A young man owns a canister vacuum cleaner marked 531 W at 150 V and aVolkswagen

Question

A young man owns a canister vacuum cleaner marked 531 W at 150 V and aVolkswagen Beetle, which he wishes to clean. He parks the car inhis apartment parking lot and uses an inexpensive extension cord12.0 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 0.900 what is the actual power delivered to the cleaner?
1 W

(b) If instead the power is to be at least 521 W, what must be the diameter of each of twoidentical copper conductors in the cord he buys?
2 mm

(c) Repeat part (b) assuming the power is to be at least528 W. Suggestion: A symbolic solution cansimplify the calculations.
3 mm (a) If the resistance of each of the twoconductors in the extension cord is 0.900 what is the actual power delivered to the cleaner?
1 W

(b) If instead the power is to be at least 521 W, what must be the diameter of each of twoidentical copper conductors in the cord he buys?
2 mm

(c) Repeat part (b) assuming the power is to be at least528 W. Suggestion: A symbolic solution cansimplify the calculations.
3 mm

Explanation / Answer

   In the problem we can observe that there issame potential difference    V = 150 V    the internal resistance will be    Ri = (V)2 /P1 .........(1)    P1 = 531 W    if we take that if each of the twoconductors in the extension cord has resistance Rc    the totral resistance in the path of thecurrent will be    Rt = Ri + 2Rc ...........(2)    the current existing is    I = V / Rt    the power delivered to the cleaner is    Pd = I2Ri         = (V /Rt)2 Ri ............(3)    the resistance of copper conductor of lengthl and diameter d is         Rc = Cu(l / A)         =Cu [l / ( d2 /4)]         = 4Cul / d2    now if Rc,max is the maximumallowed value of Rc the minimum acceptable diameterof the    conductor will be    dmin = [4Cul / Rc,max].......... (3) (a)   as from the given Rc = 0.900    from the equations (1) and (2)    Rt = Ri + 2 (0.900)          =[(V)2 / P1] + 1.80    from the equation (3) the power delivered to thecleaner is    Pd = ......... W (b)    if the minimum acceptable powerdelivered to the cleaner is Pmin then the equations (2)and (3)    give the maximum allowable total resistanceas    Rt,max = Ri + 2Rc,max              =[(V)4 / PminP1]              =(V)2 / (PminP1)    solve for Rc,max andsubstitute in eq 3               dmin = [4 Cul / Rc,max]

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