ubie TO haige ioW.) Consider a 60-watt lightbulb connected to a 110-volt-rms pow

Question

ubie TO haige ioW.) 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 ·cm 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

a)

We know that

P=Vrms*Irms

So

Irms=P/Vrms

=60/110

Irms=0.5454A or 545.4mA

b)

for 1kWh they are charging 13.7,

so if you leave the bubl open for 8 hours,

Power dissipated will be

60*8

480Wh

or 0.48kWh

So for 0.48kWh they will charge

0.48*13.7

=6.576 cents

c)

It will be not be same as the applied voltage will interfere with the ohmeter's measurment, and it will give us a wrong value.

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