voltage drops and hot resistor

F

fordmerc7

Member
Problem:low voltage at coil (2.6)
Voltage at left of resistor=6.1; to right of resistor =2.9 Resistor hot within 30 sec. All measurements with engine cold, not running
Question: Should the resistor get hot? Is the voltage drop across resistor supposed to be that big? (I think not!) For the record, I get similar results with two different resistors.
Harness is new a couple of years ago.

In a post on 6/16/13 koates addressed part of the issue, but not the resistor voltage drop.
Advice and comments will be appreciated.
 
To answer part of the question, yes, resistors get hot. That is where the energy goes, turned into heat.
 
voltage should only drop when points are closed. I would remove the ign wire at the coil and again check voltage--would be under a no load condition. I have seen where the ign wire has rubbed a bare spot and it is slightly arcing causing an undue load which also lowers the voltage
 
Problem resolved! Wire and resistor deliver 6.1 volts to the wire detached from the coil!
Obviously I failed the electricity part of physics class.
 
To answer your question about if a resistor gets hot. When you install a new one it'll puff a little smoke as it burns off factory coatings. Really gets ya worried the first time you do it. :)

Best if you have 3.5-4v at the coil.
 
alanwoodieman said:
voltage should only drop when points are closed. I would remove the ign wire at the coil and again check voltage--would be under a no load condition. I have seen where the ign wire has rubbed a bare spot and it is slightly arcing causing an undue load which also lowers the voltage
Click to expand...


please explain this? So after the resistor you should get 6v? Maybe I read this wrong. Should be 3.5-4v not running or after the resistor (it's only job is to reduce voltage).


Guess you can think of it as a little electric heater that burns up 2/3 volts before the coil.
 
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If you have no load(current flowing through the circuit) there will be no voltage drop over the resistor.
So unhooked from the coil or points open what you have at input will be same as output.
If you have a lower voltage at the coil then normal it can be several different problems but the most common from my experience would be...
1. Wiring issues...bad connections, worn insulation...
2. Ballast resistor...connection between resistance wire and threaded connection.
3. Short in coil...a few shorted turns in the coil raises current drawn at the same time as output of coil is lower...still have spark just less then it should be.
When the coil draws more current the voltage drop over the resistor increases and you get a lower voltage reading at the coil.
When a coil has shorted turns thoose "steal" a bit of the charge in the coil and turn it into heat so a tell tale sign is a hotter then usual coil when running.
Also doing a sparktest the spark can´t jump a gap as far as usual.
 
okay, that makes sense.


free flow voltage over amperage. Resistor only works with amps applied?
 
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If you put say 6v at one end of the resistor and no load you get the same reading at the other end.
Then you start loading it...what you do is ingrease the current flow until you ground it and all energy is burnt off as heat in the resistor and you get 0v reading at the output.
 
Tinker said:
please explain this? So after the resistor you should get 6v? Maybe I read this wrong. Should be 3.5-4v not running or after the resistor (it's only job is to reduce voltage).


Guess you can think of it as a little electric heater that burns up 2/3 volts before the coil.
Click to expand...

The resistor drops voltage only when the points are closed. When the points are open, you have full voltage available, but unused.

And you better hope your resistor doesn't drop more than 2 volts, because by the time the voltage gets to the resistor, it's already dropped a half volt through a half dozen connections, the worse of which is the ignition switch.

Actually, the resistor drop is variable, depending on the load and the incoming voltage. The drop is a percentage, not a fixed voltage value. This is the real genius of the resistor, to keep the input to the coil more or less constant between high voltage drops from the starter motor and spikes in voltage from the generator.

Now, as I'm not an Electrical Engineer, I probably misstated the facts, so take what I said with a grain of salt until a real EE chimes in!
 
flatheadmurre said:
If you put say 6v at one end of the resistor and no load you get the same reading at the other end.
Then you start loading it...what you do is ingrease the current flow until you ground it and all energy is burnt off as heat in the resistor and you get 0v reading at the output.
Click to expand...


thank you. But doesn't connecting the neg and positive circuits, completing the circuit engage the resistor to work? As the resistor is the amp draw. like a light bulb? or not
 
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Cutout is just a switch that closes when the generator is turning fast enough to be able to charge your battery.
If the cutout wasn´t there the generator would discharge your battery at idle or when car is turned off.
Voltage regulator has the added feature of being able to feed the generatorfields more or less current for magnetizing to keep a constant output voltage.
 
sorry i edited my question you answered. So a cutout responds to voltage, but a resistor responds to amperage not voltage.
 
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Tinker said:
thank you. But doesn't connecting the neg and positive circuits, completing the circuit engage the resistor to work? As the resistor is the amp draw. like a light bulb?
Click to expand...

Lightbulb is a resistor just instead of creating only heat it emits light to.
Resistor is restricting the current flow...just more or less electrons passing it.
The start and end of the currentflow is the source (battery,generator).
 
flatheadmurre said:
Lightbulb is a resistor just instead of creating only heat it emits light to.
Resistor is restricting the current flow...just more or less electrons passing it.
The start and end of the currentflow is the source (battery,generator).
Click to expand...


so that's my point, the resistor works without amp draw. so with a grounded volt system and a under dash resistor inline, you will see a voltage drop after the resistor at the coil without an amp draw.
 
Tinker said:
sorry i edited my question you answered. So a cutout responds to voltage, but a resistor responds to amperage not voltage.
Click to expand...

They are all connected.
Ohms law tells us that pushing current through a resistor gives us a voltage reading(drop) over it.

You change voltage current or resistance in the circuit the others will respond.

But without current flow there´s nothing happening...like turning the switch off.
 
I get it. i guess tomorrow i'll go out and test my underdash resistor with a volt meter....
 

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