#### Kam1

##### Member

In a 3-wire 220V circuit with a balanced load, what is the load of the neutral?

I usually picked 0V as my answer, but I think that is not correct? :happysad:

Some help in explaining this to me would be great. I got my exam coming up soon. Thanks.

#### Smart \$

##### Esteemed Member

In a 3-wire 220V circuit with a balanced load, what is the load of the neutral?

I usually picked 0V as my answer, but I think that is not correct? :happysad:

Some help in explaining this to me would be great. I got my exam coming up soon. Thanks.
It's a poorly worded question: 220V is not a standard voltage... 208V and 240V are; also a load does not exist on one wire (e.g. "load of the neutral").

Nevertheless, the desired answer is likely 0A (not 0V). The question leads me to assume this is a split single phase circuit, two line conductors, one neutral conductor, and states a balanced load... meaning no current on the neutral conductor.

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#### Besoeker

##### Senior Member
It's a poorly worded question: 220V is not a standard voltage...
Yes, if in fact that's actually how the question was worded.

#### fmtjfw

##### Senior Member
Yes, if in fact that's actually how the question was worded.
Since it is a balanced load, the voltage makes no difference, the neutral home run always carries 0A.

In providing the balance the non-home run parts of the neutral can carry anything up to the highest phase current.

Just being pedantic.

#### jumper

##### Senior Member
Since it is a balanced load, the voltage makes no difference, the neutral home run always carries 0A.

In providing the balance the non-home run parts of the neutral can carry anything up to the highest phase current.

Just being pedantic.
Er,,,, you might wanna check Besoeker"s profile before you lecture him. Senior Power EE with like 40 years experience.

#### Smart \$

##### Esteemed Member
Since it is a balanced load, the voltage makes no difference, the neutral home run always carries 0A.

In providing the balance the non-home run parts of the neutral can carry anything up to the highest phase current.

Just being pedantic.
Er,,,, you might wanna check Besoeker"s profile before you lecture him. Senior Power EE with like 40 years experience.
Additionally, since standard voltage was not used, it could be similar to a 120/208V 3-wire circuit, where the rms neutral current magnitude is equal to the either line's magnitude.

#### Besoeker

##### Senior Member
Er,,,, you might wanna check Besoeker"s profile before you lecture him. Senior Power EE with like 40 years experience.
Ta mate!
Actually, that's something the English might say and I'm not English.
So I wouldn't........

Back to the topic.......
My simple point was that the way the OP stated the question might be based on memory.

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#### Besoeker

##### Senior Member
In providing the balance the non-home run parts of the neutral can carry anything up to the highest phase current.
Just being pedantic, it could be more.

#### fmtjfw

##### Senior Member
Just being pedantic, it could be more.
Still trying to understand. If I have A = 20A, B=20A, and C=0A, I get sqrt(400+400+0-400-0-0) = 20

Am I missing something (phase angles or harmonics)?

#### Dennis Alwon

##### Moderator
Staff member
Still trying to understand. If I have A = 20A, B=20A, and C=0A, I get sqrt(400+400+0-400-0-0) = 20

Am I missing something (phase angles or harmonics)?
This is correct. Harmonic can add more but with linear loads this is correct

#### roger

##### Moderator
Staff member
Still trying to understand. If I have A = 20A, B=20A, and C=0A, I get sqrt(400+400+0-400-0-0) = 20

Am I missing something (phase angles or harmonics)?
Well, if the OP is using slang 220 but means 230 or 240 it is single phase and your formula is for a wye. If it is in fact a wye, your formula is correct and as Dennis points out, with odd order harmonics the neutral can have more current than the two ungrounded conductors.

Roger

#### Besoeker

##### Senior Member
Still trying to understand. If I have A = 20A, B=20A, and C=0A, I get sqrt(400+400+0-400-0-0) = 20

Am I missing something (phase angles or harmonics)?
Or making it more complicated?

I suppose that the original post referred to a three wire centre tapped system.
This sort of arrangement:

If Ra and Rb are equal resistive loads the the neutral current would be zero.
Like this:

I made the currents slightly different to show some neutral current.

But suppose the loads have different power factors with Ia leading and Ib lagging you could get something like this:

In is greater than Ia and Ib.

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Indeed.

#### jumper

##### Senior Member
Thought you might like that.

OT here and a bit political. Mods please forgive. I will drop it after this.

I will not ever recognize the British Union Jack until St. David of Wales cross is included.

I do not mean this as anything towards you Bes, it is directed at the English Crown and heraldry/design of the flag.

#### jumper

##### Senior Member
Thought you might like that.

OT here and a bit political. Mods please forgive. I will drop it after this.

I will not ever recognize the British Union Jack until St. David of Wales cross is included.

I do not mean this as anything towards you Bes, it is directed at the English Crown and heraldry/design of the flag.
James I should have never accepted the crown. Scotland was a sovereign nation. Wales, isolated and small, OTOH was over run and conquered by Edward I and his hordes, never really had a chance.

It is 3:00AM and I am going off on British history, probably badly at that, on a electrical code forum. I gotta go to bed.

History 101 is over.

#### mivey

##### Senior Member
But suppose the loads have different power factors with Ia leading and Ib lagging you could get something like this:
.
I also think this thread is about a single-phase load but I recall you made a similar point when that formula for an ideal unbalanced three-phase circuit showed up in a different thread. I seem to recall resistance from some who wanted an easy answer but as you pointed out, the easy answer is hardly ever applicable.

FWIW, the correct methods are not really that hard so why not just calculate it the correct way to start with?

#### Besoeker

##### Senior Member
I also think this thread is about a single-phase load but I recall you made a similar point when that formula for an ideal unbalanced three-phase circuit showed up in a different thread. I seem to recall resistance from some who wanted an easy answer but as you pointed out, the easy answer is hardly ever applicable.

FWIW, the correct methods are not really that hard so why not just calculate it the correct way to start with?
I think the real shame is that it gets taught in the first place.

#### iwire

##### Moderator
Staff member
I think the real shame is that it gets taught in the first place.
I am not sure why that is a shame.

They teach electricians what is expected of electricians, they teach engineers what is expected of them.