Second attempt:
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
another element, the brake lights. Therefore the current way have been
reduced somewhat by the increase in resistance in the switch, and the
switch now carries a somewhat reduced current, but at a greater power
consumption so it was heating up, as power = I*I*R.
Bob Reisse
76 MGB
NAMGBR 8-3559
69 MGC-GT coming together.
At 07:23 AM 5/4/99 -0400, Fred Pixley wrote:
>Perhaps I'm missing something fundamental here but I can't understand why
>increased resistance would cause heat and increased current draw. Ohm's Law
>states V=I*R. If the resistance increases the current and heat should drop.
>I'm assuming a constant voltage supply.
>
>>Date: Mon, 03 May 1999 18:47:11 -0500
>>From: Rick Huber <rickhuber@home.com>
>>Subject: Re: Brake Switch
>
>>Kelvin,
>
>>You were right, the contacts in the brake switch were corroded,
>>therefore drawing a lot of current, thereby making the switch hot and
>>not sending enough juice to the brake lights. Took it apart, cleaned
>>the contacts, and put it back in this afternoon, and the lights work
>>fine. Thanks for the help. The reason the fuel pump slows down when
>>the brake lights are on with the car not running is the extra power draw
>>to the lights causes the voltage in the whole circuit to drop just a
>>little bit, therefore slowing down the fuel pump. I'm learning more all
>>the time.
>
>>Safety Fast,
>
>>Rick Huber
>>75 V8 B Daily Driver
>>65 B undergoing lengthy restoration
>
>Fred Pixley
>Napanee, Ontario, Canada
>
>
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