David's great explanation reminds me of a good book on 12V circuitry, Weems & Plath The 12 Volt Doctor's Practical Handbook.

http://www.amazon.com/Weems-Plath-Doctors-Practical-Handbook/dp/B000IMWV34

It helped me figure out some things, and even though it is geared to boat owners, I think it is very useful for vanagon owners.

alistair

On 2011-11-03, at 7:55 PM, David Beierl wrote:

> At 09:44 PM 11/3/2011, Phil Zimmerman wrote:
>> Again, I assert here, conventional wisdom in the application of
>> Ohm's law does not explain your burnt contacts.
>> Ohm's law, E=IR describes the current decreasing as the resistance
>> increases (assuming a constant and nominal 12dc). So, how does this
>> burn the wire junction?
> 
> Same way a fuse blows.  A fuse is a deliberate high-resistance spot
> in a circuit, designed to "steal" a small amount of power from the
> load and waste it as heat in a concentrated area (most fuses also
> have deliberately low melting points, but that doesn't matter for
> this illustration).  The most common reason for "false blowing" of a
> fuse is a high-resistance contact at one end creating excessive heat
> for the amount of current in the circuit, thus prematurely melting
> the fuse element.
> 
> Say you've got a 12 amp resistive load at 12 volts.  R = E/I =
> 12V/12A = 1R.  P = EI = 12V*12A = 144 Watts.  We'll pretend the
> wiring is perfect.
> 
> Now you have a poor connection with a tenth of an ohm.  Since total
> resistance is now 1R1, current goes down to  12V/1R1 = 10.9A and
> power to 12V*10.9A = 131W.  However, the connection is now
> dissipating P = I^2R  = 10.9A^2 * 0.1R = 118.8 * .1 = 11.9W which it
> wasn't before (and since we're pretending our other wiring is
> perfect, the load is actually developing 119W).  Twelve watts is as
> much as a very small soldering iron.  If you concentrate it in a
> small area that's plenty to start bothering things.
> 
> As the load currents go up the problem gets worse.
> 
> Yours,
> David