Lightbulb A is marked \"40.0 W 120 V,\" and lightbulb B is marked \"60.0 W 120 V

Question

Lightbulb A is marked "40.0 W 120 V," and lightbulb B is marked "60.0 W 120 V." These labels mean that each lightbulb has its respective power delivered to it when it is connected to a constant 120-V source. (a) Find the resistance of each lightbulb. R(40.0 W 120 V) = R(60.0 W 120 V) = (b) During what time interval do 2.00 C pass into lightbulb A? (c) Is the charge different upon its exit versus its entry into the lightbulb? Explain. (d) In what time interval do 1.70 J pass into lightbulb A? (e) By what mechanisms does this energy enter and exit the lightbulb? Explain. (f) Find the cost of running lightbulb A continuously for 30.0 days, assuming the electric company sells its product at $0.110 per kWh.

Explanation / Answer

Resistance R = voltage divided by current
in amperes, so R = E/I and P = E times I so P = E²/R.
(a) If we know the power (P), we can find R by: R = E² / P. If E = 120, E² = 14400,
so R = 14400 / P
For the 40 watt bulb, R = 360 ohms and for the 60 watt bulb, R = 240 ohms.
(b) The current flowing in bulb A the 40-watt bulb, is 1/3 ampere or, 1/3 coulomb
per second. So the time t = charge Q divided by I (current in amperes) = 2.0 / (1/3)
The time t = 6.0 seconds for 2.0 coulombs of electrons to pass through bulb A
(c) No; the exiting charge is the same amount as the entering charge.
(d) Bulb A uses 40 watts or 40 joules per second. So it takes 1.70/40 or 0.0425
second for 1.7 joule of energy to pass through bulb A
(e) The bulb is connected to the electrical energy source by wires, usually a metal
like copper, which carry the current of electrons into and out of the bulb.
(f) If the bulb runs continuously for 30 days at 86,400 seconds per day, that is
2,592,000 seconds. At 40 J/sec, that is 103,680,000 joules over the 30 days.
One kilowatt hour is 3,600,000 joules, so we use 103,680,000 divided by
3,600,000, or 28.8 kilowatt-hours. At 11¢ per kwh, that is $3.168 total cost.

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