. Consider a 60-watt lightbulb connected to a 110-volt-rms power supply. (a) Cal

Question

. Consider a 60-watt lightbulb connected to a 110-volt-rms power supply. (a) Calculate the rms current in the lightbulb. (b) How much will you have to pay for one night of negligence if you leave the light on for 8 hours? Assume that the electric company is ripping you off at 13.7 cents/kWhr. (c) The filament of an incandescent lightbulb is made of tungsten. Tungsten has a resistivity of 5.5 ·cm at room temperature and 99.4 .crn at operating temperature. What should you expect the reading to be if you use an ohmmeter to measure the resistance of a 60-watt incandescent lightbulb? How does this compare to the resistance that you would calculate from the rms voltage and current above? Explain why the ohmmeter reading and the calculation should be different.

Explanation / Answer

Given,

P = 60W ; V = 110 V

a)P = V I

I = P/Vrms = 60/110 = 0.545 A

Hence, Irms = 0.545 A

b)P = 60 x 8 = 480 W hr = 0.48 K Whr

13.7 cents /k W hr

cost = 13.7 cents/k Whr x 0.48 k W hr = 6.58 cents

cost = 6.58 centes

c)R = V/I

R = 110/0.545 = 202 Ohm

we know that,

R = rho L/a

R1 = 5.5 L/A ; R2 = 99.4 rho L/A

R1/R2 = 0.055

R2 = R1/0.055 = 202/0.055 = 3672.7 Ohm

The resistance is very high at opertaing temperature.

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