Re: Testing for Leaking DC Current

[Larry Mittell <lmittel@ibm.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

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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
>
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