Man wishes to vacuum his car with a canister vacuum cleaner marked 535 W at 120

Question

Man wishes to vacuum his car with a canister vacuum cleaner marked 535 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.  
(a) If the resistance of each of the two conductors of the extension cord is 0.900 , what is the actual power delivered to the cleaner?
W
(b) If, instead, the power is to be at least 525 W, what must be the diameter of each of the two identical copper conductors in the cord the young man buys?
mm
** Suggestion: a symbolic solution can simplify the calculation **

Explanation / Answer

giveen that

P =535 W

V = 120 V

L = 15 m

(a) Nominal resistance of Vacuum cleaner:

Rvc = V^2 / P

Rvc = 120 * 120 / 535

Rvc = 26.91 ohm

Resistance including cable : R = 26.91 + 0.90 + 0.90 = 28.71 ohm

from ohm's law

Current flow = V/R = 120 / 28.71 = 4.17 A

Power delivered = I^2 Rvc = 4.17^2 * 26.91 = 467.93W

(b)   At 525 W,

I^2 = P / Rvc = 525 / 26.91

I^2 =19.51 A

I = 4.41 A

R(total) = V / I = 120 / 4.41 = 27.21 ohm

Therefore R of cables = 27.21 -26.91 = 0.30 ohm

So    0.15 ohm per conductor

Resistivity of copper= 1.7 * 10^(-8) ohm* m

Using R =rho* L/ A

0.15 = 1.7*10^(-8) * 15 / A

Area, A = 170*10^(-8) m^2 = 1.70 mm^2

Area = pi*r^2
r^2 = 1.70 / 3.14

r^2 = 0.54 mm^2

r = 0.73 mm

diameter = 2* 0.73

diameter = 1.46 mm

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