electricman2
Senior Member
- Location
- North Carolina
- Occupation
- Retired Electrical Contractor
electricrw said:I have two different answers from this forum. which one is correct from the question?
If a heater is rated at 4000watts at 240volts single phase.
How many watts will the same heater produce at 208volts
single phase?
I=P/E so 4000/240=16.67 amperes & P=Ice so 16.67 x 208 = 3,467 watts
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I = P/E
I = 4000/240= 16.67 amps
R = E/I
R = 240/16.67 = 14.4 ohms
P = (E * E)/R = 43264/14.4 = 3004.4 watts
LarryFine said:Okay, wiseguy!The heater will likely not be "rated" for 4000 watts at 208v. After all, the question was: I bet there is a 208v rating somewhere in the documentation, if not directly on the unit itself.
Wiseguy!![]()
smile
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infinity said:The correct answer is below the dashed line. (At least that's how I calculated it).
Apologies. I thought you were saying that, regardless of the output power, the "rating" itself wouldn't change.nakulak said:sorry, wasn't trying to be a wiseguy, I just thought it was a trick question![]()
tonyou812 said:wouldnt you want to find the resistance value first with E sq Divided by P
than use that resistance value with the new voltage by using E sq Divided by
R ?
Yeah, but I'll betcha the resistance would change quite a bit if you exceeded the wattage rating :grin:480sparky said:...When you buy a resistor, it does not have different ohms values depending on voltage.
Sure... but where is the rule that says once an ohmage is established, it cannot change?An ohm is an ohm is an ohm.
Smart $ said:Yeah, but I'll betcha the resistance would change quite a bit if you exceeded the wattage rating :grin:
Sure... but where is the rule that says once an ohmage is established, it cannot change?
Explain a thermistor.
Resistance changes any time you change the amount of current. Current creates heat in the conductor which in turn changes the resistance.
[NOTE: For all but engineering level tests, the change in resistance is usually ignored.]
That's exactly what I did in my first post: Find the resistance, then apply the new voltage.tonyou812 said:wouldnt you want to find the resistance value first with E sq Divided by P
than use that resistance value with the new voltage by using E sq Divided by R ?
Not enough info provided for even a fairly accurate guess!!! If I actually had the heater, I wouldn't even waste time trying to calculate it. I'd simply plug it in, turn it on, and start taking current measurements.480sparky said:So your answer is.....?