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Hi everyone, does anyone know how to draw this? I understand how to get the answer, but honestly, I do not know what this looks like.

Are you sure you have to decrease the phase voltage of system A and system B by 30 degrees?

What does 40MVA//(1,000kVA/4%) mean? I think I am confused as to what the // mean?

Another way to solve this problem: formula: P (tan (Phi1) - tan (Phi2) ) S=Sqrt(3)VI* = Sqrt(3) x 480 x 34 Ang (acos(.75)) |S| = Sqrt(3) x 480 x 34 = 28.3 kVA P = PF x S = 28.3 kVA x .75 = 21.2 kW => 21.2 kW x (tan(-acos(.75) - tan( - acos(.9)) = -8.4290 kVAR I used Phi1 and Phi2...

In the question, it mentions the main distribution panel is 480 V, however, chapter 9, table 9 lists a voltage for 600 Volt cables. Why do we use the table for 600 V when the question mentions 480 V?

Is this for NCEES 125. The question for NCEES 126 asks about a grounded industrial power system and what would be least likely to improve

Hi, Is this formula correct Since Vab = Van + VaA however, shouldn't it be Vab = Vna + VaA where Van = - Vna (or does reversing little a with the neutral matter?)

Hi everyone, is this the correct formula to use for questions 110 and 132? 110) Sp=ILL*VLL and 132) S3p = sqrt(3) * ILL*VLL

Hi all, I am still struggling to understand this question. I'm not sure I understand what a 3-phase, 3-wire, ungrounded wye connected system vs. a 3-phase, 3-wire grounded system looks like. Does anyone have a diagram of these, and how they would be connected to the corner of a balanced delta...

Same question. I am also confused on this problem. If anyone can point me in the right direction it would be much appreciated. Thanks

Using S=(Va-Vb)^2/X = (500*1j-500)^2/(50) = 10,000

Does anyone have the link above?

Does anyone have the updated links?

The thread above no longer works. Does anyone have the new link?

Hi, I also can't understand this problem. I don't understand why Ia1 = 1.0/(2*Z1+Z0). Can anyone explain why? A phase-to-ground fault occurs on the 33-kV transmission system shown in the figure...

Hi, I tried to use the formula from the NCEES study guide, 5.1.3 And upon doing this I get S= sqrt(3) * 480 *34 ang (acos(.75)) = 28.3 ang (41.4 deg) kVA Preal = S * .75 = 21.2 kW 21.2 kW * (tan(acos(.75))-tan(acos(.90))) = 8.4290 kVAR Moreover, my answer is positive and matches the method...

Thanks for the help, and I understand your solution, but I am uncertain when I am supposed to use the other formula... or when do I use that kind of logic to solve a problem? I hope I am making sense, please let me know.

I put Z,new = Z,old * (S,new/ S,old)*(V,old/V,new)^2 and since Z,new = .045*(2000/1000) = .09 However, I can't understand the formula Load, T1/ Load, T2 = Z, T2 / Z, T1 Does anyone know where this formula came from? I also tried to understand Art and Grover's solutions, but I just can't...

Why is this not S=3*Vp*Ip... instead we used S=3*VL*IL