Balanced Neutral

[grantcool]

i asked this question in code class and couldn't get an answer (dissapointing).
on a balanced 20 amp circuit, 2 equal 120 volt loads, the shared neutral carries no current. but if the loads are rated at 120 volts, and the current doesn't return on the neutral, does it return on the other phase? wouldn't that put 240 volts across each 120 volt load?

 

[dereckbc]

on a balanced 20 amp circuit, 2 equal 120 volt loads, the shared neutral carries no current. but if the loads are rated at 120 volts, and the current doesn't return on the neutral, does it return on the other phase? wouldn't that put 240 volts across each 120 volt load?

Yes it returns on the other phase but the voltage is 120 across each load. If the two loads are equal, meaning resistance, you have a simple series circuit and the voltage divides evenly across the load or 120 across each load.

For example let's say each load is 12 ohms. So you have 2 12 ohm loads in series which gives you a combined load of 24 ohms (12 + 12) With 240 volats applied you have 10 amps of current flow. 10 amps of current flowing through each of the 12 ohm loads equal 120 volts across each load.

 

[ray cyr]

Yes it returns on the other phase but the voltage is 120 across each load. If the two loads are equal, meaning resistance, you have a simple series circuit and the voltage divides evenly across the load or 120 across each load.

For example let's say each load is 12 ohms. So you have 2 12 ohm loads in series which gives you a combined load of 24 ohms (12 + 12) With 240 volats applied you have 10 amps of current flow. 10 amps of current flowing through each of the 12 ohm loads equal 120 volts across each load.

Not exactly...in your example if the loads are not equal then the voltage dropped across each load will not be equal. We know this isn't the case. The voltage dropped across each load will be the same regardless of the size of the load because the voltage from each line is in reference to neutral, not in reference from line 1 to line 2. The direction of current flow on the neutral will be determined by the size of the loads due to how current is flowing through the entire circuit, but the voltage across either line to neutral will be 120 volts.

 

[roger]

Not exactly...in your example if the loads are not equal then the voltage dropped across each load will not be equal. We know this isn't the case. The voltage dropped across each load will be the same regardless of the size of the load because the voltage from each line is in reference to neutral, not in reference from line 1 to line 2. The direction of current flow on the neutral will be determined by the size of the loads due to how current is flowing through the entire circuit, but the voltage across either line to neutral will be 120 volts.

Dereck's post is correct, the conversation is dealing with balanced loads which would mean the neutral conductor could be removed from the now 240 v series circuit with no effect on the operation of the loads.


Roger

 

[ray cyr]

Dereck's post is correct, the conversation is dealing with balanced loads which would mean the neutral conductor could be removed from the now 240 v series circuit with no effect on the operation of the loads.


Roger

Understood and agreed :smile: I simply thought that since the OP asked "wouldn't that put 240 volts across each 120 volt load?" it was important to clarify that would be the case only if the loads were equal.

 

[dereckbc]

Ray your point is taken and noted. My answer was based purely on theory which only exists on paper. I qualified my statement with; balanced or equal loads which is practically impossible in the real world. Under the paper world of theory, the neutral conductor is not needed.

Note to the OP:

A real world application to your question is an Open or High Impedance Neutral connection. I will let you think about this as it will be a real world application down the road when you get a service call with dim/bright lights, burned up appliances, etc.

Hint: You measure at the service voltages and let's say 100 volts on L1-N, and 140 Volts L2-N. What is wrong?

 

[dereckbc]

Grantcool, I don't want to alarm you but if your instructor could not answer this simple question, you need another instructor who knows electrical therory.

I hope your instructor did not answer because he/she wanted you to find the answer for yourself. If otherwise seek another instructor.

 

[realolman]

That is not my understanding of the situation at all.

The two currents are out of phase and you add them vectorally. At any instantaneous moment, the two out of phase currents cancel each other, which "adds up" to the currents being canceled



Please explain to me how I am wrong. ( Like I could stop you )

 

[roger]

Realolman, notice in the illustration below that equal loads cancel the neutral current, this would mean the Neutral conductor is not a part of the circuit and the loads are now in series.

You can also see this by operating the interactive outside switches in Ronalds program below.


http://home.comcast.net/~ronaldrc/wsb/ax.htm


Roger

 

[LarryFine]

wouldn't that put 240 volts across each 120 volt load?

No, it would put 240 across the total of both loads. If the loads are balanced in impedance, the midpoint voltage between the loads would match the neutral, i.e., be 0 volts, with or without that neutral connection.

