Watts and heat question

At 1/2 the voltage you will get 1/4 the power because the current also falls to 1/2. P=I x E

One quarter. You can get half-power by using a diode from 220. You still get 1000W but for half the time on average, so 500W. A diode from 110 will give you 125W. If you have two 1000W heaters you can get even more combinations with diodes:

2000/1500/1250/1125/1000/750/500/250/125W

Using diodes this way at high power levels is a bit naughty.

If the resistance of the heating coil remains the same, yes, the amperage will be halved as well and so you'll get 1/4 power.

In actuality, a high-power heating coil is likely non-linear enough that it'll have noticeably lower resistance at 110V than 220V, so I'd wager you'll end up with a bit more than 1/4 power. To really figure this out you'd have to look up with coefficient of resistance for nichrome or whatever it is the heating element is made out of.

---Joel

This depends on how the heating plate is implemented. If it is a *simple* resistance that doesn't vary with temperature, etc. and has no fixed "overhead" (controls sometimes require power), then the power is quartered when the voltage is halved.

Note, however, that your hotplate might *claim* to consume

1000W yet not actually do so -- the manufacturer can err on the high side in the specification (1000W at 220VAC nominal? or, "high line", "low line", etc.).

If you want to dot the i's, you would look at the actual electrical model for the load (but, for the purpose of a forum, "quarter" is probably closer to actual than "half" -- regardless of the "little details")

I've wondered what a lot of DC will do to an electric meter.

--

John Larkin         Highland Technology, Inc

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Thanks everyone. OK, time to eat a little crow and slink back to the forum and make my apologies. :)

Dave

Another way to figure it is P = E^2/R

Assuming R is the same at 110 and 220, P is 4 times greater at 220 than 110. The quarter theory is correct.

Ed

Robert you are so right. So often these SIMPLE questions from people with a very limited knowledge (no crime in that )of the subject get dragged off into the stratsophere by some people wanting to display their grasp of theories (often of limited practical use)when the OP needs a simple answere to his simple question.

--
John G

Start with the false assumption that the heater is linear and of fixed resistance over temperature. Then forget stupid theories and use MATHEMATICS to derive facts. Start with specification: 220VAC,1000W and calculate resistance: P=E*E/R or (1000W) = (220V)*(220V)/(Rohms) or R=48.4 ohms. Then calculate power at new voltage: P=E*E/R or (Pwatts) = (110V)*(110V)/(48.4ohms) or P=250 watts. THEREFORE, given the assumptions above, the FACT is that the power dissipated at one-half voltage will be approximately one-quarter. NO STUPID THEORIES!

Neither. If R remained constant it would be 1/4 the power, but R doesn't remain constant over temperature, it varies a fair bit. So it'll eat more than 1/4 rated power.

NT

The concept is valid, but I bet it makes a difference what kind of heater you have.

At the risk of tainting a philosophical discussion with experimental data. I measured

120V >> 1505W 60V >> 387W This on a low temperature forced air portable space heater.

I suspect that for a radiant heater, where the element temperature variation could be significantly greater, you'll see two symptoms.

1)The power at half volts will be greater than 1/4 the full power. 2)The heater won't work worth a crap because it's not radiating and convection don't get the heat where you want it.