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Electric Heater Theory

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Ztrain1985

Electrical
Dec 18, 2007
34
Maybe its just me, but im having some issues wrapping my head around this.

I was told that in a resitive heater, if you decrease voltage, you increase current draw. However, I can not figure out how.

Say we have a 240 V heater, drawing 10 A, meaning its a 2400 Watt unit.

If we half the voltage, P=(1/2 * V)^2/R. R should be constant, so using V=IR, 240=10*R, R = 24 Ohms. Since Voltage is squared though, halving the voltage means reducing the power output to 1/4 its value. Or, 120^2/24 = 600 Watts. Again using the Power Equation, 600=120V*I meaning current is 5A.

How exactly is current supposed to increase with a voltage drop? Unless Power doesnt change when voltage does (or the resistance somehow changes), what am I missing?
 
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At 1000W, 240V resistance heater will draw half as much current as a 1000W, 120V resistance heater. A 1000W, 240V resistance heater operated at 120V will draw 50% of the current it does when operated at 240V and produce 250W rather than 1000W.
 
So am I right to think that a heater WILL NOT draw more current if the voltage is decreased?
 
A given heater element will draw current proportional to voltage. A heater element is a constant impedance device and will always draw less current as voltage decreases.

Constant power devices, such as motors (over limited ranges of voltages), draw more current as voltage decreases.
 
You were told wrong. V=IR, R is a constant, so if V goes up then I goes up. If V goes down then R goes down. This is true for any constant-resistance load.

In contrast, motors do what you state, as motors are not constant-resistance, but rather constant-power. To maintain the constant P=IV power, if voltage goes down then current must go up. But that's just for motors, not heaters.
 
Depending on how hot the element runs you may not get a linear reduction in current with voltage but it will still be a reduction.

(Probably more info then you wanted.)

Keith Cress
kcress -
 
To say that motors are constant power devices is gross oversimplification. It is not so, and saying that without adding the necessary details an constraints only creates confusion.

Gunnar Englund
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
 
That's why I included "over limited ranges of voltages". Generally considered close enough that for load flow studies (voltage [±]5% or less) collections of motors can be assigned to a block of constant power loads. I'd never dream to suggest that going from 240V to 120V a motor would be constant power.
 
Alright, thanks guys. Im gonna have to go back and ask him to show me how he calculated this.
 
I think davidbeach had the right idea in his initial post. What the guy meant was that if you use a 240 volt heater instead of a 120 volt heater, you will draw less current to get the same heat. This is correct because now we're not talking about reducing the voltage on a constant resistance element. We're talking about using a higher resistance element to get the same power with a higher voltage.

 
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