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Mashy

  • 2 years ago

i have an odd question.. in some videos regarding making your own batteries.. i saw some homemade batteries produce high voltage but do not light up LEDs.. now.. why is that.. i mean ohm's law says v= IR.. so if any other battery of same voltage works then this should give out same current for the same led.. right? :-/.. or is it the internal resistance that comes into play?

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  1. mathmate
    • 2 years ago
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    Certain batteries have high voltage, but cannot sustain a current. They are great for measuring the voltage, but lights no bulb!

  2. Mashy
    • 2 years ago
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    i dunno why!?? i mean.. how?? what happens to ohms law??

  3. Mashy
    • 2 years ago
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    the terminal pd very less?? compared to emf?? so is it the internal resistance?

  4. mathmate
    • 2 years ago
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    Depending on the nature of the battery. With the high voltage, once there is a current, the energy of the battery could be depleted before the bulb is lit. Ohm's law applies, but there is probably not enough energy to light a bulb long enough for you to see. Ever tried to use a 9 volt battery and two AA batteries to start a car?

  5. Mashy
    • 2 years ago
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    well even then .. there should be a FLASH atleast.. like how a capacitor works.. it gets depleted instantly creating a flash so definitely there cannot be any depletion cause there is not current flowing and as for car.. they need higher voltage.. 200 to 250 volts.. !!

  6. mathmate
    • 2 years ago
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    A car battery is 12 volts, = 9+2*1.5. Would you expect the engine turn before the batteries are exhausted?

  7. Mashy
    • 2 years ago
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    can you do some math to explain it? :P..

  8. Mashy
    • 2 years ago
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    so you mean.. the moment you close the circuit.. the potential difference instantly decreases?? and becomes almost zero?

  9. mathmate
    • 2 years ago
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    The math part depends a lot on the nature of the battery, and the characteristics associated with it. @mashy yes, that's what happens when you draw a large current from a small energy source. I would not say it becomes zero, but decreases exponentially to zero, like: |dw:1358008592637:dw|

  10. Mashy
    • 2 years ago
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    oh wait is it like.. the battery doesn't get enough time to make its red ox reactions.. and maintain the required EMF??.. so that would mean.. the moment i open circuit.. the battery comes back to normal??

  11. mathmate
    • 2 years ago
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    The redox reaction is choked when you close the circuit (Energy=Vit). It doesn't supply fast enough. Yes, most batteries recuperate when you give it time, but not a 100%.

  12. Mashy
    • 2 years ago
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    aha.. that makes sense now!!>. thanks you should name yourself.. math and physics mate.. cheers :)

  13. mathmate
    • 2 years ago
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    Thank you for the honour! :)

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