A hair dryer consumes P=1600 Watts of power from a 120 V wall plug. a) What curr

Question

A hair dryer consumes P=1600 Watts of power from a 120 V wall plug. a) What current does the hairdryer draw? b) The current is carried from the breaker box to the power outlet by a 14-gauge copper wire, with diameter D1=1.628 mm. The wire runs a distance L=10.0 m from the box to the outlet and the same distance back. The resistivity of copper is =1.680×10-8 m. How much power is consumed due to the resistance of the wire? c) Suppose the above ratio of power consumed by devices to power consumed by wires is typical in the house. If the house spends $1,200 per year in electricity bills, how much money can be saved per year by rewiring the house with 12-gauge wire, with diameter D2=2.053 mm? (Give your answer in dollars, but do not write '$')

Explanation / Answer

power = V*I

current I = P/V = 1600/120 = 13.33 A

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part(b)

power consumed = I^2*R1

R = Resistance = rho*L/A1 = rho*L/(pi*D1^2/4) = 1.68*10^-8*10/(pi*(1.628*10^-3)^2/4) = 0.081 ohms

power consumed P1 = 13.33^2*0.081 = 14.4 W

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for D2

P2 = I^2*R2

R2 = rho*L/A = rho*L/(pi*D2^2/4) = 1.68*10^-8*10/(pi*(2.053*10^-3)^2/4) = 0.051 ohms

P2 = I^2*R2 = 13.33^2*0.051 = 9.06 W

P2/P1 = x/1200

9.06/14.4 = x/1200

x = 755

amount saved = 1200-755 = 445 dollars

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