# Calculating Symmetric Components in a Three Phase 4 Wire System

• CasterSkux
In summary, this student is trying to find out the symmetrical components of the current in a three-phase system with a phase-to-neutral voltage of 230 V when one phase has a blown fuse. The student first calculates the balanced current and then uses Kirchoff's law to find the neutral current.
CasterSkux
Hello fellow engineers!
I am a student doing a simple course in Electrical Engineering.

I've got an enquiry regarding this question

"
1. In a three phase 4 wire system with phase to neutral voltage of 230V, a balanced set of resistive loads of 8 ohms are connected between each phase and neutral.

If the fuse cartridge of phase “C” blows, what will be the currents in the corresponding phases and the symmetrical components of the load currents?
""

We use a program to calculate the symmetric components. I believe I need to construct the resultant phasor of this system and derive the symmetric components

This is my approach:

1) Assume this is a star connected load, therefore the phase voltage is Vp = Vl / sqrt(3) = 230/sqrt(3) = 132.7906 V

2) Calculate the balanced current that flows through each resistor I = V/R, I = 132.7906V/8 = 16.5988.

3) If Phase C is eliminated from the fuse, I think that the line/phase voltages stay the same (as there is a neutral) and hence the current is the same.

4) Therefore Phase A is 16.5988 amps and so is Phase B. Phase B is offsetted by 120 degrees taking A as the relative point.

5) Phase C is 0 and it's on the 240 degree mark on the phasor diagram.I was wondering if this is all right in approach?

CasterSkux said:
This is my approach:

1) Assume this is a star connected load, therefore the phase voltage is Vp = Vl / sqrt(3) = 230/sqrt(3) = 132.7906 V
Didn't you say phase to neutral voltage was 230 V?

CasterSkux
Yes, oh have I overlooked something?

You should show by using symmetrical components what is new voltage phasor between phase B and star neutral after phase C is gone. Looks like a typical homework problem .

CasterSkux
zoki85 said:
Didn't you say phase to neutral voltage was 230 V?
Ah I ,maybe somehow I've interpreted "phase" as the "line", so the current is simply 230/8 = 28.75 amps.

Indeed It's a typical homework problem, the problem is my course is an online one: they didn't go through the basics as through as I would've liked and we hardly do math problems, so I'm not as well practiced unfortunately.

Zoki85, do you mean the new Phase A?

I was under the impression I work backwards to find the symmetrical components, not the other way around?

CasterSkux said:
Ah I ,maybe somehow I've interpreted "phase" as the "line", so the current is simply 230/8 = 28.75 amps.

Indeed It's a typical homework problem, the problem is my course is an online one: they didn't go through the basics as through as I would've liked and we hardly do math problems, so I'm not as well practiced unfortunately.
230 V is a common phase voltage (phase to neutral) in LV networks in Europe, so I guess it isn't mistake. This is simple situation and can be quickly solved even without method of symmetrical components. Phase to phase voltage is 230×√3 ≈ 400 V. Since phase C is open, current flowing through phases A and B is the same current that flows through series resistance 8+8=16 Ω. Current has magnitude 400/16=25 A. Note that in a phase diagram currents in phase A and phase B have same magnitude but are in phase oposition (angle difference between phasors is 1800 ).

Cheers

Last edited:
CasterSkux
Hello Zoki85,

Thanks for the reply. I was wondering if we should've considered current flowing to the neutral? would the phase C being broken, cause the load to be unbalanced and because it's (I assume a star connection): would cause current to flow into the neutral?

Cheers,

Indeed, I forgot you were referring to the case with connection between neutral points of two stars. In that case you were right! Current through the phase A: IA= VA/R=230/8=28.75 A, and current through the phase B: IB=VB/R= 28.75 ∠1200 A.
By Kirchoff's law, current through the neutral is I0=IA + IB. That would give I0= 28.75 ∠60° A (if I didn't make mistake without pencil&paper). For excersise, find from that, for both cases, direct, inverse, and zero sequence components of currents.

CasterSkux
I started my calculation when nobody was on the web. Now some of the correspondents already answered [partially in my opinion].

230 V it is the voltage phase-to-neutral [your declaration].So phase-to-phase will be sqrt(3)*230=400 V [as usual].

The current in phase A [let's take it as origin of the angles] will be IA= 230/8=28.75<0;

IB=28.75<-120=28.75<240 and IC=0.Neutral current will be IN = - (IA+IB+IC) = -28.7*{[cos(0)+cos(240)]+j*sin(240)}

cos(0)=1; cos(240)=cos(pi()/180*240)=-0.5; sin(240)=-0.866

IN=-28.75*(0.5-0.866j)=-14.38+24.9j

CasterSkux
Hello Zoki and Babadag!, Thanks for the help! This was vaguely what I thought I was supposed to do.

I got Zoki's answer of 28.75 ∠60° by converting everything to the complex number and adding, but that's only because I used the convention that all the phase angles are referenced from Phase A and rotate counterclockwise. Babadag you're also right and I know that convention of everything anticlockwise of A is negative and clockwise is positive.

You're both legends!

The symmetric components are (if I take 28.75 arg(60)) for this question are:

Zero =19.2 arg(60)
Positive = 19.2 arg(-60)
Negative = 9.58 arg(0)

The symmetric components are (if I take 28.75 arg(240)) for this question are:

Zero = 0
Postive = 0
Negative = 28.8

Interesting differences.

