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sakebu
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Homework Statement
A lightbulb marked "75W at 120V" is at one end of a long extension cord in which each of the two conductors has a resistance of 0.800Ω. The other end of the extension cord is plugged into a 120V outlet. Find the actual power of the bulb.
Homework Equations
P = V^2/ R
V = IR
P = I^2R
The Attempt at a Solution
The answer according to the book is 73.8W
I found the Resistance of the bulb to be 192Ω
by doing R = V^2/P R = 120^2/75 = 192Ω
Then I added on the TWO extra resistors .8 x 2 = 1.6Ω
The total resistance therefore becomes 193.6Ω
I used that along with the 120V to find the P.
120^2/193.6 = 74.4W ..
What am I doing wrong please?