# Symmetrical Components



## Machiavelli999 (Sep 27, 2010)

So, I am trying to really master symmetrical components and a lot of the problems on symmetrical components that I have seen basically have you find the line currents given unbalanced voltages at the source.

But can you use symmetrical components to find line currents in the following situation

Typical ABC, Wye Source - Wye Load system. Voltage is 480Vl-l / 277 Vl-n with all the appropriate 120 degree phase shifts between the phases. However, the loads are not balanced. Let's say 20 ohms for phase A-N, 30 ohms B-N, and 10 ohms C-N. All loads are purely resistive for simplicity's sake.

How would you use symmetrical components to determine the line currents in this unbalanced load situation?

Thanks.


----------



## cableguy (Sep 28, 2010)

You'd open up Grainger and Stevenson's Power Systems Analysis to page 460 and look at their formulas on how to do this.  It's in there.

It's a pretty big mess, but the essence is

Vaa=1/3Ia0(Za+Zb+Zc)+1/3Ia1(Za+a^2Zb+aZc)+1/3Ia2(Za+aZb+a^2Zc)

etc (follow through similarly for Vbb and Vcc)

I don't think this "long" of a question would be on the exam. Something I notice from the sample exam, if it takes you more than 10 or so lines to solve a problem, you're doing it wrong.


----------



## mull982 (Sep 28, 2010)

Machiavelli999 said:


> So, I am trying to really master symmetrical components and a lot of the problems on symmetrical components that I have seen basically have you find the line currents given unbalanced voltages at the source.
> But can you use symmetrical components to find line currents in the following situation
> 
> Typical ABC, Wye Source - Wye Load system. Voltage is 480Vl-l / 277 Vl-n with all the appropriate 120 degree phase shifts between the phases. However, the loads are not balanced. Let's say 20 ohms for phase A-N, 30 ohms B-N, and 10 ohms C-N. All loads are purely resistive for simplicity's sake.
> ...


Why do you need symmetrical componenets to solve this problem? If you have a wye load and you know all of the L-N load impedances why not just use the L-N system voltage to solve for the individual neutral currents?


----------



## LMAO (Sep 30, 2010)

When you have unbalanced loads (instead of balanced ones), the only thing changes is that you can't just solve for one phase and use the calculated values for all phases. You'll have to do each phase separately. Pretty simple.


----------

