Power

[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?

[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...:wink:

[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

OK, I'm stuck. Please help!

[dextercioby]

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


Daniel.

[thisisfudd]

Huh. Awesome. Thanks for your help!