lighting problem

[gar]

091228-1602 EST

Bobhook149:

Any power source has an internal impedance. Suppose we have an idea voltage source of 10 V (this means the voltage is invariant with load current) and to this we add an internal series impedance of 1 ohm before the terminals of this power source.

Put a short circuit load on this source and the load current is 10 A and zero voltage across the load (the short) and the source terminals. Next remove the short and put a 9 ohm resistor across the terminals as a load. Now the current is 1 A in the load and the voltage across the load is 9 V.

Next consider the characteristics of a tungsten filament incandescent lamp. Lamp intensity is a function of applied voltage. To see a plot of intensity vs voltage visit my website http://beta-a2.com/EE-photos.html and scroll down to photo P9. Note: the intensity drop from 100% at 120 V is to about 15% at 60 V.

Resistance of this type of lamp varies as you change voltage, but for the following illustration assume the resistance is constant. In the above circuit with the 10 V source with a 1 ohm internal impedance place a lamp of 9 ohms at 9 V as the load. Classify this as 100 % intensity. In parallel with the lamp place another load of resistance of 7.199999 ohms. The equivalent resistance of 9 ohms in parallel with 7.19999 is 4 ohms. The load voltage is now 10*4/5 = 8 V. This change produces about 25% change in light intensity.

You can run an experiment with a 15 W bulb and a 1500 W heater. The 15 W is so it is not too bright to look at. Pick any outlet some distance from the main panel. Plug the light in and turn it on. Set your heater in its high power position. Plug the heater in and the bulb will dim. Unplug the heater and the light brightens.

Next, find an outlet on the opposite phase. Doesn't matter where this outlet is relative to the main panel, or for that matter where the lamp is from the panel for this experiment. When you plug the heater in the lamp should slightly brighten. Can you see the change? I can not at my house because my source impedance is quite low.

If I plug the heater in at the main panel, about 12 A, on the same phase as the lamp I see about 0.3 V drop at the bulb. On the opposite phase I get 0 V change. This means most of the drop is a result of transformer internal impedance, very little from the neutral between the pole and main panel.

With the test lamp plugged into my work bench, and the heater plugged into the same outlet the change in voltage is 123.1 to 119.4 or 3.7 V. The light change is noticeable. Roughly the source impedance is 3.7/12 = 0.31 ohms.

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[gar]

091228-1712 EST

Bobhook149:

With a standard transformer supplying a standard tungsten filament lamp the the transformer input current will drop as the voltage is reduced. This is no different than if the transformer was not present, and you looked directly at the lamp current.

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[Bobhook149]

Gar,

Thank you i appreciate you taking time to explain in detail.

Bob

 

[gar]

091228-1735 EST

Bobhook149:

I sense that you need some good reference books to study, but I am too far removed from knowing what is out there that would be appropriate to provide any recommendations. Others can help you with suggestions.

Do experiments on your own, like the light bulb I mentioned above. Get two light bulbs and extension cords to do experiments at your father in law's and look at different circuits. Use a 1500 W heater or 1500 W hair dryer as a variable load, 120 V, (ON-OFF) load changer. Generally at 120 V I can see about a 1 to 2 V change with a lamp bulb.

If you do not have a good Fluke that can resolve 0.1 V at 120, then put that on your list of tools to get sometime.

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[gar]

091230-0857 EST

Following is a PM question I received:

With a mwbc if one heavy load kicks on say a toaster or microwave, the other phase will have an increase in voltage? And what you are saying is this is created because of impedance the source (the transformer at the street)??? Is this impedance eddy currents in the transformer?

I will start backwards on this question.

Eddy current losses in a transformer are heat and therefore appear as an equivalent shunt resistance across the primary in an equivalent circuit. These are also relatively independent of load current. Therefore, do not affect the output voltage.

Transformer impedance is primarily a result of three items.

1. Leakage flux. Not all flux from the primary links with the secondary. This you could easily see if you take air core coils and space them some distance apart. Quite obviously not all of the primary flux lines from the primary link to the secondary. Adding an iron core and making the primary and secondary very close together makes a very large percentage of the primary flux link the secondary, but not all.

