Recent content by CasterSkux

Hello again, I've got another question that involves symmetrical components, this time I've attempted it. The Question follows: 1. In a 3 phase circuit with resistance earthed and neutral and Earth fault in phase “A” which produces a fault current of 800 amps. Load currents may be taken as...

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 I can do the maths involved, but I don't know how the phasors are 180 degrees apart and I don't...

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...

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

I really do appreciate your help though Zoki!

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:(

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...

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

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 positive = 0 Negative = 28.8 Interesting differences. Cheers,

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...

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,

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

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...

Yes, oh have I overlooked something?