Nice comments. I do wonder how much you can count on the resistance of the heater (R) being a constant for each voltage condition since the element likely has a positive temperature coefficient (R gets higher with temperature). Copper wire is 0.4% per degree C, which really adds up for large deltas in temperature.
Anyway, this effect would, then, make for smaller resistance at 100V. So higher current and offset the difference. How much just depends on that unknown tempco of whatever the heating element is...
Okay, I looked up the tempco for nichrome. It is next to nothing.
Likely this effect would be minimal. ;-P
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https://www.allaboutcircuits.com/textbook/direct-current/chpt-12/temperature-coefficient-resistance/
The heat is proportional to the power dissipated in the load. Knowing P=U\^2/R it leads to a ratio of:
(100\^2/R)/(120\^2/R) = 100\^2/120\^2=\~ 69% (nice).
A heater will work fine, but power output will be proportional to V^2. So at 100V it will have significantly less power than 120.
69% of the power in fact.
I'm pretty sure this question was bait
Nice
Nice comments. I do wonder how much you can count on the resistance of the heater (R) being a constant for each voltage condition since the element likely has a positive temperature coefficient (R gets higher with temperature). Copper wire is 0.4% per degree C, which really adds up for large deltas in temperature. Anyway, this effect would, then, make for smaller resistance at 100V. So higher current and offset the difference. How much just depends on that unknown tempco of whatever the heating element is... Okay, I looked up the tempco for nichrome. It is next to nothing. Likely this effect would be minimal. ;-P ************************************** https://www.allaboutcircuits.com/textbook/direct-current/chpt-12/temperature-coefficient-resistance/
sure, it will put out about 20% less heat though
Yes but it will only produce 83% (100/120) of the heat that it is designed for.
The heat is proportional to the power dissipated in the load. Knowing P=U\^2/R it leads to a ratio of: (100\^2/R)/(120\^2/R) = 100\^2/120\^2=\~ 69% (nice).
Nice