An extension cord made of two wires of diameter 0.129 cm (no. 16 copper wire) and of length 2.6 m is connected to an electric heater which draws 17.0 A on a 120 V line. How much power is dissipated in the cord?
2007-02-20 08:46:48 · 4 answers · asked by Minamoto 1 in Science & Mathematics ➔ Physics
You need to know the resistivity of the copper wire. My reference book shows 4.0 Ohms per 1000 ft, which is 13.1 milliOhms per meter.
Your teacher might also want you to look up the resistivity of copper in your book, and multiply it by cross-sectional area of the wire, but it will be close to 4 Ohms per 1000 ft.
13.1 milliOhms per meter * 2.6 meter = 34 milliOhms
the Power law: P = I^2 * R
P = 17^2 * 0.034 = 9.8 Watts or about 10 Watts.
FINALLY: There are TWO wires in the cord, so double that number to 19.7 Watts (about 20 Watts).
That's enough to get the wire a little 'toasty'.
2007-02-20 08:48:57 · answer #1 · answered by tlbs101 7 · 4⤊ 0⤋
2 w
2007-02-20 08:52:39 · answer #2 · answered by Anonymous · 0⤊ 0⤋
P=current squared time resistance
P = Square the current running through the cord then multiply by the resistance of the cord. This assumes you don't care about any other insignificant losses such as eddy currents, etc.
2007-02-20 08:53:23 · answer #3 · answered by Anonymous · 0⤊ 0⤋
a few milliamps
the cord is like 1/2 Ohm or so
2007-02-20 08:48:48 · answer #4 · answered by agropelter 3 · 0⤊ 1⤋