heat helps, I can’t find a resistance/temperature curve for a soldering iron…
In my defense, I’ve been helping a friend with an EVSE install where the load (electric vehicle) is smart. In that context, it’s just voltage X current capacity of the line = power. The rest of the story is true as far as I know.
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.
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.
Shoudn’t it be 25%?
Current is not controlled here, resistance (aka the soldering iron) and voltage are.
Power = Voltage ^ 2 / Resistance. Double the voltage, that quadruples the power. So you only want to plug in 25% of the time to get the equivalent power of 120V.
But it might not melt at double power? Maybe the extra heat helps, I can’t find a resistance/temperature curve for a soldering iron…
Source: EE dropout.
nnnNNEEEEEEERRRRRRRRRd!
Right you are. Oops.
If only the oop was here to see this 😔
In my defense, I’ve been helping a friend with an EVSE install where the load (electric vehicle) is smart. In that context, it’s just voltage X current capacity of the line = power. The rest of the story is true as far as I know.
What are you lookin at
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.
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.
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!
Oh man I was going through it in my head too…
Also, welcome to Lemmy!
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.
I sure hope someone will be fired for this obvious blunder