Question

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 75.0 ¢, 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.

Answer 1

A)

for compact bulb the power user for 3 years \(P_{c m p}=23 W * 4 * 365 * 3\) \(\Rightarrow P_{c m p}=100740 k W h\)

Total cost of electricity \(=1.007 * 10^{5} \mathrm{kWh} * 9 c=5439960 \mathrm{cents}=9066.6\) dollers Number of bulbs \(=\) total hours \(/\) bulb life time \(=4 * 365 * 3 / 10000=0.438\) hence 1 bulbs needed and cost for 1 bulb \(=12\) dollers Total cost \(=9066.6+12=9078.6\) dollers

B)

for Incandescent bulb the power user for 3 years \(P_{c m p}=100 * 4 * 365 * 3\) \(\Rightarrow P_{c m p}=438000 k W h\)

Total cost of electricity \(=1.007 * 10^{5} \mathrm{kWh} * 9=3942000\) cents \(=39420\) dollers Number of bulbs \(=\) total hours \(/\) bulb life time \(=4 * 365 * 3 / 750=5.84\) hence 6 bulbs needed and cost for 6 bulb \(=75 * 6\) cents \(=4.5\) dollers Total cost \(=39420+4.5=39424.5\) dollers

C)

savings \(=\) -cost of incandescent-cost of campact fluorescent \(=39424.5-9078.6=\) 30345dollers

D)

for power of \(100 \mathrm{~W}\) and Voltage of \(230 \mathrm{~V}\) the resistance \(\mathrm{R}\) can be found \(\mathrm{by} \mathrm{P}=\mathrm{V}^{2} / \mathrm{R}\)

\(\mathrm{R}=230^{2} / 100=529 \mathrm{ohms}\)