1755 Parking brake light

Genz

Member
This switch is activated when parking brake is on . Isn't this switch a single pole switch. I hook leads up as show in picture and it smokes the leads. On the ground end, is says normal closed. What am i not understanding. Doesn't there have to be a constant hot going to this switch?
mvphoto50835.jpg
 
IRC the switch is open when the parking brake is off and then closes when you put the brake on. The circuit is a grounding one not a hot one IRC. The hot wire is on the light and the ground wire goes to this switch.

What your doing is dead shorting the hot wire you running to the switch.

Many common circuits control the ground side of a circuit not the power side. Interior lights on vehicles is one ( before body controllers) and fuel sending units are another that ground to work.
 
If I understand correctly, there shouldn't be any hot going to this switch, all the switch is doing is breaking the ground..correct? the pigtail goes to frame ground and the empty terminal goes to the light ground side..correct?
 
Your correct Genz. I am sorry for using the term IRC wrong. If I remember correctly.

I owned several Oliver 1755 and 1855s in the mid 1980s. I liked my Oliver 1655 so much that I wanted one with a little more HP and bought a 1755 next. I did not like it as well as my 1655 so I only kept it for a few years and sold the 1755.

The park brake was one of the things I disliked about the 1755. If you did not apply it really hard it was easy to drive off with it engaged and ruin the bake. The parking brake light was hard to keep working on the one I had. I finally replaced the entire harness and light socket to finally fix it.
 
Got it..thanks JD Seller...that's main reason I want to get this fixed, we've already smoked the brakes twice prior. Parking brake is really handy when backing and hooking up implements by yourself, once hook up, it is really easy to forget the park brake is engaged, mile down the road you see smoke cloud behind ya!..not good..thanks again
 
Think back to school days and ?resistors in series?. This is a fairly classic case for that scenario.

You have (theoretically) two resistors in circuit - the lamp and the switch. When the switch is open, its resistance is close to infinity, when closed
its resistance is close to zero. So two options for the resistance/current value of the line. In both cases the voltage drop across each series
resistance will be shared according to their individual resistance values.

That means the voltage drop across the switch, while open circuit, will be supply voltage and current will be zero. As soon as the switch closes
(zero resistance) all the voltage drop will be across the lamp and the current will be limited by the lamp filament according to Ohm?s Law.
 

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