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Mobile connector flashes 2x - “ground loss”

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Please explain how a meter will correctly determine resistance on a circuit if a portion of said circuit has a voltage offset.

The meter won’t be correct, it isn’t accurate enough to be correct. It will read zero ohms, even if there is in actually A few hundred milliohms of resistance. With the power off, the meter will also show zero ohms. Result of the test will not change.


As you state above, a shared neutral will have a voltage drop of some level:

Or is it:

?
One is the max it can be per nec, but nec is the minimum, the other is is what it should be ideally.
 
The meter won’t be correct, it isn’t accurate enough to be correct. It will read zero ohms, even if there is in actually A few hundred milliohms of resistance. Result of the test will not change.



One is the max it can be, the other is is what it should be
One step at a time then. How much voltage offset will you accept as reasonable on the shared subpanel neutral feed? Is 200 mV acceptable?
 
At that point you should in increase the wire size for the circuit. You should keep voltage drop to less than 3 percent.
The 3% guideline is typically used for the branch circuit only. With a 2% allowance for a feeder and a 5% total allowance. [Unless I've reversed the 2% / 3%, which is possible.]

At 120v to ground that is 3.6v since the meter will not be sensitive enough to detect the difference.
Not an expert on the specs of a "typical" DMM, but I would certainly expect one to be able to distinguish between 0 VAC and 3 or 6 VAC.

I just took my $40 DMM and measured the voltage N to G at a bathroom receptacle. It read 20 mV. Then I turned on a hair dryer (nominal 1800W) plugged into that receptacle, and the voltage N to G jumped to 920 mV. I'm dropping the last figure because I doubt the DMM has true mV accuracy. But that 900 mV difference is real and measurable. Probably with a $10 DMM, too, although I don't have one handy.

Again, none of this will affect the test that was conducted.
Based on how the DMM measures resistance, an outside source inducing a voltage difference between the leads would on the face of it seem to interfere. So again, unless you can explain how the details of how the DMM is making the voltage measurement and will filter out the AC voltage offset without damage to it, I don't believe you're in a position to say that.

That is not why motors have a 230v name plate. The reason they have a 230v nameplate is so they can be used in 208 and 240v.
That is absolutely why motors say 230V instead of 240V. A motor that just states 230V is only for use on a 240V system, not a 208V system.

If it's for use on a 208V system, it will say 208-230V. And that's a bit of a compromise, as it's really optimized for some intermediate voltag and often won't work well on a 208V supply that's at 5% below nominal. Whereas a motor that is designed for a 208V supply will say 200V on its nameplate and should work fine on 208V - 5% = 198V.

Cheers, Wayne
 
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As I already stated, for the test the OP was doing, whether power was on or off did not affect the result.
"Did not" or "would not" or "could not"? OP only made one measurement.

My issue is with these claims :
The potential between neutral and ground should be zero. From the meters point of view the circuit isn’t live. The fact that the neutral is carrying current doesn’t matter. If carrying current affects the reading, there is a problem. High resistance path for neutral to source or from egc bond to neutral.

For there to be potential(voltage that would affect the reading) between neutral and ground there would need to be resistance somewhere, which would also show up when you read the meter. It should not affect the reading, especially if the reading is zero resistance.
We've already established the potential would likely not be zero. So what figure shall we use? How about we cut your ideal voltage drop of 200mV in half to 100mV.
I'll continue, jump in where you disagree:
How does the meter measure resistance?
Most apply a current (I) and read the resultant voltage (V).
What calculation would the meter then do?
R=V/I
What will the meter report if there is an additional offset (A)?
R=(V+A)/I
Like I said, I'd be warry.
 
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To spell out mongo's last post, say the meter injects 0.2 mA DC as suggested, and you are measuring the resistance of a 400 ohm resistor. It will see (measure) 80 mV DC of voltage, do the math, and read 400 ohms on the display.

Now suppose you take 10 of those resistors in series and hook them up to a 1.5V alkaline battery. Each resistor should drop 150 mV across it (assuming negligible internal battery resistance, not sure if that's accurate here), and the battery current would be 0.4 mA. Now try to measure the resistance of one resistor. The meter will see either 230 mV or 70 mV of DC voltage, depending on the polarity of the battery vs the meter, so it should read either 1150 ohms or 350 ohms, respectively. [Actually for the case of 70 mV DC voltage, it could tell the voltage polarity is wrong, and conceivably throw an error. Not sure.]

Now for the case of the OP, it's possible the saving grace here is that the external source is AC, not DC. So if the AC grid voltage has no DC offset of its own, and if the multimeter is sampling the voltage fast enough and then averaging those samples over a long enough time period , that process will cancel out the AC voltage. So the meter will just see the DC offset resulting from its injected current, and maybe it would work.