However, as the load impedances differ, the shared neutral carries current as it forces the neutral point to remain at 0 volts. When it loses that battle, the result is the open-neutral syndrome: bright and dim lights.

 

[gar]

080915-2000 EST

Larry that is a good description and I am glad you included impedance.

Also note if you use the parallel equivalent circuit for each impedance to make it easy to see what happens, then there are two separate components of the current in the neutral. Now if you use two phase sensitive demodulators of the neutral current, one in phase with the voltage source, and the other with a 90 degree shift, then you can separate the resistive component from the reactive component.

This was the basis of a patent I got in the mid 60s for a linear unbalanced capacitance gage for liquid level controls. In this case I was only concerned with the reactive component, and this circuit was adequately insensitive to the shunt resistance in parallel with the capacitor. The sensor was typically a rod covered with Teflon.

.

 

[K8MHZ]

Why the talk about phase differential in a 120/240 volt single phase system?

 

[gar]

080915-2123 EST

K8MHZ:

Put a load on a voltage source that is not a pure resistance and you can separate the current into two components, resistive and reactive. The resistive component times the source voltage is the power being consumed by the load. This in phase current component and the applied voltage allow you to calculate the equivalent shunt resistance of the load.

An inductor in series with a resistance can be replaced by a pure inductor in parallel with a shunt resistance. Here it is easy to see what the resistive current component is and the inductive component.

Now suppose you have two impedances with equal equivalent shunt resistances, but different reactances, then in the center tapped transformer circuit the neutral current contains no resistive component, but has a reactive component.

.

 

[grantcool]

I still don't get how that 20 amps from phase A & phase B get back to the source if they don't travel on the neutral. If it travels on the opposite phase, it's taking 240 volts and the 120 volt load burns up, right?

 

[Minuteman]

I still don't get how that 20 amps from phase A & phase B get back to the source if they don't travel on the neutral. If it travels on the opposite phase, it's taking 240 volts and the 120 volt load burns up, right?

Why not go out and try it. I'm not trying to be rude, but try it. Take 3 conductors and connect two equal loads. See if something smokes.

 

[Strahan]

I still don't get how that 20 amps from phase A & phase B get back to the source if they don't travel on the neutral. If it travels on the opposite phase, it's taking 240 volts and the 120 volt load burns up, right?

The key to this is "balanced load" = balanced impedances. At first I was where you're at now after making some basic drawings and putting ohm's law to work it's proven. Now you wouldn't want to go and try it with un-balanced loads.

 

[LarryFine]

I still don't get how that 20 amps from phase A & phase B get back to the source if they don't travel on the neutral. If it travels on the opposite phase, it's taking 240 volts and the 120 volt load burns up, right?

First read this: http://forums.mikeholt.com/showpost.php?p=708650&postcount=4

Then I'll tell you more that you'll understand.

 

[roger]

If it travels on the opposite phase, it's taking 240 volts and the 120 volt load burns up, right?

No, if the loads are balanced all you have is a 240 v series circuit supplying two loads, each load will drop 120 v for a total of 240. Current would return to the source the same as it would on any leg to leg circuit.

Play with the switches in the link I provided in post #9 to see how this would work when the center switch is open and both of the other switches are closed making believe the current flow is flip/flopping 60 times a second

Larry's link is a writen narrative of this same example.

Roger

 

[Minuteman]

The key to this is "balanced load" = balanced impedances. At first I was where you're at now after making some basic drawings and putting ohm's law to work it's proven. Now you wouldn't want to go and try it with un-balanced loads.

Why not? With an unbalanced load, the common neutral will just carry the imbalanced amperage. Typical MWBC.

 

[dereckbc]

080915-2000 EST

Larry that is a good description and I am glad you included impedance.

Also note if you use the parallel equivalent circuit for each impedance to make it easy to see what happens, then there are two separate components of the current in the neutral. Now if you use two phase sensitive demodulators of the neutral current, one in phase with the voltage source, and the other with a 90 degree shift, then you can separate the resistive component from the reactive component.

This was the basis of a patent I got in the mid 60s for a linear unbalanced capacitance gage for liquid level controls. In this case I was only concerned with the reactive component, and this circuit was adequately insensitive to the shunt resistance in parallel with the capacitor. The sensor was typically a rod covered with Teflon..

Gar I don't think that is what the OP was asking. I am under the impression he is just asking a simple theory question using balanced resistive loads.