I'm about to give up on my car.

[phucheneh]

It was sarcasm. ;P Fluke's pretty much the standard for electrical work of this scale.

The 'magic' comment was also some mild sarcasm. It's not magic at all; it's physical law.

If you're looking for mV losses, I would assume you're dealing with small, sensitive circuits. You likely do not encounter what I'm describing because the small ohmmeter current is not in such stark contrast to the amount of power the circuit may carry during normal operation.

I really can't put it much more clearly than the water analogy. For most people, equating electricity with water works pretty well, since one of them is something they're familiar with that they can see.

If you don't feed the circuit enough electricity to actually test it in a real-world kind of way, you're not going to get real-world results.

Example: I have an 18ga wire with 3/4 of the strands cut at a particular location. Guess what? That 18ga wire is now ~28ga or so. The resistance of that bad spot in the wire will be directly proportional to the amount of current it tries to carry. Feeding a 2 ohm load with 100v? Your shit's not gonna work. 100 ohm load with 2v? You'll still get plenty of electricity.

I think the problems comes from people thinking of resistance as a fixed value. That's fine when you're talking about the load in a working circuit. But points of extraneous resistance are not going to be constants. The more power you attempt to run through them, the more voltage the trouble spot will consume (and heat it will produce- because it's basically a resistor now).

 

[T2urtle]

My 98 had a miss at idle in middle front cylinder. I forget what number. I threw coils and plug at it. I swap injectors as well. Ran compression all good.

Turned out the intake plenum gasket had a leak. Cheap quick fix drove me nuts for over 2 months.

 

[JCH13]

It was sarcasm. ;P Fluke's pretty much the standard for electrical work of this scale.

The 'magic' comment was also some mild sarcasm. It's not magic at all; it's physical law.

If you're looking for mV losses, I would assume you're dealing with small, sensitive circuits. You likely do not encounter what I'm describing because the small ohmmeter current is not in such stark contrast to the amount of power the circuit may carry during normal operation.

I really can't put it much more clearly than the water analogy. For most people, equating electricity with water works pretty well, since one of them is something they're familiar with that they can see.

If you don't feed the circuit enough electricity to actually test it in a real-world kind of way, you're not going to get real-world results.

Example: I have an 18ga wire with 3/4 of the strands cut at a particular location. Guess what? That 18ga wire is now ~28ga or so. The resistance of that bad spot in the wire will be directly proportional to the amount of current it tries to carry. Feeding a 2 ohm load with 100v? Your shit's not gonna work. 100 ohm load with 2v? You'll still get plenty of electricity.

I think the problems comes from people thinking of resistance as a fixed value. That's fine when you're talking about the load in a working circuit. But points of extraneous resistance are not going to be constants. The more power you attempt to run through them, the more voltage the trouble spot will consume (and heat it will produce- because it's basically a resistor now).

Are you trying to explain V=IR to me?

I think what you're trying to say is: "continuity does not always mean functionality". But you seem to be trying hard to confuse the matter with imprecise analogies and language.

The resistance of that bad spot in wire of your example might change resistance with temperature, a little bit. The amount of resistance of the 'bad spot' doesn't change when the circuit is under load, but the amount of voltage lost across the 'bad spot' will increase with current draw. This might mean that a device would see less than the ~12V required to operate it. However, the resistance of that spot is not increasing.

You might have the right basic concept in there somewhere, but you're interchanging terms and phenomena in a hap-hazard way that will make it very confusing for anyone to rectify your statements with real definitions and principals.

 

[phucheneh]

Are you trying to explain V=IR to me?

I think what you're trying to say is: "continuity does not always mean functionality". But you seem to be trying hard to confuse the matter with imprecise analogies and language.

The resistance of that bad spot in wire of your example might change resistance with temperature, a little bit. The amount of resistance of the 'bad spot' doesn't change when the circuit is under load, but the amount of voltage lost across the 'bad spot' will increase with current draw. This might mean that a device would see less than the ~12V required to operate it. However, the resistance of that spot is not increasing.

You might have the right basic concept in there somewhere, but you're interchanging terms and phenomena in a hap-hazard way that will make it very confusing for anyone to rectify your statements with real definitions and principals.

