Assuming equal voltage, will it burn-out an AC light bulb if I put it on a DC circuit?
What do you mean by equal voltage?
Equal RMS: probably not.
Equal peak: depends on how hot the filament gets
Equal average: won't turn on.
HINT: how is power calculated in an AC circuit?
RMS volts, yes. Since a lightbulb is basically a resistor it should be fine. RMS AC is the same voltage that it takes to make the same amount of heat in a resistor driven with the same DC voltage.
Ohm's Law: volts (E) = amps (I) * ohms (R)
Watts Law: watts (P) = amps (I) * volts (E).
A useful derivation is: P = (I^2) * R = current squared * resistance
Let's try a 100-watt lightbulb. And 120 VAC.
AC: 100 watts = .833^2 amps * 144 ohms
120 volts = .833 amps * 144 ohms
I believe your previous correspondent, that a lightbulb is a resistor. So, maybe we know from above that a 100-watt lightbulb is a 144-ohm resistor. Let's try with a 12VDC car battery.
DC: 12 volts = .0833 amps * 144 ohms
1 watts = .0833^2 amps * 144 ohms
Does this show that the lightbulb will not burn out? Does it also show that the bulb won't dissipate enough power to glow?
Who said anything about using the light bulb on 12 volts? The question was whether it was AC or DC.
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