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Old 10-12-2007, 07:06 AM   #1
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Default 120/240 single phase neutral

Purely a thought experiment to help see if I understand.

120/240 single phase
Each leg has the exact same amps of incandescent lights burning, and nothing else.

Is the only reason half the amps are returning to the transformer through the service neutral because the neutral is an equally attractive path as the phase conductors?

If the neutral was lifted under the above conditions the incandescent lights would keep burning like normal?

Trying to think this through in a way that is easy to wrap my un-formidable brain around

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Old 10-12-2007, 07:50 AM   #2
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With the exact same amperage on each leg the neutral is not necessary, there will be no current on the neutral. Change that load and the voltage on one leg will rise (Without the Neutral) if the current change is significant and the voltage exceeds the safe/permitted voltage of the utilization equipment SMOKE occurs.

Formula for neutral current in a 240/120 VAC distribution L1I-L2I=NI

Basically with balance loads the loads are in series on a 240 VAC system, the neutral need not be in the equation (IN THEROY)

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Old 10-12-2007, 05:33 PM   #3
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Quote:
If the neutral was lifted under the above conditions the incandescent lights would keep burning like normal?
They would burn very bright for a very short time @240V
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Old 10-12-2007, 05:56 PM   #4
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They would burn very bright for a very short time @240V
Disagree. If the load is exactly the same on each leg, you would still have 120v each side.

When the legs are unbalanced is when the voltages get screwy.
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Old 10-12-2007, 06:56 PM   #5
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Disagree. If the load is exactly the same on each leg, you would still have 120v each side.

When the legs are unbalanced is when the voltages get screwy.
Agreed.

If the current was balanced on each leg the voltage would still be 120 volts at the lights.

In a perfectly balanced system there would be 0 current on the neutral, therefore removing the neutral would have no effect.

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Old 10-12-2007, 07:11 PM   #6
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They would burn very bright for a very short time @240V
Draw it out and apply Ohms Law. 240 volts applied on two identical (Ohms) loads in series, the two loads will each have 120 volts across them.
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Old 10-21-2007, 08:41 AM   #7
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The lights should no longer work if you were to "lift" the neutral. Also the neutral has a better "attractiveness" because it eventually goes to ground, which of course is where all electricity wants to go.
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Old 10-21-2007, 09:07 AM   #8
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Other than safety GROUND has nothing to do with anything. you can operate any voltage distribution system without a ground. "Electricity" wants to return to it's source and mother earth has nothing to do with it.


As noted several times above if you have 240/120 distribution L1 has a 150 watt bulb, L2 has a 150 watt bulb (1.25 amps per leg), the neutral will have ZERO "0" amps and can be lifted the bulbs will continue to operate they operate in series, lose one bulb and the other will go out.
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Old 10-21-2007, 10:05 AM   #9
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The lights should no longer work if you were to "lift" the neutral. Also the neutral has a better "attractiveness" because it eventually goes to ground, which of course is where all electricity wants to go.

Could you explain "attractiveness"? In a residential, single-phase transformer, there are three connections. None of them 'attract' any electricity. Electricity is powerful, yes, but it is also very dumb. Electrons cannot flow through a load and think 'Hey, this is a 120 volt load, I need to find a white wire!' Electricity will simply flow where it can.
Two equal loads, rated 120 volts each, will operate perfectly when in series on a 240v circuit. A neutral, by definition, carries only the unbalanced portion of the circuit. If the loads are balanced (ie, equal in resistance), there is no current on the neutral, so it doesn't matter if it's there or not. There are many loads out there that do not use or even need a neutral. Your air conditioner compressor sitting outside is a prime example. 240V baseboard heaters, 480volt motors & lighting....

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Old 10-21-2007, 11:33 AM   #10
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I was simply using the terms the original message had. "an equally attractive path"
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Old 10-21-2007, 12:27 PM   #11
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The key word is "series". In series circuit the path is not broken until a bulb/load goes out or a wire is disconnected. Add another equal load to the drawing above and you have half the voltage across each load.

Nothing electrical will work without two wires. In todays world most everything is wired in parrallel except special situations like controls. Drop a neutral on a receptacle it/they will not work.
That is why I never connect receptacle to receptacle. Pig tails at every location.
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Old 10-31-2007, 02:37 PM   #12
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In parallel circuits the voltage is constant and the current varies. In a series circuit, the current is constant and the voltage varies. If you lift or disconnect the neutral on two lamps (as described above) the lamps change configuration and become wired in series, but will continue to light. Depending on what the actual wattages are, their brightness may vary somewhat (if they are unequal).

Do the math = Lets say that you were able to find 120 Watt bulbs and you wired them across a 120 - 240 Volt 3 wire system. Each circuit (bulb) would draw 1 amp [I=P/E] 120 watts divided by 120 volts = 1 amp
The bulb is rated at 120 Watts because P is = EXI or 120v x 1 amp = 120 Watts, also Ohms is equal to E / I or 120V / 1 amp = 120 Ohms resistance ( not addressing any inductive component)

In this example, the two bulbs are in parallel across each side of the 120-240 v power source. If you lift the neutral, then the bulbs are wired in series. The resultant total resistance of the bulbs combine [r1 + r2 = Rt]to equal 120 ohms + 120 ohms = 240 Ohms.

I = E / R or 240 volts / 240 Ohms = 1 amp

In series, current is constant. therefore at the first bulb - The voltage drop across the bulb = E = I x R or 1 amp x 120 ohms = 120 volts and Watts =
I X E or 1 amp x 120 Volts or 120 Watts (same brightness as before) and at the second bulb - the remainder of the source voltage [120 volts] is dropped across that bulb. or 120 volts x 1 amp = 120 watts (same brightness).

However, change one of the two bulbs to a 60 watt instead of the 120 watt and the remaining 120 watt bulb would try to run at 159.6 watts and the 60 watt bulb would try to run at 79.8 watts and would probably be short lived.
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Old 10-31-2007, 02:57 PM   #13
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If you have PowerPoint:

http://code-elec.com/userimages/Lost Neutral.ppt
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Old 10-31-2007, 03:16 PM   #14
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If you have PowerPoint:

http://code-elec.com/userimages/Lost Neutral.ppt

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