2. Primary resistance and series inductance.

3. Secondary resistance and its series inductance.

If there are two secondaries and only one is loaded, then the transformer secondary impedance for that one side is the primary impedance reflected to the secondary plus the secondary impedance plus the effect of flux leakage.

Now suppose the voltage of the unloaded secondary is monitored while the loaded secondary is loaded and unloaded. This voltage will change in the same direction as the loaded secondary, but to a smaller extent. In this case only by the change in apparent primary voltage from the loaded secondary current reflected to the primary and the primary impedance.

What do I mean by reflected? If we have a 1 to 1 transformer, then a change of 1 A in the secondary produces a 1 A change in the primary. If the turns ratio is 10 to 1, then the 1 A secondary reflects to 0.1 A in the primary. Note: impedances reflect as the square of the turns ratio.


Now to the MWBC part of the question. It is not the transformer impedance that causes the one phase to increase when the other decreases, but rather the neutral impedance.

Create a very simple DC circuit like Larry Fine has illustrated somewhere. Two 1.5 V cells in series so as to add to 3 V. Connect a 1 ohm resistor from the midpoint of the cells. This is your neutral wire impedance. Using a high impedance voltmeter measure the voltage from the output end of the 1 ohm to each of the hot ends from the voltage sources. One reads +1.5 V, the other -1.5 V.

Next put a 14 ohm resistor from the +1.5 V to the neutral output. The load resistance on the +1.5 V cell is 15 ohms, and the current thru the load is 1.5/15 = 100 MA. The voltage across the load is 1.4 V and 0.1 V across the neutral.

What is the voltage from the neutral output to the -1.5 V cell? It is 1.5+0.1 V.

Change the circuit a little more. Put a 1 ohm internal resistance in each cell. Make the single load resistor 13 ohms. This is so I still have a load current of 100 MA. What is the voltage drop across the load? 1.3 V. What is the voltage from the neutral output to the -1.5 V cell? It is 1.5+0.1 V.

Does this explanation help any of you? If not, what questions do you have?

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[Bobhook149]

So if a large load were to kick on phase a of a mwbc phase b will have voltage fluctuations. So now that i have tightened the neutral in the panel, that may not have been the problem. Could it just be the fact that it is on a mwbc?? I'm going to try the experiment.. i get it better when i physically do something. Thanks alot for your continued help

 

[gar]

100108-2138 EST

Bobhook149:

You will get a voltage change. The magnitude of the voltage change is a function of the current thru and the resistance of the path over which the voltage is measured.

I assume your loose connection increased the resistance on the neutral path. Thus, after tightening the connection the voltage change from the load change should be less.

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[Limitless101]

Insects

Insects

They get in to warm places and cause ground fault search for an outlet that is seldom used in a dark location plug in you tester but don't blow your self up

 

[RETRAINDAILY]

Had some tools on me today, opened the panel up. Neutrals are doubled up ... insulation under screws.. just a mess. Fixed that problem. Made sure the service neutral was tight in the main panel. Hopefully that was the problem.. all other connections are tight. I put my voltage tester on the main lugs coming into the panel and read 243v L to L When the dryer was put on it went to 239v and then the range it went down to 237v and stayed there. Is this normal or still an indication of a loose neutral? I have not checked the neutral passed the main panel yet.

Thanks

Bob

when you did that did you turn off the main breaker and isolate your neutral from you're ground

 

[Bobhook149]

could the voltage problem also originate at a loose hot?

 

[gar]

100113-1927 EST

A loose hot is not going to cause excessive high voltage. Rather it will cause low voltage.

Arcing at a loose hot might cause some large short time high voltages from LC circuit ringing. But not a lot of sustained energy.

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[RETRAINDAILY]

if you have 120v at your fixture and 240v at your service it sounds like you have both ungrounded conductors.

 

[RETRAINDAILY]

this is the only fixture that has a problem no other lights dim or brighten when you turn the oven on?

 

[TT009]

I dont know if it has come up yet, but the way i understand it, if a MWBC has a loose neutral with two 120V circuits and the "feed" neutral becomes disconnected, 240V would go through all loads connected. Depending on what was connected it, may become a series circuit.

If the transformer is the smaller load compared to the other load(s) it will have the same current (which may be higher than rated) but will have less votage applied.

This may cause the burned out lamps, and over time burnt out transformer.