>The short answer is that in this case the resistance of the switch went
>from essentially zero to some larger value, but the circuit still has
>anohter element, the brake lights. Therefore the current way have been
>reduced somewhat by the increase in reistance in the switch, but the switch
>now is carry a somewhat reduced current, but at a greater currect so it ws
>heating up, as power = I*I*R.
>
>Bob Reisse
>76 MGB
>NAMGBR 8-3559
>69 MGC-GT coming together.
Let's try this again. Purely theoretical:
If the brake circuit consisted only of the switch and brake lights (and a
12 volt battery), and the brakes normally drop 12 volts, and the switch
zero, than the brake lights are dropping all the voltage and therefore
dissipating all of the heat. If the switch takes on resistance, let's say
equal to that of the brake lights combined, each part, the brake lights and
the switch, will drop 6 volts, assuming they are in series. It is this
voltage drop, dissipating power, which creates heat (or motion in a motor,
or whatnot, but in the switch, it had nothing to do but heat up). Before
the switch took on resistance, it had no voltage drop, and therefore
dissipated no heat, and since power=I*E, or current times voltage,
(resistance of the component determines the voltage drop of the component,
and total resistance of the circuit determines the current in the circuit,
but resistance is not part of the formula for power), then the switch
heated up.
In the real world, wires and the contacts of a switch actually do have a
tiny bit of resistance, but that's why we use larger wires and switches for
higher power applications. The switch should have been a negligible
resistance in this case, but was not.
Corrections to my corrections, anyone?
Nevin
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