Current, Power dissipation in a cord

  • #1
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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.
 

Answers and Replies

  • #2
lewando
Homework Helper
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The heater operates at 1800 W. You would think the cord Pd would be much less. Need to determine resistance of copper wire given length and diameter. Consider both wires.
 
  • #3
191
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I thought the equation R = ρL/A considered the resistance of copper. Was this the incorrect equation to use?
 
  • #4
20,766
4,494
Your main error is that the voltage drop across the cord is not 120 volts. That's the voltage drop across the heater. The voltage drop across the cord is much less. The current through each of the two wires of the cord is equal to the current through the heater. Use ohm's law to correctly calculate the potential drop across each of the wires of the cord. Or skip this step, and use P = I2R to get the power consumed in each wire of the cord.
 
Last edited:
  • #5
191
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Found the right answer to be ~7 W. Thank you.
 
  • #6
CWatters
Science Advisor
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As a check...

Resistance of No. 16 wire is about 4 Ohms per 1000 ft so 4ft would be 0.016 ohms.

However you have to double that because the length of wire is twice the length of the cord. So 0.032 Ohms round trip.

Power = I2 * R
= 152 * 0.032
= 7.2 W
 
  • #7
191
0
Thank you CWatters. I got my exact answer to be 6.9 W which I rounded up to 7.
 

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