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Old 12-14-2008, 08:21 AM   #16
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Lights dim


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Originally Posted by jamiedolan View Post
One lamp has a 40wbulb and one has a 100w bulb. Once the service neutral is gone the 100wbulb acts as a resistor, and the 40w bulb gets much brighter than the 100w bulb was. The 100w bulb gets very dim. They burn for a while like this in my "lab" setup. (not sure how long, I only left it on for a few minutes at a time, but my test bulbs never burnt out.). I have not taken voltage readings off of the bulbs during this test, but you can make some educated guesses as to what the readings might looks like and what that would do to your electronics.

Jamie
Close, they both are resistors, not just the 100w.

My educated guess would be 171 volts across the 40w and 68 volts across the 100w.
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Old 12-14-2008, 08:37 AM   #17
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Lights dim


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Close, they both are resistors, not just the 100w.

My educated guess would be 171 volts across the 40w and 68 volts across the 100w.
Actually it is just the opposite. here how I figure it.
140 watts across 240 volts = .58 amps
.58 amp across 40 watts = 40/.58 = 68.9 volts
.58 amp across 100 watts - 100/.58 = 171.4 volts

Of course since filament resistance varies as a light heats up the actual number will be slightly different.
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Old 12-14-2008, 09:05 AM   #18
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Lights dim


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Actually it is just the opposite. here how I figure it.
140 watts across 240 volts = .58 amps
.58 amp across 40 watts = 40/.58 = 68.9 volts
.58 amp across 100 watts - 100/.58 = 171.4 volts

Of course since filament resistance varies as a light heats up the actual number will be slightly different.

No, you can't figure it that way. If you try using the bulb wattages after the neutral is broken it won't work cause the wattages will change. You need to first find the resistance of each filament, then the current in the series circuit using the R total, and lastly you can use the total I to find how the voltage will divide between the lamps.

Try it this way, if you can't get it I'll show you the math.

Last edited by Silk; 12-14-2008 at 09:59 AM.
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Old 12-14-2008, 10:27 AM   #19
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Lights dim


Unplug everything that contains electronics or ballasts or might contain electronics or ballasts. This includes practically everything other than incandescent lights and portable electric heaters with primitive knob controls.

Does this brightening occur everywhere in the house. If only parts of the house are affected you can start investigating and checking outlet boxes for loose connections yourself (if you wish).

Some of us with more DIY electrical experience are daring enough to check the breaker panel for loose connections.
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Old 12-14-2008, 10:57 AM   #20
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Lights dim


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Originally Posted by 220/221 View Post
Worst case, the neutral cionnection goes completely out and you end up with 240 volts at all of your 120 volt receps and lights. That is when it gets expensive.
Worst case, a 10A, 120v load (12Ω, let's say a toaster) on one side of the neutral,
and a 4w analog clock (3600Ω) on the other side,
gives 239v to the clock and 1v to the toaster.

It's all good news: you use the clock to mask other odors,
and the voltage on the toaster side of the neutral is very safe.

There'd be anywhere from ≈0.1v up to 119.9v across a bad neutral connection. The midrange neutral connection voltages would make for a very hot connection, so if you happen to have one of these $60 IR meters you can find the hot connection inside your panel without touching any conductors.

Removing a panel cover presents a shock danger
and a danger due to
http://en.wikipedia.org/wiki/Arc_Flash

Last edited by Yoyizit; 12-14-2008 at 11:26 AM.
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Old 12-14-2008, 11:07 AM   #21
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Lights dim


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the voltage on the toaster side of the neutral is very safe.
Right up until the moment that the bulb burns out.
Jamie
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Old 12-14-2008, 11:43 AM   #22
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Right up until the moment that the bulb burns out.
Jamie

Not so.

If the bulb burns out, the circuit will be open and there will be zero current flow.
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Old 12-14-2008, 11:48 AM   #23
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Lights dim


The bulb or clock burns out and the toaster is then in series with 120v and a bad neutral connection.

With 1kw into the house, a 75Ω neutral connection should give maximum heating in the connection, ~50w. 50w dissipated in heavy, thermally-conductive busbars is probably warm or hot to the touch.

I never realized how complicated a simple two-current-loop circuit could be.

Last edited by Yoyizit; 12-14-2008 at 11:52 AM.
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Old 12-14-2008, 12:02 PM   #24
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Lights dim


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The bulb or clock burns out and the toaster is then in series with 120v and a bad neutral connection.

With 1kw into the house, a 75Ω neutral connection should give maximum heating in the connection, ~50w. 50w dissipated in heavy, thermally-conductive busbars is probably warm or hot to the touch.

I never realized how complicated a simple two-current-loop circuit could be.

It's not complicated
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Old 12-14-2008, 12:06 PM   #25
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Lights dim


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Not so.

If the bulb burns out, the circuit will be open and there will be zero current flow.
Ok. Yes, only if that bulb was the last decive / bulb on that side of the panel and it burns, then there is no flow (but this isn't likely to be the only thing left in most homes).
If any other load like a larger bulb is present on that side of the panel, then the toaster on the other leg of the panel begins to TOAST. Correct?

Jamie
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Old 12-14-2008, 12:16 PM   #26
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Lights dim


correct
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Old 12-14-2008, 12:18 PM   #27
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Lights dim


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Originally Posted by jamiedolan View Post
Ok. Yes, only if that bulb was the last decive / bulb on that side of the panel and it burns, then there is no flow (but this isn't likely to be the only thing left in most homes).
If any other load like a larger bulb is present on that side of the panel, then the toaster on the other leg of the panel begins to TOAST. Correct?

Jamie
According to the original setup, it was a toaster and wall clock or bulb. So, there is no other bulb in the circuit, by definition.

A circuit must be complete. No circuit = no toast. A circuit is a circuit, intentional or not. There is not, nor can there ever be, a current flow on an incomplete circuit. So, if anything electrical works, even a little bit and very dim, you KNOW there is a complete path back to somewhere. The laws of physics still apply, even in your kitchen.
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Old 12-14-2008, 12:31 PM   #28
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Lights dim


Sorry to make this even more complex, but incandescent light bulbs have non-linear resistance curves. 120v incandescent is nearly a dead short at 10volts, but at 120v they are the resistance you expect for their wattage.

Sorry.
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Old 12-14-2008, 12:33 PM   #29
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Sorry to make this even more complex, but incandescent light bulbs have non-linear resistance curves. 120v incandescent is nearly a dead short at 10volts, but at 120v they are the resistance you expect for their wattage.

Sorry.

Don't be, you just made it more fun.
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Old 12-14-2008, 12:36 PM   #30
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Lights dim


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Originally Posted by InPhase277 View Post
According to the original setup, it was a toaster and wall clock or bulb. So, there is no other bulb in the circuit, by definition.

A circuit must be complete. No circuit = no toast. A circuit is a circuit, intentional or not. There is not, nor can there ever be, a current flow on an incomplete circuit. So, if anything electrical works, even a little bit and very dim, you KNOW there is a complete path back to somewhere. The laws of physics still apply, even in your kitchen.
Thanks, I follow. I was confused about the context. I thought we were speaking of the above items being in the home with the loose neutral. So I thought there were other items that would be completing he circuit.

But I understand now. Thanks for the clarification.
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