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Suppose you have 2 filament lamps which each have a resistance of 1 ohm and each need 3V.. If you connect them in parallel with each other and a 6V supply, you will get a current of 12A in the whole circuit and 6A to each lamp, because combined resistance would be only 0.5 Ohms. But, because the current to each lamp is high, wouldn't the resistance of the filament lamp increase(because of temperature), reducing current again? So will the resistance of the lamp keep increasing till current is 0?

When I say current is high, I mean higher than if we connected everything in series.

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    $\begingroup$ "But, because the current to each lamp is high, wouldn't the resistance of the filament lamp increase(because of temperature..." Perhaps yes. Have you worked out what would happen for a given lamp, whose resistance versus temperature is some known function? $\endgroup$
    – DanielSank
    Oct 29, 2023 at 5:06
  • $\begingroup$ No, I don't have a function for that. This is based on a homework question about why the resistance of a filament lamp in parallel is greater than its resistance in series. But, if this is possible, how do the lights in our home stay on for over 24 hours? $\endgroup$
    – KitKat
    Oct 29, 2023 at 8:03
  • $\begingroup$ Such an easy experiment. Why not just do it and see what happens? $\endgroup$
    – John Doty
    Oct 29, 2023 at 11:21

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wouldn't the resistance of the filament lamp increase(because of temperature), reducing current again? So will the resistance of the lamp keep increasing till current is 0?

If the current was zero, then the power dissipated must also be zero. (Power equals current times voltage, right?) If no power is dissipated, then what will happen to the temperature?

I don't know the math (I'd guess it's a second order ODE) but what happens when you switch on a real light bulb is, there's initially a very high current, but as the filament heats up, the resistance increases, and the current and power decrease until an _equillibrium_temperature is reached. The heat carried away from the filament by radiation and convection becomes equal to the power dissipated, and the temperature and current and power all settle to constant values.

For most light bulbs, when operated under normal conditions, the whole process happens within a fraction of a second.

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  • $\begingroup$ Ohh ok makes sense. Thank you! $\endgroup$
    – KitKat
    Oct 29, 2023 at 18:34
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Just for the sake of argument the resistance of a bulb is $1\Omega$ when there is a potential difference of $3\,\rm V$ across it and $3\Omega$ when there is a potential difference of $6\,\rm V$ across it.

With $3\rm V$ across two bulbs in parallel the power dissipated is $2\times \frac{3^2}{1} = 9\, \rm W$ and with $6\rm V$ across two bulbs in parallel the power dissipated is $2\times \frac{6^2}{3} = 24\, \rm W$.
Thus, although the resistance of a bulb increases with increased voltage across it, the power dissipated in a bulb still increases as the voltage is squared.

You can do a similar analysis for the series case perhaps with the resistance of a bulb being $0.5 \,\Omega$ if the potential difference across it is $1.5\,\rm V$.

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lol no, the resistance of the lamp won't keep increasing till current is 0. that would mean the lamp would stop glowing, which is not what happens. the resistance of the filament lamp increases with voltage because the filament gets hotter and the atoms vibrate more, making it harder for electrons to flow. but there is a limit to how much the resistance can increase, because the filament can only get so hot before it melts or burns out. so the current will never be 0, but it will be less than what you would expect from ohm's law. this is why the graph of current vs voltage for a filament lamp is curved, not straight. if you connect two filament lamps in parallel with a 6V supply, each lamp will get 3V, but not 3A, because their resistance will be higher than 1 ohm. the exact current will depend on the temperature and material of the filament, but it will be less than 3A. hope this helps :)

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