Current flowing in neutral of a 3 phase wye circuit of a Herater in Circuit Breaker

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Hii
I have a question regarding Heater circiut of circuitbreaker control cubicle
The circuit is fed with 380/220V with having resistors of 800 Ohms in each leg.
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If circuit is healthy then the current we get inline B1 is 0.

Now my doubt is,
what happens if any of those resistors is failed or if any of those legs is opened.
what will be the current flowing in the line B1. is it same as the current in one of those healty legs i.e. 0.137 or will it be any other?

your suggestions with detailed explanations are appriciated.

Thank you very much
 

Besoeker

Senior Member
Location
UK
Odd drawing but assuming that the conductors marked 1, 3, and 5 are the three phases and B1 is the neutral, the current in B1 with one of the loads resistors open circuit would be the same magnitude as in each of the other legs.

Explanation:
The current flowing in the neutral, call it In, is the difference between the current in the other two legs. Ia and Ib which are equal in magnitude but different in phase.

There are a few ways to derive this.
Take Ia to be the reference phase. Magnitude = Ia. It has an in phase component only.
Ib is 120deg shifted from Ia and has two components.
The in phase component is Ia*cos(2*PI/3) or 0.5*Ia.
The quadrature component Ia*sin(2*PI/3) or sqrt(3)/2*Ia.

Thus the resulting components from taking the difference:
The in phase component is Ia-0.5*Ia or 0.5*Ia

The quadrature component is just that of Ib.
In^2 = (0.5*)^2*Ia^2 + (sqrt(3)/2)^2*Ia^2
In^2 = 0.25*Ia^2 + (3/4)*Ia^2
In^2 = 1*Ia^2
In = Ia
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
Karthik, welcome to the forum! :)

In simple words, as you know, there's no neutral current when the loads are balanced. Any current removed from one line results in an equal increase in the neutral.

So, the neutral current is equal to the current in the two intact lines (when they're equal.)
 

G._S._Ohm

Senior Member
Location
DC area
IMO, in the case of an open 800 ohm resistor, I think the current through each of the two resistors due to the line to neutral voltage is summed and passes into the neutral.

This is [220/800] amps [call it 0.28 A] at an angle of zero and 0.28 A at an angle of 120 degrees.

0.28 + j0 + [-0.14 + j0.24]
= 0.14 + j0.24
= 0.28 A at 60 degrees.



For a shorted 800 ohm resistor, perfectly conducting wires and zero source impedance the current in the neutral is infinite, by inspection.
 
Last edited:

Besoeker

Senior Member
Location
UK
IMO, in the case of an open 800 ohm resistor, I think the current through each of the two resistors due to the line to neutral voltage is summed and passes into the neutral.

This is [220/800] amps [call it 0.28 A] at an angle of zero and 0.28 A at an angle of 120 degrees.
There are two 800 ohm resistors in each leg which is where the 0.137A comes from.
 
Odd drawing but assuming that the conductors marked 1, 3, and 5 are the three phases and B1 is the neutral, the current in B1 with one of the loads resistors open circuit would be the same magnitude as in each of the other legs.

Explanation:
The current flowing in the neutral, call it In, is the difference between the current in the other two legs. Ia and Ib which are equal in magnitude but different in phase.

There are a few ways to derive this.
Take Ia to be the reference phase. Magnitude = Ia. It has an in phase component only.
Ib is 120deg shifted from Ia and has two components.
The in phase component is Ia*cos(2*PI/3) or 0.5*Ia.
The quadrature component Ia*sin(2*PI/3) or sqrt(3)/2*Ia.

Thus the resulting components from taking the difference:
The in phase component is Ia-0.5*Ia or 0.5*Ia

The quadrature component is just that of Ib.
In^2 = (0.5*)^2*Ia^2 + (sqrt(3)/2)^2*Ia^2
In^2 = 0.25*Ia^2 + (3/4)*Ia^2
In^2 = 1*Ia^2
In = Ia

Thank You very much for your detailed explination.
But is In= Ia+Ib or Ia-Ib ?
Because in the above derivation cos(2*Pi/3) we get -Ia/2 so if we consider difference, then it leads to In= Sqrt3 Ia and if we consider sum then the right answer Ia.

Ps: sorry for posting two thrds. As i am new to this forum, i was bit confused in posting thrds :)
 

Smart $

Esteemed Member
Location
Ohio
Thank You very much for your detailed explination.
But is In= Ia+Ib or Ia-Ib ?Because in the above derivation cos(2*Pi/3) we get -Ia/2 so if we consider difference, then it leads to In= Sqrt3 Ia and if we consider sum then the right answer Ia.

Ps: sorry for posting two thrds. As i am new to this forum, i was bit confused in posting thrds :)
IMO, it should be In= ?(Ia + Ib)...

KCL-1.gif


...but it depends on the convention you adopt.

Even so, In = Ia ? Ib would be very unconventional :roll:
 
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