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Thread: Test Capacitors?

  1. #1
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    Test Capacitors?

    I have a 1farad cap for my car, and I keep get dimming lights and dead batteries under heavy use. I have a feeling my cap has gone bad. I've had it for 5 years under heavy daily use. Is there a safe way to test my cap?<br /><br />

  2. #2
    Maximum Bitrate Megalomaniac's Avatar
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    i hate to burst your bubble. but the cap is doing more harm. A cap is being used as a bandaid. its covering up your electrical voltage problem. While your eyes are seeing dimming with the cap out, you can clearly see you are having a problem, but with the cap its covering up fooling you to believe you have solved the problem. In the long run, your batteries will die out sooner, and your alternaot is taking a beating from trying to keep up the charge for the demand of power. my suggestion to you is to start by getting a deepcycle battery, then upgrade your wires(with a thicker guage) coming from your alty to bat(as well as grounds). then if problem still occurs, upgrade to a higher amp alternator.

    so long story short solve the problem, dont cover it

    edit: i realize i didn't answer your intial question sorry about that. but my response should still help your other problems

  3. #3
    MySQL Error scott_fx's Avatar
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    safe way to test the cap...unhook it, see if the problem persists.

    i agree with magalo, upgrade the big three, remove the cap and you'll probably solve the dimming issues. (taking for granted you're using the correct gauge of wire from your bat. to amp and amp to ground
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    I believe a way to test it out is by first calculating the RC time constant needed to charge or discharge the cap through a load resistor.
    Time constant (Tc in seconds)=Resistance(R in ohms)*Capacitance(C in farads)
    Tc=R*C

    example(a different resistor can be used but keep in mind that the smaller the resistor value is, the higher the current will become):
    resistance=50ohms, Cap=1Farad
    Tc=50*1
    =50sec

    Each time constant increases the charge by approx 37% of the remaining difference of the voltage. So after the 2nd time constant the voltage of the cap will be approx 61% of the full charge, and 76% after that and so on. With that you can calculate the total time needed to fully charge or discharge the cap. Five time constants equals aprox 99% of supplied voltage when charging or 99% drop from supplied voltage when discharge.
    5*Tc=~full charge or discharge

    example from above:
    50sec*5=250sec=4min 10sec for full approx charge or discharge

    After calculating, test the caps charging or discharging time through the same valued load resistor as calculated. (a meter should be used to check the voltage differential to monitor progress)
    1 The charge and discharge time should be equal
    2 the actual time should equal the calculated time.
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    MySQL Error scott_fx's Avatar
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    daaaaayyuuuummmm!
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  6. #6
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    Yeah can you tell I go to school for electrical engineering
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  7. #7
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    that's awesome dude, good to have 'wicked smaht' people on the forums
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  8. #8
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    Quote Originally Posted by scott_fx View Post
    that's awesome dude, good to have 'wicked smaht' people on the forums
    Thanks man, I do what I can to help out.
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