R = Power/I^2 = 40/(0.33)^2 = ___ ohms
That is the resistance at the operating temperature, with 120 V applied. The resistance will be less as the bulb filament heats up. You could verify that with an ohmeter
What is the resistance of a 40 watt light bulb if it is it to full brilliance by a current of .33amps?
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