A baseball heater is connected to a 120V outlet, the kinds of the
heater is adjusted to half its minimum resistance, 25ohm, how much
current flows to the heater?  How much power is required to run the
heater?  If the heater runs for 12hrs. (all night), how many kilowatts
hours does the heater run?  If the cost of electricity is $0.20/KWh,
how much does it cost to run the heater all night?

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Request for Question Clarification by maniac-ga on 22 Feb 2005 17:59 PST

Hello Daisy001,

The question you ask is pretty straight forward except for the following part:
  ...the heater is adjusted to half its minimum resistance, 25ohm....
If the "mimimum resistance" is 25 ohms, you cannot adjust it to half
that value. Do you mean instead that the resistance is doubled (to 50
ohms) so the current is cut in half?

Can you please explain the proper interpretation of this so the
question can be answered?

  --Maniac

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Clarification of Question by daisy001-ga on 22 Feb 2005 18:32 PST

The question stated:  A baseboard heater is connected to a 120V
outlet, the kind of heater is adjusted to half its minimum resistance,
25ohm, how much current flows to the heater?
I am not sure what the question is really asking that could be right
that the resistance is doubled and the current is then cut in half?? 
The question stated what I had down but if that is not possible then
it must be what you were thinking.
Thanks!!

I can't help wondering "What the heck is a baseball heater?"

My best guess is that it's a baseboard heater, but if not, I'd like to
know what it is.

"kinds"???
"adjusted to half its *minimum*"???  If you can do that, it wasn't its minimum!

OK - that and the baseball heater aside, here is my take on it.

120V through 25 ohms gives a current of 4.8A 
120V @ 4.8A = 576 W (there's your required power)
576W x 12 h = 6.912 kW.h

At $0.20 per kW.h, this would cost $1.38.

Note that the SI prefix "kilo" is abbreviated "k", not "K".  "K" is
the symbol for the Kelvin, a unit of temperature.

Just a small P.S.

The calculations I gave were really for a DC circuit.  Strictly, power
is not just the product of voltage and current in an AC circuit as
they can be out of phase.  That is why you will often see AC power
ratings given as "VA", not as "W".  There is something called a "power
factor" that must be applied to take into account any phase
difference.  For a heating circuit such as the one you describe, this
is likely to be very close to unity, so it makes little practical
difference.