# Power factor correction question - PPI Exam 1 Question 10



## akyip (Nov 12, 2020)

Hey guys,

PPI Exam 1 Question 10 is supposed to be a simple power-factor correction problem that asks for the capacitance needed to correct the p.f.

The problem statement does not state whether the system is Y-connected or delta-connected. So when I first did this problem, I just thought Z = V LL ^2 / S-3ph* (in phasor form) and then determine X = Z * sin(theta) and then C = 1/(2 * pi * f * XC). Doing that, I got C = 9 microfarads... which is incorrect according to the solution.

The solution for this problem states that "since this is a three-phase system, each phase will correct for one-third of the total reactive power, or -50 VAR". Then later on, it solves for XC for each phase using V LL^2/Q = 208^2 / 50. This means that the system is delta-connected... but that was not stated in the problem statement. That was what got me in this question. I had to handwrite "delta system" next to this question for subsequent practice attempts on this problem.

So if during a power-factor correction problem the system's connection type is not stated, do we just assume a delta-connection?

Thanks for any input on this!


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## Sam_ (Jul 9, 2021)

The second chart has an error to me unless I'm not seeing something.

200 W & 150 VAR does not give 200VA. It gives 250VA-angle(36.9).

3uF is the capacitor rating in each phase. Total capacitor rating would be 9uF to get unity PF for 3 phases I think. I'm not sure delta or wye matter.

150VAR / (2 * pi * 60* 280^2) = 9uF total --> 3uF rating per phase


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## evanfus (Jul 21, 2021)

Was the problem supposed to say if it's looking for total capacitance or capacitance per phase? I interpreted as total capacitance and got the same answer akyip got.


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