A hair dryer consumes P = 1600 watts of power from a 120 V wall plug. What curre

Question

A hair dryer consumes P = 1600 watts of power from a 120 V wall plug. What current docs the hairdryer draw? The current is carried from the breaker box to the power outlet by a 14-gauge copper wire, with diameter D 1 = 1.628 mm. The wire runs a distance L = 19.0 m from the box to the outlet and the same distance back. The resistivity of copper is q = 1 680 times 10^8 Ohm m. How much power is consumed due to the resistance of the wire? Suppose the above ratio of power consumed by devices to power consumed by wires is typical in the house. If the house spends $1, 300 per year in electricity bills, how much money can be saved per year by rewiring the house with 12-gauge wire, with diameter D2=2.053 mm? (Give your answer in dollars, but do not write '$')

Explanation / Answer

a)Power is given by:

P = I^2 R = V I

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

Hence, I = 13.33 A

b)R = rho L/A

R = 1.68 x 10^-8 x 19/pi (0.000814)^2 = 0.153 Ohm

I = V/R = 120/0.153 = 784.31 A

P = I^2 R = 784.31^2 x 0.153 = 9.41 x 10^4 W

Hence, P = 9.41 x 10^4 W

c)R = 1.68 x 10^-8 x 19/pi (0.0010265)^2 = 0.084 Ohm

I = 120/0.084 = 1428.6 A

P = 1428.6^2 x 0.084 = 1.71 x 10^5 W

P = 9.41 x 10^4 ----- 1300 dollars

P' = 1.71 x 10^5 W ---- 1.82 x 1300 dollars

diff = 2366 - 1300 = 1066 dollars

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