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455 resistor resistance

455 resistor resistance

Old June 7th, 2018, 12:26 PM
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455 resistor resistance

Hello everyone,
While diagnosing a no spark issue, I noticed that my ballast resistor is reading 12 volt at both ends, so I guess it needs replaced. I am looking at them on Summit, and they are available in different resistance values such as 0.8 and 1.6 ohms, so I'm not sure which one to get. It is a 1968 Delta 88 with a 455.
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Old June 7th, 2018, 12:47 PM
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Is it a resistance wire or a ballast resistor? Voltage is measured with a load on it, it will usually be around 9v. There is no way for either to short, the failure mode is an open.

What distributor is installed, points, points replacement, HEI, etc?
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Old June 7th, 2018, 01:26 PM
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It's a ballast resistor. So it is probably still good? I didn't put a load on when testing it, I'm not sure how to do that. The ignition system is all stock, points are a new economy set I just put in but haven't set yet, the old ones were pitted.

Last edited by CrazedCountryRebel; June 7th, 2018 at 01:31 PM.
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Old June 7th, 2018, 01:58 PM
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If you have not set your dwell (gap), the points may be too far open or not open at all. Set dwell first, then timing. Dwell effects timing settings, but timing settings do not effect dwell.
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Old June 7th, 2018, 02:02 PM
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Originally Posted by oldcutlass View Post
Is it a resistance wire or a ballast resistor? Voltage is measured with a load on it, it will usually be around 9v. There is no way for either to short, the failure mode is an open.

What distributor is installed, points, points replacement, HEI, etc?
Having a load on the circuit will not change the voltage (unless the battery is draining down). Measure the voltage after the resistor.
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Old June 7th, 2018, 02:18 PM
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The reason he's reading 12v on both sides is because the points are open and there is no load on the primary side of the coil. I assure you the ballast resistor needs to have a load on it to show a difference of voltage before and after.
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Old June 7th, 2018, 02:39 PM
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Originally Posted by oldcutlass View Post
The reason he's reading 12v on both sides is because the points are open and there is no load on the primary side of the coil. I assure you the ballast resistor needs to have a load on it to show a difference of voltage before and after.
Not according to the laws of physics...........
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Old June 7th, 2018, 02:53 PM
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Originally Posted by oldcutlass View Post
If you have not set your dwell (gap), the points may be too far open or not open at all. Set dwell first, then timing. Dwell effects timing settings, but timing settings do not effect dwell.
Yes, I recently bought a dwell meter and timing light so I could do this, but I will need to get the car running first.
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Old June 7th, 2018, 02:58 PM
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as to your debate on whether or not the voltage changes under load, I just checked it on my 77 Ford truck. It reads 12v at both ends when not running, and under 11v on the output end when it is running. After seeing these results, I have to agree with oldcutlass.

Last edited by CrazedCountryRebel; June 7th, 2018 at 03:06 PM.
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Old June 7th, 2018, 03:35 PM
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Originally Posted by CrazedCountryRebel View Post
Yes, I recently bought a dwell meter and timing light so I could do this, but I will need to get the car running first.
You can set dwell cranking the engine. Or you can just set the point gap to .016 with a feeler gauge and then fine tune the dwell.
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Old June 7th, 2018, 05:26 PM
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Originally Posted by RandyS View Post
Not according to the laws of physics...........
I disagree.

If you have a resistor connected to +12 Volts on one end and the other end is not connected to anything, the voltage measured on both ends of the resistor will be 12 Volts. Since there is no current flowing through the resistor, there is no voltage drop across it.

If you then connect the open end of the resistor to ground, current will flow through the resistor, causing a voltage drop, and the measured voltage on that end will now be less than 12 Volts.
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Old June 7th, 2018, 05:26 PM
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Originally Posted by oldcutlass View Post
You can set dwell cranking the engine. Or you can just set the point gap to .016 with a feeler gauge and then fine tune the dwell.
How would I go about setting it with the feeler gauge, do I need to find TDC?
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Old June 7th, 2018, 06:17 PM
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No, take the cap and rotor off. Rotate the engine until you get the points wear strip on the tip of one of the 8 high spots on the distributor lobes. Then adjust the points by inserting a .016 feeler gauge between the 2 points to where it just rubs on both sides. Reinstall the cap and rotor, hook up your dwell meter and start the engine. Then fine tune your adjustment so your dwell is at 30*.


