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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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mathmateBest ResponseYou've already chosen the best response.1
Certain batteries have high voltage, but cannot sustain a current. They are great for measuring the voltage, but lights no bulb!
 one year ago

MashyBest ResponseYou've already chosen the best response.0
i dunno why!?? i mean.. how?? what happens to ohms law??
 one year ago

MashyBest ResponseYou've already chosen the best response.0
the terminal pd very less?? compared to emf?? so is it the internal resistance?
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
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?
 one year ago

MashyBest ResponseYou've already chosen the best response.0
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.. !!
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
A car battery is 12 volts, = 9+2*1.5. Would you expect the engine turn before the batteries are exhausted?
 one year ago

MashyBest ResponseYou've already chosen the best response.0
can you do some math to explain it? :P..
 one year ago

MashyBest ResponseYou've already chosen the best response.0
so you mean.. the moment you close the circuit.. the potential difference instantly decreases?? and becomes almost zero?
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
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
 one year ago

MashyBest ResponseYou've already chosen the best response.0
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??
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
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%.
 one year ago

MashyBest ResponseYou've already chosen the best response.0
aha.. that makes sense now!!>. thanks you should name yourself.. math and physics mate.. cheers :)
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
Thank you for the honour! :)
 one year ago
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