Compact fluorescent bulbs are much more efficient at producinglight than are ord

Question

Compact fluorescent bulbs are much more efficient at producinglight than are ordinary incandescent bulbs. They initially costmore, but last far longer and use much less electricity. Accordingto one study of these bulbs, a compact bulb that produces as muchlight as a 100W incandescent bulb uses only 23W of power. Thecompact bulb lasts 1x10^4 hours, on average, and costs $12.00,whereas the incandescent bulb costs only $0.76, but lasts just 750hours. The study assumed that electricity cost $0.08/(kW*hr) andthat the bulbs were on for 4 hours per day. (Careful with kW*hrcalculations it is NOT kW/hr. Electric companies charge for energywhich is power*time) (a) What is the total cost (including the price ofthe bulbs) to run each type of bulb for 3 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

Explanation / Answer

(a) What is the total cost (including the price ofthe bulbs) to run each type of bulb for 3 years?

incandescent bulb:

t = 3 * 365.25 * 4 = 4383 hours

number of incandescent bulbs is neede = ni = 4383/750 = 5.844

energy = Ei = P t = 100 * 4383 = 438300 W.hr = 438.3 kW.hr

cost = ci = Ei*$0.08 + ni*$0.76 = 438.3*0.08 + 5.844*0.76 = $ 39.50544

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

compact bulb:

t = 3 * 365.25 * 4 = 4383 hours

number of incandescent bulbs is neede = nc = 4383/1e4 = 0.4383

energy = Ec = P t = 23 * 4383 = 100809 W.hr = 100.809 kW.hr

cost = cc = Ec*$0.08 + nc*$12 = 100.809*0.08 + 0.4383*12 = $13.32432

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

39.50544 - 13.32432 = $ 26.18112

(d)What is the resistance of a “100W”fluorescent bulb? (Remember the actual rating is only 23W of powerfor a 120V circuit)

P = V2/R >>>> R = V^2/P = 120*120/100 = 144 ohms

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