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

Power, P = 540 W

Voltage, V = 120 V

length of the cord, l = 15 m

So, length of the conductor = 2*l = 30 m

Resistance , R = 2*0.95 ohm

a)

We know, P = V^2/R

So, R = V^2/P

So, resistance of cleaner, R' = 120^2/540 = 26.7 ohm

Net resistance of the circuit, Rnet = R' +R = 26.7 + 0.95*2 = 28.6 ohm

So, net current in the circuit , I = V/Rnet = 120/(28.6) = 4.196 A

So, potential drop across the conductors = I*R = 4.196*2*0.95 = 7.97 V

So, Voltage across the cleaner = V- 33.45 = 120-7.97 = 112.03 V

So, power delivered to the cleaner = 112.03*4.196 = 470.1 W

b)

Power delivered = 530 W

So,current, I' = sqrt(P/R') = sqrt(530/26.7)

=4.46 A

So, voltage across the conductors = I'*R = 4.46*2*0.95 = 8.47 V

We know, R = r*l/A

where r = resistivity of copper = 1.68*10^-8

l = 30 m

A = area of cross section of wire = pi*d^2/4 <----- area of circle

where d = diameter

So, I'*(rl/A) = 8.47

So, 4.46*(1.68*10^-8*30/(3.14*d^2/4)) = 8.47

So, d = 5.81*10^-4 m <---------------answer

c)

here I' = sqrt(537/R') = sqrt(537/26.7) = 4.48 A

So, potential difference across the conductors = 4.48*2*0.95 = 8.51 V

So,

4.48*(1.68*10^-8*30/(3.14*d^2/4)) = 8.51

So, d = 5.81*10^-4 m <---------------answer

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