Question

# Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They...

Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost much more, but last far longer and use much less electricity. According to one study of these bulbs, a compact bulb that produces as much light as a 100 W incandescent bulb uses only 23.0 W of power. The compact bulb lasts 1.00×104 hours, on the average, and costs \$ 12.0 , whereas the incandescent bulb costs only 76.0 ¢, but lasts just 750 hours. The study assumed that electricity cost 9.00 ¢ per kWh and that the bulbs were on for 4.0 h per day.

A.)What is the total cost (including the price of the bulbs) to run incandescent bulbs for 3.0 years?

B.What is the total cost (including the price of the bulbs) to run compact fluorescent bulbs for 3.0 years?

C.)How much do you save over 3.0 years if you use a compact fluorescent bulb instead of an incandescent bulb?

D.What is the resistance of a "100 W" fluorescent bulb? (Remember, it actually uses only 23 W of power and operates across 120 V.)

PLEASE solve whole problem out. Many just explain how to do it and I still end up confused. I would greatly appreciated it as this problem has been giving me troubles.

A.

An incandenscent bulb runs 750 hours, or 750/4 = 187.5 days.

Thus, you need 3*365/187.5 = 5.84 light bulbs in 3.0 years.

Now,

Cost = cost of bulb + cost of consumption

=5.84*(\$0.76) + (4 hrs/day * 3*365 days)(0.100 kW)(\$0.0900/kWh)

= \$43.86

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B.

A compact fluorescent runs 1.00E4 hours, or 1.00E4/4 = 2500 days.

Thus, you need 3*365/2500 = 0.438 bulbs. That means you need 1 bulb. [You cannot buy 0.438 bulbs.]

Now,

Cost = cost of bulb + cost of consumption

=1*(\$12.0) + (4 hrs/day * 3*365 days)(0.023 kW)(\$0.0900/kWh)

= \$ 21.07

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C.

Savings = \$43.86 - 21.07

= \$22.79

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D.

Note that the power lost is given by

P = V^2/R

--> R = V^2/P

The power lost here is Plost = 100 W - 23 W = 77 W. Thus,