Erm. I apologize if I'm not the best teacher? It would appear you're the one not familiar with the principles.

If a circuit uses (or attempts to use) significantly more power than a DMM passes through it, and that causes a large vdrop across a partially-broken wire....how is that not an increase in resistance? Supply voltage is a constant, current is reduced. If you can't figure out the third part of that, maybe I should explain Ohm's law to you.

 

[jolancer]

sorry again - you guys arguing doesnt bother me you can if you like, but honestly i think your both right to an extent. resistance is resistance but if you go to extremes even the laws of physics start to change :/

sorry phucheneh, i honetly didnt read your post lol i glanced over it, Im a poor reader but even if your both right i doubt very much it will help Kroze out, especialy the probability of the coil source wire being faulty(unless he physically damaged it) cause its not high current at all.

 

[JCH13]

Erm. I apologize if I'm not the best teacher? It would appear you're the one not familiar with the principles.

If a circuit uses (or attempts to use) significantly more power than a DMM passes through it, and that causes a large vdrop across a partially-broken wire....how is that not an increase in resistance? Supply voltage is a constant, current is reduced. If you can't figure out the third part of that, maybe I should explain Ohm's law to you.

For the sake of argument, lets say the DMM pushes 1mA through a wire with a 10ohm 'bad spot' in it. The voltage drop across that bad spot will be:

V=IR -> 0.001A*10ohm = 0.01V

The DMM measures the 0.01V difference in its test leads and determines the resistance is 10ohm in that circuit.

If a motor tries to pull 1A through the same wire the voltage loss will be:

V=IR -> 1A*10ohm = 10V

Note that the resistance of the 'bad spot' never changed, but the voltage drop increased proportionally with current. A DMM will still measure '10ohms' across the affected area, and the resistance doesn't jump up when the circuit is under load, but the voltage loss does increase.

Current carrying capacity is reduced compared to a perfect wire like you say, but the resistance doesn't change.

Like I said, I think you have the right idea, but are just getting a little mixed up in the details.

/threadjacking

 

[phucheneh]

[morethreadjack]

Okay, so yes, we both obviously understand Ohm's law. And I could accept your math/explanation (or rather, I do, but there's more to it), if it wasn't for the fact that I have personally been bitten in the ass by resistance checks. As in, I deemed a circuit good and moved on, only to chase my tail and eventually conclude that I needed to return to said circuit and check it some more...yep, I'm not crazy, little to no resistance...let me try my other meter...yup, still good...

...then I load test and discover how much time I've wasted. I know it's the exception rather than the rule; but that exception is common enough that I generally default to a load test instead of a continuity check.

Now, let me go outside my realm of expertise and ask you a technical question that I may very well not understand as well as I think I do:

If you have a rated resistor (say something from Radioshack), yes, it should be about the same value at any voltage (within its working limits). I agree with that.

But a real resistor is made of...well, a resistive material. Carbon or some such, surrounded by ceramic to dissipate heat. There's no such thing as a resistor with a direct copper connection inside, right?

That's what I understand to be the difference. The thinnest copper wire is still an efficient conductor at the right power level. A very small circuit designed to carry just a few milliamps does not have any real resistive value...no matter how small the wire, there is a level at which it is not impeding the flow of electrons. Right? So how could a damaged wire on a car (typically anywhere from 18-20ga down to 4 or so for a typical starter cable) automatically provide an out of spec resistance reading?

Like my above example with a standard 18ga wire having a majority of its strands broken...why would that register as high resistance? Would you expect the equivalent wire (I believe I was somewhere in the ballpark when I said 18ga minus 75% of its strands should be around 28ga), undamaged, to also have a high resistance reading?

 

[JCH13]

Every wire has a measurable resistance, 10awg is around 1mOhm/ft for example. A decent DMM can measure that resistance. Copper is efficient, yes, but have you ever seen a circuit board trace that has jogs in it or that seems excessively long? That's usually to increase the trace's resistance in a meaningful way.

For your example: 18awg is ~6mOhm/ft whereas 28awg is ~65mOhm/ft, more than 10x the resistance of 18awg, and quite measurable. Not to mention that if the insulation were punctured the copper would likely be oxidized, cracked, or both, which would further increase resistance in the wire.