neutral load ?

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SmithBuilt

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I understand 120/240 systems having 180deg phases a 240 volt load would not need a neutral.

What about a 120/208 v system. The phases are 120deg off. How does a single phase 208 v load work or 2 120 volt loads from the 208 system. Do the phases which are not 180 deg off cancel each other out completely?

I'm now this is not a very good description. Feel free to clean it up.

Thanks,
Tim
 
SmithBuilt said:
What about a 120/208 v system. The phases are 120deg off. How does a single phase 208 v load work or 2 120 volt loads from the 208 system. Do the phases which are not 180 deg off cancel each other out completely?
Click to expand...

No, in fact, on a 208/120 system, on a two-pole multiwire branch circuit (two hots and a neutral), if there are 15 Amps on each leg there will be 15 Amps on the neutral. There is a portion that "cancels," but it is not 100%.
 
The neutral connected to two phases of a Wye system will carry the same or close to the same current as the heaviest loaded phase.

SQRT (IA^2) + (IB^2) - (IA + IB)

Roger
 
Phaors ad nauseum:

Phaors ad nauseum:

The magnitude and phase angles of these currents are described by complex numbers called "phasors".

In the case of 180 degree separation, the phasors "add" to zero. That is, they cancel each other.

In the case of 120 degree separation, the phasors add vectorially to half the value of their algebraic sum.

Read up on complex numbers, and you will begin to understand.
 
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Was the original question asked regarding 208 single phase loads, like a heater, and the fact that a neutral conductor is still not used even though the two ungrounded conductor currents do not cancel?
 
SmithBuilt said:
I understand 120/240 systems having 180deg phases a 240 volt load would not need a neutral.

What about a 120/208 v system. The phases are 120deg off. How does a single phase 208 v load work or 2 120 volt loads from the 208 system. Do the phases which are not 180 deg off cancel each other out completely?

I'm now this is not a very good description. Feel free to clean it up.

Thanks,
Tim
Click to expand...


Tim
It is not necessarily the "system" would not need the grounded[neutral] conductor, it would be 2 phase conductors may not require a neutral conductor. In the 240v system, the two phase conductors installed as a branch circuit would cancel each other out. Think of a 240v A/C installation.



For 3phase, 4-wire systems that are solidly grounded, see 310.15(B)(4)(b).
 
Thanks for the answer. It brings up another question.

Take Pierre's example. If the AC was a 208v single phase, no neutral. How do the two phases cancel?


Or even in a heater as Ron suggests?

Tim
 
If my memory serves me right, but always don't,

I alway believed that on a 3 phase wye using two phase legs you had to use the 1.73 factor.

In other words if you had a load of 10 amps on line #1 to neutral and a 10 amp load on Line#2 to neutral then your neutral load would have a 1.73 amp.
 
SmithBuilt said:
Thanks for the answer. It brings up another question.

Take Pierre's example. If the AC was a 208v single phase, no neutral. How do the two phases cancel?

Tim
Click to expand...

The phase currents are equal and are displaced from the phase voltages by +/- 60 degrres.


For example, let Va = 120V @ 60 and let Vb = 120V @ -60, then Vab = 208V @ 90. Ipa = Iload @ 90, and Ipb = Iload @ -90.
 
Here's a simple explanation of why the neutral of a 208/120v 3-wire MWBC carries the same as the two phase wires:

Picture a balanced 3-phase circuit with 20a on each phase. The neutral carries no current. Now, reduce the load on one phase by 5a. What happens? The neutral current increases by the same 5a.

Increase the decrease (sorry!) to 10a. Now, the neutral carries 10a. Keep going; at 5a on that one phase, the neutral current is 15a. Finally, drop the current to zero on that phase: 20a of neutral current.

In other words, whatever current is removed from a phase is imposed on the neutral in order to maintain the neutral at zero volts.
 
There are two situations being discussed here.

1 - Two 120 Volt loads connected up to a 208/120 multiwire branch circuit

2 - A single 208 Volt load connected up to two legs of a 208 Volt system

They are not the same.

In case one, there are different loads that are receiving voltages from two different time continuums. They don't add up (without complex mathematics) because they, well, don't add up. There is an imbalance that flows back to the source through the neutral. Now, disconnect the neutral and watch what happens. Now you have changed the voltage on each load from 120 to ... what???? If the loads were equal, than the voltage would go to 104 on each load. You now have a totally different system.

In case two, the load is simply a load that responds to the Voltage impressed upon it, namely 208 Volts. There are no currents to "cancel" because it is a single load on a single circuit with a common current flowing through it.
 
eric stromberg said:
There are two situations being discussed here.

1 - Two 120 Volt loads connected up to a 208/120 multiwire branch circuit

2 - A single 208 Volt load connected up to two legs of a 208 Volt system

They are not the same.

In case one, there are different loads that are receiving voltages from two different time continuums. They don't add up (without complex mathematics) because they, well, don't add up. There is an imbalance that flows back to the source through the neutral. Now, disconnect the neutral and watch what happens. Now you have changed the voltage on each load from 120 to ... what???? If the loads were equal, than the voltage would go to 104 on each load. You now have a totally different system.

In case two, the load is simply a load that responds to the Voltage impressed upon it, namely 208 Volts. There are no currents to "cancel" because it is a single load on a single circuit with a common current flowing through it.
Click to expand...

Eric, the math isn't all that complex.

Case one:

If we define all currents as leaving the neutral node and let the magnitude of the load currents be "Im", then:

|In| = Im x cos(60) + Im x cos(-60) = Im

Case two:

If we follow the same conventions as Jim D. preaches (and rightly so), then,

Ia and Ib are negatives of each other as they are in the single phase case and they cancel.

I agree though that it is easer to point out that there is no neutral involved in case two, therefore there is no neutral current.
 
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