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Burnout bulb

Discussion in 'LEDs and Optoelectronics' started by Environ, Jan 5, 2016.

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  1. Environ

    Environ

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    0
    Jan 5, 2016
    Hello,
    New to electronics with a beginner question: Playing around with a electronics kit I burnt out a 2.5 v .3amp filament bulb. I did it with a 5v motor in series with it using a 9 volt battery source. Okay, what's the current and volts involved in this destructive endeavor?

    How do I figure out the current and volts across the lamp? .3 amps is 300 milliamps, right? So, did I exceed the voltage and the current or just one?

    9v - 5 volt motor = 4 volts. Did I fry it because of this voltage left for the lamp, and what was the current?

    Thanks,
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    The lamp failed because it got too hot. It got too hot because too much current was passed through the filament. Too much current could flow because the voltage across it was too high.

    You can't simply add and subtract voltages like you did because (in a very basic sense) the resistance of the motor is less than that of the lamp so a higher voltage shoreward across the lamp.

    Both the lamp and the motor exhibit effects where the initial current through them is much higher than when they are running. This makes the outcome difficult to predict without more information, but in your case you have told us the outcome.
     
  3. Environ

    Environ

    2
    0
    Jan 5, 2016
    Thanks Steve,

    Why will an LED light not burnout when connected directly to the 9volt battery? The LED is rated at 3 volts. Filament bulbs are more sensitive?
     
  4. Martaine2005

    Martaine2005

    3,309
    905
    May 12, 2015
    An LED will without a current limiting resistor.

    Martin
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,480
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    Jan 21, 2010
    A 9V battery may not be able to supply sufficient current to instantly damage a LED, but it might still damage it (a high power LED would likely not be damaged).

    If you measure the voltage across the LED you'll find it is way less than 9V in this case. Because the battery can't supply sufficient current its voltage drops to the point where it can (lower voltage means lower current). The extra voltage is "lost" in the battery due to the internal resistance of the battery. In extreme cases the battery can get quite hot, and with some battery types you could damage the battery or even cause it to explode.
     
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