Question

An air conditioner draws 19A at 220-V ac. The connecting cord is copper wire with a diameter of1.450mm .

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

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

B)    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.

C)    Assuming that the air conditioner is run 15

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.

I got 14 W for A and 6.5 W for B but It is showing as not correct. I don't know what I'm doing wrong.

Answer 1

Resistivity of copper = 1.68*10^-8

Resistance per meter of wire = resistivity/area = (1.68*10^-8) / (pi * (7.25*10^-4)^2)

= 0.01017 ohms/m

Total length of wire = 3.5*2 = 7m

Total resistance of wire = 7*0.01017 = 0.0712 ohms

Power dissipated = I^2 * R = 19^2 * 0.0712 = 25.7 Watts

If wire is changed,

Resistance per meter of wire = resistivity/area = (1.68*10^-8) / (pi * (1.0265*10^-3)^2)

= 0.00507 ohms/m

Total resistance of wire = 7*0.00507 = 0.0355 ohms

Power dissipated = I^2 * R = 19^2 * 0.0355 = 12.82 Watts

Svaed watts = 25.7 - 12.82 = 12.88 W

Energy saved per day = 15 * 12.88 /1000 = 0.193 kWh

Energy saved per month = 5.796kWh

Money saved per month = 12*5.796 = 69.55 cents