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Old June 7th, 2018, 06:24 PM
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Originally Posted by oldcutlass View Post
No, take the cap and rotor off. Rotate the engine until you get the points wear strip on the tip of one of the 8 high spots on the distributor lobes. Then adjust the points by inserting a .016 feeler gauge between the 2 points to where it just rubs on both sides. Reinstall the cap and rotor, hook up your dwell meter and start the engine. Then fine tune your adjustment so your dwell is at 30*.


Thanks! I will try that.
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Old June 7th, 2018, 09:24 PM
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Originally Posted by Fun71 View Post
I disagree.

If you have a resistor connected to +12 Volts on one end and the other end is not connected to anything, the voltage measured on both ends of the resistor will be 12 Volts. Since there is no current flowing through the resistor, there is no voltage drop across it.

If you then connect the open end of the resistor to ground, current will flow through the resistor, causing a voltage drop, and the measured voltage on that end will now be less than 12 Volts.
In a circuit with a resistor attached to a voltage source, (V+) the voltage on both sides of the resistor cannot be the same. The resistor will reduce the voltage all the time, as that is the nature of what they do. You have to test it relative to V- (ground in this case) and there will always be a difference. Your DMM may not be able to measure it if it is not sensitive enough.
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Old June 8th, 2018, 12:21 PM
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Originally Posted by RandyS View Post
In a circuit with a resistor attached to a voltage source, (V+) the voltage on both sides of the resistor cannot be the same.
Yes it can, if one end of the resistor is not connected to anything and therefore there is no current flow through it.

Originally Posted by RandyS View Post
The resistor will reduce the voltage all the time, as that is the nature of what they do.
That happens only when there is current flow through the resistor. There is no voltage drop across the resistor if there is no current flow through it, which is the case when one end is not connected to anything. This is what happens when the points are open.
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Old June 8th, 2018, 12:31 PM
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Originally Posted by Fun71 View Post
Yes it can, if one end of the resistor is not connected to anything and therefore there is no current flow through it.

That happens only when there is current flow through the resistor. There is no voltage drop across the resistor if there is no current flow through it, which is the case when one end is not connected to anything. This is what happens when the points are open.
I don't think we are on the same page here. A resistor can only 'resist' when there is voltage flowing through it. There has to be voltage applied to measure the drop caused by the resistor. A DMM places a small load on the circuit so it can measure voltage drop. I mentioned the laws of physics earlier, which cannot be changed. A resistor must always resist.
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Old June 8th, 2018, 12:35 PM
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There is no current flowing across an open circuit. Take a light bulb with 12v on one side but the other side is not connected to anything. You can measure the same voltage on both sides yet the bulb is not lit, no current flowing.

Here I found this, specifically the 3rd paragraph and illustration:
https://www.allaboutcircuits.com/tex...-1/resistance/

Last edited by oldcutlass; June 8th, 2018 at 12:40 PM. Reason: added link
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Old June 8th, 2018, 01:34 PM
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Technically speaking, voltage does not flow, current does. As for checking voltage, it is "potential" and will not change under a load. The only way it can change is if the circuit has a fault and goes to ground, which is really what the points do. Every time the points close it discharges the coil to the distributor. So technically speaking you both are kinda correct and kinda incorrect.

By the way, to check current you need to be in series and is measured in amps.

Regards,

John
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Old June 8th, 2018, 03:23 PM
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Originally Posted by oldcutlass View Post
There is no current flowing across an open circuit. Take a light bulb with 12v on one side but the other side is not connected to anything. You can measure the same voltage on both sides yet the bulb is not lit, no current flowing.
Yep, you and I are on the same page. Translated to the automotive circuit we are discussing, the resistor is not connected to ground when the points are open.

Originally Posted by friesjh View Post
By the way, to check current you need to be in series and is measured in amps.
Yes, that is the direct way to do it, but may times current is checked by measuring the voltage drop across a known resistor. That is the way we design the circuitry in some of our ICs to measure current consumption (I work in a microelectronics design / manufacturing / failure analysis area).
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Old June 11th, 2018, 04:17 PM
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Yes, I was trying to be direct. A lot of people do not understand electricity. I've been an electrician for 30+ years and just graduated in 2015 with a bachelor's degree for Engineering Management and Electrical Engineering. My training is geared more towards the power generation side of the business, but this is circuits 101. Currently I am managing solar installation projects for a locally based company.

