Will an AC Light Bulb Burn Out on a DC Circuit?

AI Thread Summary
Using an AC light bulb on a DC circuit can yield different results depending on the type of voltage compared. If the voltage is equal in RMS terms, the bulb is unlikely to burn out, as it behaves like a resistor. However, if the voltage is equal in peak terms, the filament's heat may affect its longevity. When using a 100-watt lightbulb rated for 120 VAC, calculations show it operates as a 144-ohm resistor, which means at 12 VDC, it won't produce enough power to glow. Ultimately, while the bulb may not burn out, it won't function effectively on DC without sufficient voltage.
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Assuming equal voltage, will it burn-out an AC light bulb if I put it on a DC circuit?
 
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pgoyer said:
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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