Basic (stupid?)Electrical Question

[Brett Dikeman]

At 12:49 PM -0500 11/2/02, Forhan, Thomas wrote:

>I then removed and replaced all the fuses, one by one, expecting
>that when I pulled the one on the guilty circuit, the multimeter
>would drop to zero. It never did -including the one fuse on the knee
>bolster aux relay board. It just continued to read 13 volts.

Multiple systems in the car utilize at least a tiny amount of current
when off.  The radio, the ECU, the dash clock, etc.  If any one of
them is connected to the battery, you'll see almost battery voltage
between chassis and - or cable + and battery +.


>Can anyone point me in the right direction?

Yes.  The multimeter needs to go in serial with the battery, just
like you have it, but you need to switch to current measuring, not
voltage.

Make sure your meter is rated to at least an amp or two, just so you
don't blow the internal fuse(or the meter itself), and make sure
everything is off.  I believe drain current is usually much lower
than that, but clearly something is drawing more than the normal
amount of current.

If your meter lacks current measuring capabilities, go to Rat Shack
and buy a sand resistor- ie those white ceramic bar shaped ones.  Put
it inline(instead of the meter), and measure the voltage differential
between the legs of the resistor.

The meter reading, in volts, will exactly equal the amount of
current(current = voltage / resistance.  If resistance is 1 ohm,
current  effectively equals voltage.)

BTW- two resistors in parallel don't 'just' double their ability to
handle current.  It also changes the resistance.

(1/Rp) = (1/R1) + (1/R2)

By the way, to size the resistor(they're marked in watts), P = V x I,
so at 13.8v a 15 watt resistor would be able to handle 1.08 amps.

Side note- ever wondered why corrosion on battery terminals can have
a huge effect on starting?

V = 50A x .1 Ohm

V= 5 volt drop!

Hope this helps,
Brett
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