Confused about the ballast resistor

[mysavioreigns]

I went to bypass the ballast resistor tonight and when I took off the guage panel, I couldn't find a big pink wire. I looked and looked, but didn't see a thing.

I went to the cab-side of the connector and looked, still didn't see anything pink.

Soo that got me to thinking... When I got the car, it had points/condenser. I then put in a Pertronix II unit with coil, but never touched anything to do with the ballast. Would it be possible that someone had already done it? I don't have anything electrical hooked up to the car at the moment, but if you guys will let me know what voltage should be where, I'll be sure to check it when I get all of the crap back in the car.

thanks

edit: Oh, and I forgot to mention...I'm putting in a new engine with a DSII w/ HEI module, so that is the reasoning behind it.

 

[kukm66]

According to the 65 Mustang diagram, the pink wire comes from a red-green wire that is attached to the C pole on the ignition switch and feeds into the firewall plug where it splits off on the engine bay side into 2 wires. One goes to the coil and is red-green while the other goes to the starter relay and is black-yellow.

Here are 2 links to view the diagram:

http://www.tocmp.com/tOCMP/wiring/5765w ... 65-235.jpg

http://www.tocmp.com/tOCMP/wiring/5765w ... 65-234.jpg

 

[Anonymous]

I just got installing the DS2 over the weekend. I can tell you that my pink wire didn't look so pink after 42 years. You are looking for the red wire that is coming out of the ignition switch. About three to four inches down the wire is a bullet connector. That's the pink wire that's connected to it. Hope this helps.

 

[Caballo Diablo]

mysavioreigns,

The pink wire IS your resistor, there will not be a seperate ballast resistor. I just went though this a couple weeks ago when I installed the DSII.

--mitch

 

[CoupeBoy]

mysavioreigns,

The pink wire IS your resistor

But what he said was

I couldn't find a big pink wire. I looked and looked, but didn't see a thing.

I went to the cab-side of the connector and looked, still didn't see anything pink.

I'd guess somebody already removed it for him. Best way to check though would be with a voltmeter and some more eyeball time under the dash.

-ron

 

[mysavioreigns]

Thanks guys,
I understand what it is and what it should look like, but I'm not sure what the voltage should read (and where i should read it). If I know that, I can check it to see if it's already bypassed.

 

[woodbutcher]

I THINK that the resistor cuts the voltage to 6 volts.If I`m wrong,someone please correct me.
Leo

 

[ludwig]

I THINK that the resistor cuts the voltage to 6 volts.If I`m wrong,someone please correct me.

Righto. Read the voltage across the positive to the coil. It will be 6 V if the resistor wire is intact., 12 if not.

 

[mysavioreigns]

Awesome...is this when the key is in some certain position?

 

[ludwig]

Awesome...is this when the key is in some certain position?

On

 

[Anonymous]

The voltage won't necessarly be 6 volts on that wire, if the coil circuit isn't complete then it will be 12 volts. If this was on a points system the points would need to be closed then you would have current flow and then a voltage drop will occur across the resistor wire. If there is no current flow then your voltage will be the supply voltage.
You can always unhook your negative side of your coil wire and connect a wire from there to ground to do the test.

 

[jamyers]

Awesome...is this when the key is in some certain position?

With the key in the "Run" position, the resistor will drop the voltage at the coil to 6V.

... it splits off on the engine bay side into 2 wires. One goes to the coil and is red-green while the other goes to the starter relay and is black-yellow.

That other wire is to supply full 12V when starting - When the starter relay closes (to crank the engine), it also routes full 12V through that wire to the coil for extra starting ignition power. Not to argue with Donero, but the points being open or closed only affects the current flow (amperage), not the voltage. Voltage through the resistor will ALWAYS be less, because it's going through a resistor.

 

[61Futura]

"Voltage through the resistor will ALWAYS be less, because it's going through a resistor"


Not always. Current must be flowing for any voltage to be dropped. That is why you need to ground the negative side of the coil before checking the voltage on the positive side of the coil with the ignition switch on "run"

With the engine running, the voltage should be somewhere between 14 volts and the lowest voltage the coil sees when energized.

 

[jamyers]

"Voltage through the resistor will ALWAYS be less, because it's going through a resistor"


Not always. Current must be flowing for any voltage to be dropped. That is why you need to ground the negative side of the coil before checking the voltage on the positive side of the coil with the ignition switch on "run"
...

Then why could I unhook that wire from the coil altogether, and my multimeter showed 6V with the key on? You don't need current flowing to check voltage (well, except for the tiny bit that'll flow through your voltmeter...).

 

[ludwig]

Let's see:

Voltage = water pressure

Amperes = rate of flow (gallons per minute)

Watts = amount of water (total gallons used)

Isn't house water pressure still (nominal) 70 psi whether the faucet is on or off?
So, with the key on run, isn't the voltage the same across the hot and the terminal at the coil with the wire connected or not (i.e. faucet on or off)?

Then why could I unhook that wire from the coil altogether, and my multimeter showed 6V with the key on? You don't need current flowing to check voltage (well, except for the tiny bit that'll flow through your voltmeter...).

 

[jamyers]

My point exactly. Voltage is potential difference between two points, and it matters not if current is actually flowing between those two points.

Voltage = "Pressure" = Effort
Amperage = Flow = Work (movement)

Like Effort vs. Work, you can have a lot of Effort and have no Work, like me trying to push my Buick.

 

[jamyers]

...Isn't house water pressure still (nominal) 70 psi ...

30-40 psi is standard. 70 psi would likely have your commode going airborn when you flushed.

 

[AZstang66]

Calm down fella's!
If you get confused, consult Mr. Ohm.

He states: V = I*R

That being said, if there is resistance, there MUST be a current if there is a voltage. And vice versa. And voltage is the potential difference between to nodes in a circuit, but determining voltage across an open circuit is impossible, without infinite resistance. So you do in fact need current flowing to check a voltage.
The equation also shows that the voltage and resistance do not necessarily depend on each other- you can keep a constant voltage, even accross a resistor, if only the current is changing.

 

[Bort62]

What AZ said, sort of.

Because (between two points) V = IR, for the voltage to drop, there needs to be a current.

In an open circuit (key on but coil not grounded) the current is 0, so the voltage = source voltage, independent of resistance.

However, once you close the circuit and start to draw current through the resistor, the voltage on the output side of it will drop.

The other important equation here is P = VI.

Power = Voltage X current.

So, as an example - you have a 15 ohm ballast resistor hooked up to a 12v feed(Vin). With the circuit open and drawing 0 current, Then your voltage across the resistor is Vin - (I*R) or Vin - (0 * 15). Obviously, Vin - 0 = Vin. Also, P = VI = 0*0 = 0.

So in this case, you will measure 12v on the output of the resistor.

However, if you now put a .1 amp load on the output of your circuit, you equation's change...

V = I*R = .1 * 15 = 1.5v

So Vout = Vin - 1.5 = 10.5v

Since V = 1.5, then P = 1.5 * .1 = .15 watts


Clear as mud.

 

[AZstang66]

I remember it by:
"Twinkle twinkle little star, voltage equals I time R,
Up above the world so high, power equals V times I"

Heaven forbid there's a capacitor or inductor in there... no easy math for that!