I was thinking someone would get confused if we started talking voltage drops across a resistor.
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Old June 11th, 2018, 08:47 PM
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Originally Posted by friesjh View Post
Yes, I was trying to be direct. A lot of people do not understand electricity. I've been an electrician for 30+ years and just graduated in 2015 with a bachelor's degree for Engineering Management and Electrical Engineering. My training is geared more towards the power generation side of the business, but this is circuits 101. Currently I am managing solar installation projects for a locally based company.

I was thinking someone would get confused if we started talking voltage drops across a resistor.
Voltage DOES drop across a resistor!!!!!!!!!! Current also drops across a resistor. I don't have an EE Degree, but I've been tinkering with electronics for 60 years.

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Old June 12th, 2018, 04:52 AM
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My apologies Randy. Kirchkoff proved this in the 1800's. It's been my experience that most people do not know much about electricity and how it works. It is worth noting that this applies to series circuits only.
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Old June 12th, 2018, 05:55 AM
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Thank You.
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Old June 12th, 2018, 06:00 AM
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It does not apply to an open circuit.
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Old June 12th, 2018, 06:19 AM
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Originally Posted by oldcutlass View Post
It does not apply to an open circuit.
That is not my point.

Last edited by RandyS; June 12th, 2018 at 06:21 AM.
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Old June 12th, 2018, 09:29 AM
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Originally Posted by RandyS View Post
That is not my point.
But we were discussing this in relation to open points (pun intended!).
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Old June 12th, 2018, 09:34 AM
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Originally Posted by friesjh View Post
I was thinking someone would get confused if we started talking voltage drops across a resistor.
How about the time it takes to charge a capacitor?
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Old June 12th, 2018, 10:38 AM
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Originally Posted by Fun71 View Post
How about the time it takes to charge a capacitor?
Would that be a Flux Capacitor???????
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Old June 13th, 2018, 10:57 AM
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I think some of you guys need to brush up on Ohm's law!
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Old June 13th, 2018, 11:07 AM
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Originally Posted by marty59 View Post
I think some of you guys need to brush up on Ohm's law!
Is that V = I R


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Old June 13th, 2018, 12:17 PM
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Ohms law does not apply to an open circuit.
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Old June 13th, 2018, 12:27 PM
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This thread is feeling like an open circuit..............
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Old June 13th, 2018, 12:28 PM
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Look at it this way you have 12.6v and an 1156 bulb which is .9 ohms. How many watts does it pull when its off? If your switching the negative side of the bulb and the switch is open you will read 12.6v on both sides of the bulb. The bulb is not pulling any current nor using any voltage because its off. The bulb is a resistor.
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Old June 13th, 2018, 02:12 PM
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With a sensitive enough DMM, you WILL see a difference on the different sides of the bulb. You have to, resistance doesn't just go away, it is always there (as in, a constant). Again, this is in reference to V-. The discussion is moot if there is no circuit..........

Last edited by RandyS; June 13th, 2018 at 02:51 PM.
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Old June 13th, 2018, 03:37 PM
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Originally Posted by RandyS View Post
With a sensitive enough DMM, you WILL see a difference on the different sides of the bulb. You have to, resistance doesn't just go away, it is always there (as in, a constant). Again, this is in reference to V-. The discussion is moot if there is no circuit..........
Yes, when you connect a voltmeter you are completing the circuit path to ground, so there is a small amount of current flow that is dependent upon the input resistance of your meter. But when you disconnect the meter, the voltage is exactly the same on either side of the resistor.
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Old June 13th, 2018, 04:22 PM
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Originally Posted by Fun71 View Post
Yes, when you connect a voltmeter you are completing the circuit path to ground, so there is a small amount of current flow that is dependent upon the input resistance of your meter. But when you disconnect the meter, the voltage is exactly the same on either side of the resistor.
There is NO voltage at all because there is not a complete circuit.........
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Old June 13th, 2018, 04:38 PM
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Come on Randy, we're splitting hairs here. We all understand what your saying, however the voltage drop is miniscule with a meter. It certainly is not the same voltage drop as it would be if the circuit is closed. The original question was why he read the same voltage on both sides of the resistor. The correct answer is … because the circuit was open due to the fact the points were open.
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Old June 13th, 2018, 05:18 PM
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Originally Posted by RandyS View Post
There is NO voltage at all because there is not a complete circuit.........
So a battery sitting on the shelf has no voltage if it is not connected in a circuit?
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Old June 13th, 2018, 09:18 PM
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Originally Posted by Fun71 View Post
So a battery sitting on the shelf has no voltage if it is not connected in a circuit?
You would not be able to tell if you only touched one post........

If a tree fell in a forest and............
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