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PeachBanana
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Homework Statement
An extension cord made of two wires of diameter 0.129 cm (no. 16 copper wire) and of length 1.2m (4ft ) is connected to an electric heater which draws 15.0A on a 120 V line. How much power is dissipated in the cord?
Homework Equations
P = IV
R = ρL / A
P = V^2/R
The Attempt at a Solution
First I found the power required by the heater.
P = (15.0 A)(120 V)
P = 1800 W
Next I wanted to find the power within the cords.
R = ρL / A
A = 6.45*10^-4 (∏)
R = 1.68*10^-8(1.2 m) / (6.45 * 10^-4)^2(∏)
R = 1.542*10^-2 Ω
P = V^2 / R
P = (120 V)^2 / (1.542*10^-2 Ω)
P = 9.338*10^5 W
I thought that was the power within the cords so next I:
9.388*10^5 W - 1800 W = 9.32 * 10^5 W.