A young man owns a canister vacuum cleaner marked 533 W at 120 V and aVolkswagen
ID: 1723901 • Letter: A
Question
A young man owns a canister vacuum cleaner marked 533 W at 120 V and aVolkswagen Beetle, which he wishes to clean. He parks the car inhis apartment parking lot and uses an inexpensive extension cord13.0 m long to plug in the vacuum cleaner.You may assume the cleaner has constant resistance. (a) If the resistance of each of the twoconductors in the extension cord is 0.700 what is the actual power delivered to the cleaner?(b) If instead the power is to be at least 523 W, what must be the diameter of each of twoidentical copper conductors in the cord he buys? (a) If the resistance of each of the twoconductors in the extension cord is 0.700 what is the actual power delivered to the cleaner?
(b) If instead the power is to be at least 523 W, what must be the diameter of each of twoidentical copper conductors in the cord he buys?
Explanation / Answer
a. Resistance ofcleaner RC = Voltage2/ power = 1202 /533 = 27.02 Resistance ofextensioncord RE = 2*0.700 = 1.40 totalresistance R = RC + RE = 27.02+1.40 = 28.42 CurrentI = V /R = 120 /28.42 = 4.22 A Powerdelivered tocleaner PC = I2*RC = 4.222*27.02 = 481.18 W b. If power isPC' = 523 W, PC' = V2 /R' => R' = 1202/523 = 27.53 Resistance ofnewcord RE' = R' - RC = 27.53-27.02 = 0.51 Also RE' = * L / A = * L / *(d2/4) = resistivityof copper = 1.72 *10-8 -m , L = Lengthof cord = 2 *13.0 = 26.0 d = diameterof wire 0.51 = 1.72* 10-8 * 26.0 * 4 / 3.14 * d2 diameter d = {1.789* 10-6 / 1.601) = 1.06* 10-3 m diameter d = {1.789* 10-6 / 1.601) = 1.06* 10-3 mRelated Questions
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