A heater rated at 2.5KW

Dear friends,

I have a question while I'm studying power in AC circuits. What does it mean by saying 'a heater rated 2.5KW'?

Does it mean that a rms current of I=2500/120 will flow into the heater when it is plugged into a V(rms)=120V socket? And a curent of I=2500/240 will flow into when plugged into a V=240V socket? Is 2.5KW the average power that the heater consumes?

Why is that? I thohght a heater could be treated like a resistor, how can the average power consumed be known before we know about the resistance or impedance of a resistor or equipment?

Thank you!

Heidi said:

Why is that? I thohght a heater could be treated like a resistor, how can the average power consumed be known before we know about the resistance or impedance of a resistor or equipment?

Click to expand...

The heater is like a resistor. The power rating will be for a certain voltage input.

e.g. if it's 2.5kW @ 120 vac, then it will output 10kW @ 240VAC (and likely burn out).

dougy83 said:

e.g. if it's 2.5kW @ 120 vac, then it will output 10kW @ 240VAC (and likely burn out).

Click to expand...

And CERTAINLY burn out (VERY, VERY quickly)

I see. Thank you, friends!

Sorry, Nigel, I was going to click on the "+" sign, wrong click but I don't know how to change it back

dougy83 said:

The heater is like a resistor. The power rating will be for a certain voltage input.

e.g. if it's 2.5kW @ 120 vac, then it will output 10kW @ 240VAC (and likely burn out).

Click to expand...

The full calculations :

Using V=IR and P=IV etc .....

2.5kw@120v = 20.83A
R=V/I = 120/20.83=5.76R element resistance

Plugging this into a 240v source:

I=V/R = 240v/5.76R = 41.6A
P=I*V = 240*41.6=9998.4 watts

picbits said:

The full calculations :

Using V=IR and P=IV etc .....

2.5kw@120v = 20.83A
R=V/I = 120/20.83=5.76R element resistance

Plugging this into a 240v source:

I=V/R = 240v/5.76R = 41.6A
P=I*V = 240*41.6=9998.4 watts

Click to expand...

Actually, it's exactly 10,000 Watts, assuming constant resistance.

Yup - my rounding was slighty out but I did the above in a bit of a hurry

picbits said:

Yup - my rounding was slighty out but I did the above in a bit of a hurry

Click to expand...

I guess my point is, when the voltage is doubled, the power is quadrupled. P=V^2/R
You can do it in your head.