Okay,last week you helped me out a lot regarding my whole 6 volt question. I found the resistor wire coming from the fuse block. I checked the voltage at the wire while the engine was running and if read 13 volts!!! I'm sure it's the resistor wire because it's wrapped in a cloth like insulation and I measured the resistance to be 1.5-1.6 Ohms. That's a lot for a normal wire, so that tells me it IS the resistor wire.

So my question is, how can I still have full potential going to the ignition if it still goes through the resistor wire? Ohm's law and common sense says that's not possible. Also I just noticed that were the resistor wire connector is, there's a yellow wire attached to it, I traced it back, and saw that it grounds to the engine. What's that all about? Any help on this matter would be much appreciated. :confused:

You're close, Toad. You've measured 13 volts with the engine running and with a Digital Multimeter rather than an analog volt-ohmmeter. The DMM reacts faster than the analog meter.

Also, think of electricity as water. Points open, no current flow, (water turned on, thumb over end of hose, no water flow). Maximum volts (water pressure) will be read/felt, feel the hose expand in your hand?

Points closed, current flows, voltage drops, (remove thumb from hose end, pressure goes down, water flows), the hose got smaller in the hand, didn't it? Current flowing through the wire will heat it up, thus causing a reduction in voltage.

One usually measures the coil voltage after the engine been running for, say 1/2 hour, this allows the wire to "heat up," thus building resistance. The wire has a temperature coefficient. Battery voltage at start-up, and less voltage after the engine has ran. It takes less voltage to operate an engine when hot than it does when cold.

The yellow wire should be coming from the starter solenoid terminal "R", thus providing start igniton voltage when starter is engaged.

So are we saying it's possible to have 12 volts to the distributor even when the resistor is in place?

I guess it's possible, because I let my engine warm up today to (above 160*) and I kept checking the voltage...and nothing. I had 13+ volts across the resistor wire no matter how long I let it run. The voltage even increased as I revved the engine. Go figure! graemlins/clonk.gif My resistor wire never even got hot either. I could've held it all day long...cool to the touch. I'm positive it's a resitor wire too, because it doesn't look like any of the others, and it has 1.5 ohms too! Does anyone know what the deal is?

Almost sounds like your coil is getting battery from another source. Maybe the 12v lead from the starter is always hot instead of being hot just when you are starting the car for some reason...

I'm more inclined to agree with DjD on this one. If you were taking your reading with the DMM on the DC scale you should have seen more like 9 volts with the engine running hot or cold.

-dnult

You'd think so, my mine is the same way. The wiring harness has not been touched, either.

Thank god it's friday. Now that I've had my friday beer...
graemlins/beers.gif
The resitance multiplied by the current through the resistance equals the voltage drop. (E = I * R). It is possible to have 12 volts on both sides of the resitance provided the current (I) is zero. But on a running motor the voltage should be something on the order of 9 volts.

The current through the circuit will be a ringing square wave with a duty cycle proportional to the dwell setting. The ringing will not show up on a DC meter so the following holds true.

If you know the coils resistance (Rc) and the ballast resistance(Rb) you should see...
Vcoil = Ebatt * Rc / (Rb + Rc)
...with the points closed. Vcoil = Ebatt with the points open. With the engine running, the voltage on the coil (as read with any meter set on DC) will be some value in between the two extreams. If Dwell is expressed as a decimal precent (between 0 and 1) the voltage on the coil will be...
Vcoil = Ebatt * [Dwell * (Rc / (Rc+Rb)) + (1-Dwell)]
If Dwell is in degrees, divide Dwell in the equation above by 360.

Heating will have some affect. Both the coil primary winding and ballast resistance will increase due to heating. But since they both increase, the voltage change is fairly slight.

-dnult

[ 11-21-2003, 04:45 PM: Message edited by: dnult ]