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

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SUMMARY

Using an AC light bulb on a DC circuit will not cause it to burn out if the voltage is equal in terms of RMS, as the bulb behaves like a resistor. However, if the voltage is equal in peak terms, the filament's temperature may rise significantly, potentially leading to burnout. When using a 100-watt light bulb rated for 120 VAC, calculations show that at 12 VDC, the bulb will not glow due to insufficient power dissipation. The key takeaway is that while the bulb may not burn out, it will not operate effectively on a DC circuit.

PREREQUISITES
  • Understanding of AC and DC voltage characteristics
  • Knowledge of Ohm's Law and Watts Law
  • Familiarity with RMS voltage calculations
  • Basic principles of electrical resistance in light bulbs
NEXT STEPS
  • Research the differences between AC and DC voltage effects on resistive loads
  • Learn about RMS voltage and its significance in power calculations
  • Explore the thermal characteristics of light bulbs under different voltage types
  • Investigate the implications of using AC-rated devices on DC circuits
USEFUL FOR

Electrical engineers, hobbyists working with circuits, and anyone interested in the operational characteristics of light bulbs in varying electrical environments.

pgoyer
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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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