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 W incandescent bulb uses only 23.0 W W of power. The compact bulb lasts 10000 hours, on the average, and costs $ $ 11.00, whereas the incandescent bulb costs only $ $ 0.75, but lasts just 750 hours. The study assumed that electricity cost $ $ 0.070 per kilowatt-hour 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.What is the resistance of a "100 WW" fluorescent bulb? (Remember, it actually uses only 23 WW of power and operates across 120 VV.)

Answer 1

a.

p_i = price of incandescent bulb = $0.75

C_i = Cost of using incandescent bulb

t = total time of operation in 3 years = 4 x 3 x 365 = 4380 h

n = number of bulbs required = t/750 = 4380/750 = 6

P_i = Power of incandescent bulb = 23 W = 0.023 kW

c = cost of electricity = $0.070 (kwh)^-1

Cost of using incandescent bulb is given as

C_i = n p_i + P_i t c

C_i = 6 (0.75) + (0.023) (4380) (0.070)

C_i = $11.5518

b)

p_f = price of fluorescent bulb = $11

C_f = Cost of using fluorescent bulb

t = total time of operation = 4 x 3 x 365 = 4380 h

n = number of bulbs required = t/750 = 4380/10000 = 1

P_f = Power of fluorescent bulb = 100 W = 0.100 kW

c = cost of electricity = $0.070 (kwh)^-1

Cost of using fluorescent bulb is given as

C_f = p_f + P_f t c

C_f = 11 + (0.100) (4380) (0.070)

C_f = $41.66

c)

P = Power used = 23 W

V = potential difference = 120 volts

R = Resistance

Using the equation

P = V²/R

23 = 120²/R

R = 626.09 ohm