A young man owns a canister vacuum cleaner marked 533 W at 120 V and aVolkswagen

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   m   

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