sharing a neutral

[steelersman]

I am not sure I understand why esp. if they are the same phase.

BTW, I agree there is probably a loose neutral I just don't see the effect on the pot.

dennis they're not on the same leg or phase. read the post.

 

[ohm]

actually you are backwards. the higher load (watts) item will have higher voltage with loose neutral. Look at the powerpoint thin that Bill provided and you'll see.

Wrong..the higher the load the lower the resistance..the lower the voltage across them.

 

[steelersman]

Wrong..the higher the load the lower the resistance..the lower the voltage across them.

wrong because it's now a series circuit and the current stays the same. look at the powerpoint presentation and learn.

 

[LarryFine]

actually you are backwards. the higher load (watts) item will have higher voltage with loose neutral. Look at the powerpoint thin that Bill provided and you'll see.

Sorry, but you're wrong. The higher wattage device will have a lower impedance. Ohm's Law tells us that, in a series circuit, the voltage drop across each load is directly proportional to the resistance.

 

[steelersman]

Sorry, but you're wrong. The higher wattage device will have a lower impedance. Ohm's Law tells us that, in a series circuit, the voltage drop across each load is directly proportional to the resistance.

then mr.bill's powerpointpresentation is wrong. check it out and tell me what you think

 

[LarryFine]

wrong because it's now a series circuit and the current stays the same. look at the powerpoint presentation and learn.

Sorry again, but if you're reporting the data correctly, the presentation is wrong.

 

[LarryFine]

then mr.bill's powerpointpresentation is wrong. check it out and tell me what you think

Okay, I did. Except for using a W for ohms, the presentation is correct.

It is you who has it wrong.

 

[steelersman]

Sorry again, but if you're reporting the data correctly, the presentation is wrong.

so is the presentation wrong or not?

 

[steelersman]

Wrong..the higher the load the lower the resistance..the lower the voltage across them.

the resistance doesn't change just because they have now become in series.

 

[steelersman]

Okay, I did. Except for using a W for ohms, the presentation is correct.

It is you who has it wrong.

tell me where I'm wrong then please.

 

[Dennis Alwon]

If the're the same phase it's not a loose neutral. They would both feel the same high voltage ...if they were both turned on.

I took this statement to mean they were on the same phase.

Only the two coffee pots (in the same office) have 'burned up''. there has been no effect on the computers/printers/shredders/phones.

Now I assume he means the computers etc are on another phase.

The reason I thought there might be a loose neutral because of the VD on the line. Now I see they are on opposite phases and it makes total sense

 

[steelersman]

Wrong..the higher the load the lower the resistance..the lower the voltage across them.

look at it this way: the one with the higher resistance will 1) have a higher voltage drop across it because the current is the same for both since they're in series 2) will then have a higher wattage since P=E I.
So the one with more watts now since they're in series will also be the one with the higher voltage and the higher resistance.

 

[wptski]

Mr. Bill I can't recall where I heard the 80 X number but I found something in the 14th edition of "Electrical Wiring..Residential by Ray C. Mullin pg 134:
"A tungsten filament lamp draws very high momentary inrush current at the instant the circuit is energized. This is because the cold resistance of tungsten is very low. For instance the cold resistance of a typical 100W lamp is aprox. 9.5 ohms. This same lamp has a a hot resistance of 144 ohms" It says the normal operating current is 0.83A but the max. inrush could be 17.9A.

It drops to normal current in 6 cycles (0.10 sec). Therefore "T" rated switches are a good idea on loaded incandescent circuits and Ohms Law (hope my namesake doesn't hear this) could mislead you on impedance testing.

Sorry about the 80 X, I'll keep looking.

Here's a inrush capture of a typical 100W incan bulb. Each division is 100ms, so how long is that 10.9A period which is in peak not RMS lasting?

 

[ohm]

Here's a inrush capture of a typical 100W incan bulb. Each division is 100ms, so how long is that 10.9A period which is in peak not RMS lasting?

I'm impressed! Looks like one click (100ms) to me. I was just going by the text and memory but the point I was trying to make is you can't take a resistance measurement and expect Ohm's Law to help you map the circuit.

 

[LarryFine]

so is the presentation wrong or not?

No.

the resistance doesn't change just because they have now become in series.

Correct.

tell me where I'm wrong then please.

You read it wrong. It correctly shows the higher-resistance load having a greater voltage across it.

 

[wptski]

I'm impressed! Looks like one click (100ms) to me. I was just going by the text and memory but the point I was trying to make is you can't take a resistance measurement and expect Ohm's Law to help you map the circuit.

Nope, the whole screen is 1s, major divisions are 100ms and minor would be 20ms.

 

[LarryFine]

look at it this way: the one with the higher resistance will 1) have a higher voltage drop across it because the current is the same for both since they're in series . . .

That is absolutely correct.

. . . 2) will then have a higher wattage since P=E I.
So the one with more watts now since they're in series will also be the one with the higher voltage and the higher resistance.

Oh, now that you add "now" to that explanation, okay. But that load did start out as the lower-power device.

Yes, the device rated at a higher power will now be the one with the lower voltage across it.

That also explains why the higher power forced through the device orginally rated lower will let the magic smoke escape.

 

[LarryFine]

the higher load (watts) item will have higher voltage with loose neutral.

This is a very misleading statement. The "higher load (watts) item" sounds like the greater load, having the power impedance.

You should have said "With a loose neutral, the device now receiving the resultant higher voltage will absorb more power."

 

[steelersman]

This is a very misleading statement. The "higher load (watts) item" sounds like the greater load, having the power impedance.

You should have said "With a loose neutral, the device now receiving the resultant higher voltage will absorb more power."

Well I admit I didn't word it as well as you. But I still was right the whole time. I can't help it if someone misunderstands me. Anyway this is another one of those refresher things that under normal circumstances I never deal with or think about. It is fun and interesting for me to get this little refresher problem. Thanks for the powerpoint MRBILL.

 

[76nemo]

Nothing else????

Nothing else????

I am still going with the coffee pots. If you have an impedance tester, you should be able to see the circuits layout even behind the walls.

There was a thread from the day before yesterday on a breaker tripping when the phone rang. People kept writing in with suggestions after the question was answered. Thing here is, the problem hasn't been solved.

I have several questions on this. I was waiting for someone else to ask before me.

-Has there always been coffee pots at this location?
-If so, are the 2 that fried a different type?
-How big and what type of pots are these?
-How many shifts are run there? Are these pots electronically controlled?
-Are there other things plugged into this recep, or just the one pot?
-I apologize, I can't see a pot frying with a supply of, what did you say, 112V's?
-What about make and model so we can see specs?

I am willing to bet if you open up these pots, you'll find the TCO fried.


Always curious,...........ohhhhh the suspense!:grin: :grin:


More questions after your response to come:wink: