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 |
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