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

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Discussion Overview

The discussion centers around whether an AC light bulb will burn out when used in a DC circuit, assuming equal voltage conditions. Participants explore the implications of using different voltage definitions, such as RMS, peak, and average voltage, in relation to the operation of the light bulb.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest that if equal RMS voltage is applied, the light bulb may not burn out.
  • Others argue that using equal peak voltage could lead to different outcomes depending on how hot the filament gets.
  • One participant notes that equal average voltage would not turn the bulb on at all.
  • A participant provides calculations using Ohm's Law and Watt's Law to analyze a 100-watt light bulb's behavior under AC and DC conditions, questioning whether it would glow or burn out.
  • Another participant challenges the focus on a specific voltage (12V) rather than the AC versus DC distinction.

Areas of Agreement / Disagreement

Participants express differing views on the implications of using AC versus DC and the definitions of voltage. There is no consensus on whether the bulb will burn out or not, as multiple competing views remain.

Contextual Notes

Participants reference various voltage definitions (RMS, peak, average) and their effects on the light bulb's operation, but the discussion does not resolve the implications of these definitions on the bulb's performance.

Who May Find This Useful

Individuals interested in electrical engineering, circuit design, or the behavior of electrical components in different circuit conditions may find this discussion relevant.

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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