Cheers,

Just been told that there shouldn't be any zero sequences through the neutral, don't know why?

CasterSkux said:
The symmetric components are (if I take 28.75 arg(60)) for this question are:

Zero =19.2 arg(60)
Positive = 19.2 arg(-60)
Negative = 9.58 arg(0)

The symmetric components are (if I take 28.75 arg(240)) for this question are:

Zero = 0
Postive = 0
Negative = 28.8

Interesting differences.

Cheers,
?
Symmetrical components here you get from known phase currents IA,IB,IC .
Zero sequence phasor definitelly isn't 0 due to fact sum of the currents in phases A and B isn't 0.
I see that in my calcs I took oposite way of system rotation, and not the convential, counterclockwise one (like Baba did) .
Better stick to the convention to avoid confusion with arguments

CasterSkux
Sorry I made a mistake in the inputting to calculate the symmetrical values.

Start again: the resultant phasor values are (using Badabung's convention):

Ia = 28.75 arg(0)
Ib = 28.75 arg(240)
Ic from Badabung =-14.38+24.9j = 28.75 arg(-60) = 28.75 arg(300) = this is the resultant neutral phasorThe calculator for Symmetric components gave me

I0 = 19.2 arg(-60)
I1 = 9.60 arg(0)
I2 =19.2 arg(60)

CasterSkux said:
Start again: the resultant phasor values are (using Badabung's convention):

Ia = 28.75 arg(0)
Ib = 28.75 arg(240)
Ic from Badabung =-14.38+24.9j = 28.75 arg(-60) = 28.75 arg(300) = this is the resultant neutral phasor

Oh dear, Ic = 0. Phase C is dead broken (remember?) and no current can be flowing through it. You should really learn basics of this stuff...

CasterSkux
Sorry I've attached Ic as the resultant vector of A + B, it's a terminology error. I'd like to learn more but we don't get enough practice excersizes unfortunately:(

I really do appreciate your help though Zoki!

Hello again, I was wondering if anyone can confirm my answer?

IA= 28.75 A
IB= 28.75 ∠240° A
IC= 0 A
Use formulae:)

CasterSkux
Ah the take home lesson I guess is to "always allocate A B and C to their appropriate phases and you MUST do so" I made Ic = In subconsciously because I thought it would be needed in a Phasor diagram, but these Phasor Diagrams are strictly for Phase A, B and C - something that I wasn't so brought on about.

Thank you heaps Zoki85, I've learn't more from you about these diagrams than reading them bit by bit in a year!

Hello all:)
Have to revive this thread just to clarify a new question:

Now we must answer the same question, but with no Neutral, this was answered without intention in the beginning of the post

230 V is a common phase voltage (phase to neutral) in LV networks in Europe, so I guess it isn't mistake. This is simple situation and can be quickly solved even without method of symmetrical components. Phase to phase voltage is 230×√3 ≈ 400 V. Since phase C is open, current flowing through phases A and B is the same current that flows through series resistance 8+8=16 Ω. Current has magnitude 400/16=25 A. Note that in a phase diagram currents in phase A and phase B have same magnitude but are in phase oposition (angle difference between phasors is 1800 ).

I can do the maths involved, but I don't know how the phasors are 180 degrees apart and I don't know how to mathematically derive it. I'm under the impression that Phase A and B are still 120 degrees separated.

I was wondering how A and B can be 180?

With no neutral, and phase C open, the line currents add at the "centre junction" to equal zero. ( Σ current in = Σ current out )

So ##\mathbf{I_A = -I_B}## (equal and opposite means 180° apart)

The voltage across one load is ##\mathbf{V_A - V_B}## and across the other load it's ##\mathbf{V_B - V_A}##

Construct the phasor diagrams and you'll see these phase relationships.

## 1. What is the purpose of calculating symmetric components in a three phase 4 wire system?

The purpose of calculating symmetric components is to analyze and understand the behavior of a three phase 4 wire system. It helps to simplify complex systems into more manageable components, making it easier to identify and troubleshoot issues.

## 2. How do you calculate the positive, negative, and zero sequence components in a three phase 4 wire system?

To calculate the positive, negative, and zero sequence components, you will need to use mathematical formulas and equations. These formulas involve taking the phase voltages and currents and applying them to the appropriate equations. Alternatively, you can use software or online calculators specifically designed for this purpose.

## 3. What is the significance of symmetric components in power system analysis?

Symmetric components are important in power system analysis because they help to understand the behavior of a system under different fault conditions. By analyzing the positive, negative, and zero sequence components, engineers can determine the impact of different types of faults on the system and plan for appropriate protection measures.

## 4. How does unbalanced load affect symmetric components in a three phase 4 wire system?

Unbalanced loads can create asymmetrical conditions in a three phase 4 wire system, which can affect the symmetric components. This can result in increased voltage and current imbalances, leading to potential equipment damage and power quality issues. Calculating symmetric components can help identify and mitigate these effects.

## 5. Are there any limitations to calculating symmetric components in a three phase 4 wire system?

Yes, there are limitations to calculating symmetric components in a three phase 4 wire system. The methodology assumes a balanced system and does not account for non-linear loads or harmonics. Additionally, it may not accurately represent the system behavior under certain fault conditions, such as open or short circuits. Therefore, it is important to validate the results with actual measurements and consider other factors in power system analysis.

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