Code challenge question

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480sparky said:
An ohm is an ohm is an ohm.
Click to expand...

He would agree
ohm.jpg
:grin:
 
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=IxE so 16.67 x 208 = 3,467 watts
-------------------------------------

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
 
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
-------------------------------------

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
Click to expand...

The correct answer is below the dashed line. (At least that's how I calculated it).
 
LarryFine said:
Okay, wiseguy! :D 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! :rolleyes: :)smile:)
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sorry, wasn't trying to be a wiseguy, I just thought it was a trick question :D
 
The resistance will not change, so you find the resistance of the heater first.

R=E?/P : R = 240?/4000 : R=57600/240 : R=14.4 Ohms.

Now, calculate P with 208 volts and 14.4 Ohms.

P=E?/R : P=208?/14.4 : P=43264/14.4 : P=3004.44444 watts


Or, using a shortcut, W=4000/240?*208?.
 
nakulak said:
sorry, wasn't trying to be a wiseguy, I just thought it was a trick question :D
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Apologies. I thought you were saying that, regardless of the output power, the "rating" itself wouldn't change.

Now, if we had a piece of equipment that could maintain the same power with varying voltage*, that would be something!


* Switching power supplies aside, of course. :wink:
 
480sparky said:
...When you buy a resistor, it does not have different ohms values depending on voltage.
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Yeah, but I'll betcha the resistance would change quite a bit if you exceeded the wattage rating :grin:

An ohm is an ohm is an ohm.
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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.]
 
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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.]
Click to expand...

So your answer is.....?
 
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 ?
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That's exactly what I did in my first post: Find the resistance, then apply the new voltage.

I just chose to use simpler math for the sake of explanation.
 
480sparky said:
So your answer is.....?
Click to expand...
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.

As I noted, for all but engineering-level tests, ignore the change in resistance.
 
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