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