A young man owns a canister vacuum cleaner marked 535W at 120V and a Volkswagen

Question

A young man owns a canister vacuum cleaner marked 535W at 120V and a Volkswagen Beetle, which he wishes to clean. he parks the car in his apartment parking lot and uses an inexpensive extension cord 15 m long to plug in the vacuum cleaner. you may assume the cleaner has constant resistance. (a) if the resistance of each of the two conductors in the extension cord is 0.900, what is the actual power delivered to the cleaner? (b) if instead the power is to be at least 525W, what must be the diameter of each of the two identical copper conductors in the cord he buys (c) repeat part (b) asuming the power is to be at least 532W.

Explanation / Answer

Hi, Given Wattage of the cleaner = 535 W                                    voltage= 120 V.    The relation between the power and voltage isgiven by P = I * V = V2 / R       Hence the resistance ofthe cleaner = R = V2 / P = 120 * 120 / 535 = 26.92.       Given the resistance ofeach wire = 0.9 . Two such wires are now included inthe circuit and hence the total resistance          of thecircuit will be Rt = R + 0.9 + 0.9 = 28.72 .       a)  Hence the actual power delivered will bePa = V2 / Rt = 120 * 120 / 28.72= 501.39 W. (this loss of power isdissipated as heat in the wires) b)  If the actual power should be 525W, then the total power dissipation through thewires should be only 10 W.          then thetotal resistance of the system should   Rb=  V2 / Pb = 120 * 120 / 525 =27.43 .          Hencethe total resistance of two wires = 27.43 - 26.92 = 0.51 .and the resistance of each wire = 0.255 .          Giventhe length of each wire = L = 15m.          we knowthat resistance of a wire can be found by eq.   R = L / A    ( where isresistivity,                                                                                                                           and Ais its area of cross-section).             forcopper =  1.72 × 10-8 m.          Hence  A=  r2 = L/ Rb=  1.72 × 10-8 x 15/  0.255 = 101.177× 10-8 m2                   =>radius of wire r = 5.68 x 10-4 m.                   =>diameter should be d = 2 * 5.68 x 10-4= 11.36 x10-4 m                   =>radius of wire r = 5.68 x 10-4 m.                   =>diameter should be d = 2 * 5.68 x 10-4= 11.36 x10-4 m                                                                             c) Arguing as above, for the delivered power to be 532, thediameter of each wire can be calculated as 20.94 x10-4m                   Hope this helps you.

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