- #1

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.