Extension cord connected to heater - Find the power

1. Mar 1, 2005

thisisfudd

An extension cord made of two wires of diameter 0.129 cm (no. 16 copper wire) and of length 2.7 m is connected to an electric heater which draws 15.0 A on a 120V line. How much power is dissipated in the cord?

Do I just use P=IV?

But that leaves a lot of "extraneous" information.

My second thought was to find resistance:

R = p (L/A)
A = pir^2 = 1.29E-3^2 x pi = 5.23E-6

R = (1.68E-8) x (2.7/5.23E-6)
R = .00867

But then I'm not sure what to do from there.

Using P=IV I get 1800 W. But then do I have to multiply by length and area?

2. Mar 1, 2005

dextercioby

Obviously u cannot use the 1800W...That's the power dissipated by the SOURCE...For the line,u need to use another formula
$$P=RI^{2}$$

Daniel.

P.S.The potential between the ends of the chord in NOT 120V...

3. Mar 1, 2005

thisisfudd

OK, so I have found resistance, .00867. So then I can use P = RI^2?

P = .00867 x (15.0A)^2?

Your PS is intriguing but of course I don't understand. Are you saying that it draws 15 A on a 120 V line but I have to find what it draws on this line, given the voltage of this line? How would I go about finding that? Let's see what I know: resistance, and a ratio of current to voltage?

If V = IR

V/I = .00867

4. Mar 1, 2005

dextercioby

You found the power without computing the voltage on the specific portion.You're done.

Daniel.

5. Mar 1, 2005

thisisfudd

Huh. Awesome. Thanks for your help!