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*To*: "Matthew Brenengen" <qcar@hotmail.com>, quattro@coimbra.ans.net*Subject*: Re: Testing for Leaking DC Current*From*: Larry Mittell <lmittel@ibm.net>*Date*: Wed, 10 Feb 1999 17:36:32 -0800*In-Reply-To*: <19990210181102.17534.qmail@hotmail.com>*Sender*: owner-quattro@coimbra.ans.net

Disconnect the negative battery cable, temporarily connect a resistor of known value between the negative post and ground, then measure the voltage across the resistor. Current in amperes may now be calculated by dividing the voltage by the resistance in ohms. So if you were to use, say, a 10 ohm resistor and measure 2.5 volts across it, the current would be 0.25 amperes. The rub is since you've placed a resistance in series with the load, this is actually *less* than the real-life current draw (in this case, about 0.315 amperes if I've done my back-of-the-envelope math correctly), but it will give you a order of magnitude and may be useful for trouble shooting. The larger the resistance you use, the closer the current will be to the "real" current draw. Conversely, the higher the resistance the lower the measured voltage will be. If the resistance is high enough, you'll get a voltage too low to measure with your meter. Just to be safe, if I were to use a 10 ohm resistor, I'd make sure it was rated for at least 10 watts. That's if the current draw is an amp or less. The rating you'll need will be the product of the resistance and the square of the current. Makes it a guessing game, since you won't know the current until you take the measurement. If you want to go to the trouble of calculating what the current draw *would* have been if it weren't for the resistor, the following equation should do it: I = (12*v)/(r*(12-v)) Where "I" is the "would-have-been" current, "v" is the voltage measured across the resistor and "r" is its resistance in ohms. It is assumed you have 12 volts at the battery posts. If you want to gild the lily, substitute actual battery voltage in place of the number 12. For the benefit of all the sparkies hanging out here, I suppose I ought to point out that all this assumes a purely resistive load whose resistance is independent of the current passed; even this may not hold if you happen to be traveling at a significant fraction of lightspeed the moment you run the test (who said Quattros weren't fast?). Gad, how'd I get into this? I gotta get a life. Where's the door? HTDBY, Larry Mittell ============================================================================ At 10:11 AM 2/10/99 PST, Matthew Brenengen wrote: >All this battery talk has gotten to my Syncro Wagon. If I let it sit, >now, even overnight, the battery will die to the point that the car will >not start. I have been putting off fixing it by disconnecting the neg. >terminal every night. I have noticed that there is a good spark from >the terminal when I connect it, so I figure there is a significant drain >somewhere in the car. > >I have tried measuring the current with my little multimeter, but all it >does is blow the multimeter's fuse -- it is only rated to about 40mA. >In checking Radio Shack's stock, I found that none of theirs really went >much higher in DC Amps. > >So how can I measure the draw without spending hundreds on some fancy >Fluke meter? I want to start pulling fuses, but unless I know when the >current stops, it is a waste of time. Is there an creative, alternative >way to measure the current? > >Thank you >Matthew Brenengen >'87 4kq; '88 QSW; '76 '02 > >______________________________________________________ >Get Your Private, Free Email at http://www.hotmail.com > >

**References**:**Testing for Leaking DC Current***From:*"Matthew Brenengen" <qcar@hotmail.com>

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