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1989 cushman


46wl

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hello, I have a 1989 cushman truckster  and I am troubleshooting the ignition system, it  is a coil and battery system. in the circuit is a resister does anyone know what the voltage drop should be on a 12 volt system? thanks  Dave

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My guess is the resistor is there to limit the current flow through the coil to a manageable level in the order of 4 to 5 amps during normal running.  Is it electric start?  Sometimes the resistor is switched out during starting and back in when the engine is running.  Does it use points or is it a solid state module for firing the coil?  Are you familiar with using a multimeter for troubleshooting?

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yes you are correct, during cranking , with the electric starter, the coils get  full  12 volts and make spark and the engine starts and runs  then voltage switches  and  goes through the resister.  the resister on the machine drops the voltage to 3.8  to 4.0 volts, a level that coils dont make power then the motor stalls. I am trying to see if anyone knows what the resister should voltage should be? the ignition system uses points . Dave

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You can do a test for a brief interval of engine running. Place a jumper wire across the resistor and start the engine. If it continues to run after starting then the resistor is the problem. Don’t run the engine for a long time with the jumper in place. Does the resistor have two terminals on each end or just one? The resistor block with two terminals on each end has the starting jumper and running resistor in the same package.  Too much resistance in a bad resistor can limit the current flow in the coil and not produce enough high voltage to make a spark. If the run resistor is open (broken inside) then the engine dies when the key is released too.

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6 minutes ago, Mikefit said:

You could buy a hand full of resisters all for various cars and trucks I would, bet they all drop the voltage to the 6-8 volt range

Yes, the run voltage would be about 1/2 of the battery voltage in a 12v system. The ballast resistor would typically be about 1.5 ohms to get to this run voltage.   

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hello, that is what I have done I hot wired the coil with strait 12v and the cart runs, when I pull the hot wire off it stalls, and I am getting 12v to one side of the resistor, and 4v to the other side. so you think the resistor has gone bad? I would have thought that it would have just burned open. I still need to see what is up with ignition switch to see why the coils are not getting the 12v during cranking?  thanks Dave

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3 minutes ago, 46wl said:

hello, that is what I have done I hot wired the coil with strait 12v and the cart runs, when I pull the hot wire off it stalls, and I am getting 12v to one side of the resistor, and 4v to the other side. so you think the resistor has gone bad? I would have thought that it would have just burned open. I still need to see what is up with ignition switch to see why the coils are not getting the 12v during cranking?  thanks Dave

Those are wire wound resistors so they can have unusual issues. Heavy overload will burn them open.

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this does look like mopar resistor  it has one connection on each end it has a sheet metal strap for mounting . the setup is 2 coils wired in a series with one set of points. my harley has a coil with 2 plug wire connections  they both fire at the same time, one on the power stroke the other on the exhaust stroke . Dave

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hello, we got the new resistor today, and I was surprised to see when I put 12v to one side of it I got 12v on the other side. I did measure the ohms  and  the resistor had .6 or .5  ohms of resistance.  so if this is the right resistor then I will not have to add a  wire to the coils that is hot while cranking. I suppose time will tell if the coils and or points are damaged with extended running of the motor. I really  thought that there should be a voltage drop with the resistor. Dave.    

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15 hours ago, Oldtech said:

a resistor with no load will give you the same voltage at both ends.

Yes!  The ignition points have to be closed to make current flow through the resistor so you can measure the voltage drop across it. With the points open the voltage on both ends of the resistor will be the same just as Oldtech said above.

Edited by TerryB
More info (see edit history)
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