Calculating Power Output of a 580W Heater in the US

jdoggigs
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Homework Statement


A 580w heater is designed for use in Germany, where electric outlets supply 230V(rms) service. What is the power output of the heater when plugged into a 120V(rms) outlet in the U.S.? Ignore the effects of temperature on the heater's resistance.


Homework Equations



barP=I(rms)V(rms)

The Attempt at a Solution


I tried to fine I(rms) by first calculating for I in germany: I(rms)=(580w)/(230V).
Then I used this I(rms) to calaulate U.S. power output; barP=(I(rms))(120V).
This did not give the correct answer.
Would I have to consider that germany uses DC and U.S. uses AC, is there a conversion for this?
Thanks
jdoggigs
 
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Irms will be different in the US than it is in Germany. However, one out of the 4 usual variables will be the same. Question for you: which one?

(The 4 usual variables are V, I, R, and P.)
 
jdoggigs said:

Homework Statement


A 580w heater is designed for use in Germany, where electric outlets supply 230V(rms) service. What is the power output of the heater when plugged into a 120V(rms) outlet in the U.S.? Ignore the effects of temperature on the heater's resistance.


Homework Equations



barP=I(rms)V(rms)

The Attempt at a Solution


I tried to fine I(rms) by first calculating for I in germany: I(rms)=(580w)/(230V).
Then I used this I(rms) to calaulate U.S. power output; barP=(I(rms))(120V).
This did not give the correct answer.
Would I have to consider that germany uses DC and U.S. uses AC, is there a conversion for this?
Thanks
jdoggigs

Btw, Germany does use AC. No country (as far as I know) uses DC as mains.
 
Irms is what I need to find th US power outout, I have no resistance, P is the answer I'm
looking for, so voltage is the same. I'm lost here how does this help?
I assumed that Europe used DC because You need different plugs there.
 
Okay after some consideration of the question asked, I figured it out.
Thanks for all the help'
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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