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Old 6th November 2007, 03:09 AM   (permalink)
Default Capacitor Question

I'm trying to become more familiar with capacitors and whether its feasible for for a project. I found that Farad = Current * time / Voltage. In this project, we're trying to have some LED bulbs (12V, 20W) lit up for some period of time (at least 1 hour). I figured F=1.6A * 3600s / 12V = 480 Farads, which I realize is ridiculously huge for a small light system. Is my concept math right?
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Old 6th November 2007, 03:16 AM   (permalink)
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Just use a normal lead-acid battery.
A 480 farad capacitor (if it exist) will be really, but really huge.
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Old 6th November 2007, 03:43 AM   (permalink)
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A capacitor is not an ideal energy storage medium, and it's VERY non linear over time voltage and current drop at an exponential rate. A 480 farad capacitor at 12 volts is not practical, aside from the non linear nature, high farad capacitors are usually only in the 2.5 volt range and even then 480 farads would require one hell of an array. The price would probably make you crap your pants as I have 1 farad 5 volt cap's that cost 2.50 each, your equations are direct energy relation equations and don't take into account the real world nature of energy and it's conversion efficiency.
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Last edited by Sceadwian; 6th November 2007 at 03:47 AM.
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