But regardless, as the manual instructs, just turn the power off before making any resistance measurements.

Cheers, Wayne
 
To spell out mongo's last post, say the meter injects 0.2 mA DC as suggested, and you are measuring the resistance of a 400 ohm resistor. It will see (measure) 80 mV DC of voltage, do the math, and read 400 ohms on the display.

We aren’t measuring a 400 ohm resistor though.

Now suppose you take 10 of those resistors in series and hook them up to a 1.5V alkaline battery.

We aren’t doing that either.

just turn the power off before making any resistance measurements.

I never said you shouldn’t. I said in this case it won’t make a difference.
 
We aren’t measuring a 400 ohm resistor though.
Right, but I illustrated how the ohmmeter measurement process could clearly be disrupted by a concurrent applied voltage, DC in this example.

So if you are going to claim that in the OP's case it's not going to make a difference, it's going to take more than your simple assertion. I mean, you may be right, but you haven't given us any reason to expect that to be true.

If you can explain further why the DMM ohmmeter measurement won't be affected by a concurrent AC voltage, including what the DMM is doing internally, and how the AC voltage effects will average out to zero, and how close the grid's DC offset is to zero, then that would be a plausible argument.

Cheers, Wayne
 
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It did not. It would not.
So, it occurred to me that at the risk of perhaps damaging my $40 DMM, I could do the experiment easily enough. At that bathroom receptacle previously mentioned, I first measured the N-G resistance while unloaded. The DMM only reads to 0.1 ohms, and the result was fluctuating between 0.1 and 0.3 ohms, not sure why it fluctuated (contact resistance? effects of the N-G voltage difference?).

Then I plugged in the 1800W nominal hair dryer and took the same resistance measurement. It consistently read 0.0 ohms, no fluctuating. The only change in conditions was the concurrent AC load. So apparently the superimposed AC voltage signal caused whatever measuring and averaging the DMM was doing to end up swamping out the signal that was previously detectable.

Thus it is absolutely the case that measuring the resistance with the power on will affect the result. Of course, for the OP's case, the difference between 100 and 300 milliohms is probably not relevant. A more directly applicable test would have been to insert, say a 5 ohm resistor into the circuit, and see how the DMM is affected while measuring that resistance. But I didn't have one handy.

Cheers, Wayne

PS I know the branch circuit for the bathroom receptacle is otherwise unloaded, but I have a single ~30' long feeder from my service disconnect (which has the N-G bond) to my distribution panel for all my loads. So when measuring the voltage difference between N and G as in my previous discussion, I could see the effect of the house load changing. E.g. the background voltage difference was ~20 mV, but when the sump pump came on (it's stormy here), it jumped up to 60 mV.
 
Sorry, that is not correct.
You're just wrong here. Did you not see this chart that summarizes it nicely?

372CB66F-BAA5-4FAA-8C6A-9F92F33C15C0.png


Cheers, Wayne
 
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Don’t bury the lead…
That's not the lead. I didn't do the perfect experiment for the OP's original purposes, because I don't have a defective ground at the receptacle, but I did an experiment that disproves your assertion.

So unless you've got something substantive to add, rather than repeating the same misinformation without any substantiation, I think we're done here.

Cheers, Wayne
 
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That's not the lead. I didn't do the perfect experiment for the OP's original purposes, because I don't have a defective ground at the receptacle, but I did an experiment that disproves your assertion.

My assertion is that whether or not the power was on would not materially change the result of the test, eg, the grounding path is good. I don’t think you disproved that. In fact, you said the opposite.
 
Look at the NEMA column. Thanks!
From my understanding of the chart, the NEMA column shows the absolute limits a given nameplate motor would tolerate and still operate effectively. That means a 230V nameplate motor would tolerate 207-253V (at the motor). However, if you use 208V Nominal service with that motor, the Service column shows the utility can deliver down to 197.6V, which is already out of the range that motor allows. The Utilization column further accounts for voltage drop at the circuits of the end user, bringing the lower range down to 181V that the motor sees, that's a whole 26V lower than motor allows, or a further 10% reduction.

Instead when you use 240V Nominal service, the expected voltages at the end work out quite nicely for a 230V nameplate motor (lowest at 208.9V even with drops at the end wiring).
 
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My assertion is that whether or not the power was on would not materially change the result of the test
Ah, your assertion has changed. Originally it was that there would be no effect.

I agree that my test doesn't show a big effect, but given the limitations it couldn't. Your new assertion may still be true, but you have presented no support for it. And we have some theoretical and experimental reason to doubt it.

Let me know when you do the test I described with a DMM and a 5 ohm resistor (say) and a large load and what the results are.

Cheers, Wayne
 
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