1. The problem statement, all variables and given/known data 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. 2. Relevant equations P = V^2/ R V = IR P = I^2R 3. 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?