A man wishes to vacuum his car with a canister vacuum cleaner marked 540 W at 12

Question

A man wishes to vacuum his car with a canister vacuum cleaner marked 540 W at 120 V. The car is parked far from the building so he uses an extension cord 15.0 m long to plug the cleaner into a 120 V source. Assume that the cleaner has constant resistance.

(b) If, instead, the power is to be at least 530 W, what must be the diameter of each of the two identical copper conductors in the cord the young man buys?


(c) Repeat part (b) if the power is to be at least 537 W. [Suggestion: A symbolic solution can simplify the calculations.]

Explanation / Answer

First, consider the power equation
P=V*I

Then, using ohm's law,
V=I*R
or I=V/R
so
P=V^2/R

a) The nominal rating is:

540=120^2/R
R=120^2/540
=26.67 ohms

By introducing the long extension cord, the actual resistance is now:

26.67+1.900
=28.57

The new power is
120^2/28.57
504.025 W

b) If you want 530W
R=120^2/530
R=27.17 Ohms
Subtract 26.67 to find the resistance of 15 meters of wire = 0.5 Ohms

Resistance = Rho*l/A

=> pi*r^2 = Rho*l/R = Rho*15/0.5 = 30 * Rho => r = sqrt((30*Rho)/pi)

Rho -> Resistivity of wire,

L-> Length of wire

A -> Area of cross-section of wire

Ratio of Diameters between 540W and 530W => d1/d2 = 2r1/2r2 = r1/r2 = ((Rho*l)/R1)/((Rho*l)/R2)

=> R2/R1 = (27.17)/(26.67)

c) If you want 537W
R=120^2/537
R=26.81564 Ohms

r1/r3 = R3/R1 = (26.81564)/(26.67)

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