Question

An air conditioner draws 18A at 220-V ac. The connecting cord is copper wire with a diameter of1.291mm .

A. How much power does the air conditioner draw?

B. If the length of the cord (containing two wires) is 5.5

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 12

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

a)

P=V*I=220*18

P=3960 W or 3.96 KW

b)

since cord containing 2 wires

L=2*5.5 =11 m

Area A=pi*d^2/4

A=pi*(1.291*10^-3)^2/4

A=1.31*10^-6 m^2

resistivity of copper is

p=1.68*10^-8 ohm-m

Resistance

R=pL/A =(1.68*10^-8)*11/(1.31*10^-6)

R=0.1412 ohms

so

P=I^2*R=18^2*0.1412

P=45.74 W

c)

A=pi*(2.053*10^-3)^2/4 =3.31*10^-6 m^2

R=(1.68*10^-8)*11/(3.31*10^-6)

R=0.05583 ohms

P=18^2*0.05583

P=18.09 W =18.1 W (approx)

d)

Power saved

P=45.74-18.09

P=27.65 W=(27.65

running AC for 24 hours and 30 days ,energy saved

E=(27.65/1000)*24*30

E=19.91 KWH

Cost save is

C =12*19.91 =238.9 cents or $2.39