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Why does this work, and is it doing anything at all? Answered

Take a look at the pic, i have a 9v battery, two 50 volt 3300uF capacitors, and a 2.4 volt green 20ma led, when i hook up led normally i need a resistor and proper powr supply, but i was bored and started randomly hooking things together(mad science) any way i hooked the 2 capacitors together and wired a led inbetween them, plugged in the 9volt and the led started dim then got brighter then i have ever seen these crappy leds get, i left it on for 30 mins and it stayed on, i expected a fast burn out due to no resistor but no it was great, then i unplugged the 9v and the led slowly faded off which i kinda guessed would happen dimmed for 10 seconds then turned off.. ok my Q why did this happen, i have little expierence with capacitors, know they make good tazer but thats about it. and why did the led work like this with no resistor? is the capacitor pulling enough power that the 9v is working on lower volt? w/e answers please

4 Replies

lemonieBest Answer (author)2010-09-11

The capacitors take most of the current from the battery from uncharged, which is why the LED starts dim, when charged-up they don't affect it.
It's extra bright because you're overdriving it with 9V, but perhaps the battery is near the end of it's life and doesn't supply enough current to burn the LED in 30 min?
Disconnect the battery and the cap's discharge through the LED.


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awesume (author)2010-09-11

on the left it is because the energy gets trapped in the capacitors. You need to take out the double path way on the right capacitor because positive repels positive and negetive repels negetive thats why it can't get out.

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framistan (author)2010-09-11

The amperes delivered to the led are limited by the resistor.. but also can be limited by the ABILITY of the battery to deliver lots of amps or milliamps. Every battery has an "internal" resistance. batteries that are nearing the end of their life i believe have higher internal resistance. If your 9volt source was a power supply... the led would have burned out.

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