Question

An air conditioner draws 17 A at 220-V ac. The connecting cord is copper wire with a diameter of 1.291 mm .

a.) How much power does the air conditioner draw?

b.) If the length of the cord (containing two wires) is 6.5 m , how much power is dissipated in the wiring?

c.) If no. 12 wire, with a diameter of 2.053 mm, was used instead, how much power would be dissipated?

d.) Assuming that the air conditioner is run 13 h per day, how much money per month (30 days) would be saved by using no. 12 wire? Assume that the cost of electricity is 12 cents per kWh.

Answer 1

Part A
How much power does the air conditioner draw?

a) 17 x 220 = 3740 watts rounded to 3700 W

Part B

If the length of the cord (containing two wires) is 3.5 m , how much power
is dissipated in the wiring?

Resistance of a wire in Ω
R = ρL/A
ρ is resistivity of the material in Ω-m
L is length in meters
A is cross-sectional area in m²
A = πr², r is radius of wire in m
resistivity Cu 17.2e-9 Ω-m

r = 1.291/2 = 0.6455 mm = 0.0006455 m
L = 2 x 3.5 = 7 m(there are 2 wire in the cable)
A = π(0.0006455)² = 1.31e-6 m²
R = (17.2e-9)(7) / (1.31e-6)
R = 0.092 ohms
Voltage drop is 17 x 0.092 = 1.56 volts
power loss = 17 x 1.56 = 26.52 watts

Part C

If no. 12 wire, with a diameter of 2.053 mm, was used instead, how much power
would be dissipated?

Express your answer to two significant figures and include the appropriate units.

Resistance of a wire in Ω
R = ρL/A
ρ is resistivity of the material in Ω-m
L is length in meters
A is cross-sectional area in m²
A = πr², r is radius of wire in m
resistivity Cu 17.2e-9 Ω-m

r = 2.053/2 = 1.0265 mm = 0.0010265 m
L = 2 x 3.5 = 7 m(there are 2 wire in the cable)
A = π(0.0010265)² = 3.31e-6 m²
R = (17.2e-9)(7) / (3.31e-6)
R = 0.0364 ohms
Voltage drop is 17 x 0.0364 = .6188 volts
power loss = 17 x .6188 = 10.52 watts rounded to 10 W

Part D

Assuming that the air conditioner is run 13 h per day, how much money
per month (30 days) would be saved by using no. 12 wire? Assume that the
cost of electricity is 12 cents per kWh. Express your answer using two
significant figures.

3726 W = 3.726 kW
3710 W = 3.71 kW

3.726 kW * 13 h/day * 1 month * 30 days/month * $0.12/kWh = 174.38
3.71 kW * 13 h/day * 1 month * 30 days/month * $0.12/kWh = 173.63

savings 80 cents per month