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I am assuming you're serious and no offense intended but I would recommend a review of basics like this before taking exam. If I write the numerator as 35 /_ 0 times 30 /_ 90 and you multiply those by multiplying magnitude and adding angles does that help you out?

Both Wildi and Stevenson have this relationship.. from Stevenson (Elements of PSA 4th ed): P =((V sub term x E sub g) / X sub g )  typically times sin Pangle Since this is on a per-phase basis if the voltages are given in phase voltages you have 3X And to maximize in theory you set sin Pangle...

Here's the way I think about it... The phasor vectors ALWAYS rotate CCW.  Imagine you are sitting stationary on the arrow at end of +x (real) axis... So the voltage phasor comes by and then a small fraction of time later the current vector sweeps by the same place.   The current is lagging (that...

1) True 2) False... the relationship S = VI* MUST be used because positive complex power was defined a long time back for a lagging or inductive load and they wanted S in 1st quadrant.  As you could see by the video you attached and my explanation...  Whether you choose voltage reference at 0...

Maybe I'll help or make worse but I'll give it a whirl.  If you prefer to think of 1st quadrant, then simply choose the current as the reference phasor.  Therefore, I is I at 0 degrees. Then the voltage at the load has to be at POSITIVE 33 degrees (in 1st quad) (due to lagging PF) Now just...

I think the 600 just comes from combining up the efficiency, PF, √3, and KW to HP conversions into an empircal constant given in equation 13.5 on page 277. If you want to make some assumptions simialr to Kovz above, say efficiency of 0.95 and Pfactor of 0.9 the KVA in then is (500*0.746)/(0.95...

Now you've got it. To really understand this, I recommend you beg/borrow a copy of IEEE Red book and find and study the phasor diagram that creates the basis for the equation used in Table 9. It will take a few minutes to review the trigonometry and comprehend but then this will all make sense.

I'm not sure what you mean by this but think you may be making more complex than needed. The line impedance is of course distributed uniformly but is modeled with lumped. The load end of line connects to the actual load. The drop is Vsource - VLoad Some of the math above assumes the source...

look here... http://engineerboards.com/index.php?showtopic=8598 or here... http://engineerboards.com/index.php?showtopic=11550

I guess I was just looking at the text you entered and not problem number... the explanation from 107 referenced doesn't seem to match... we must have a mismatch somewhere?

Think Zbase = V base 2 / S base so you have and extra sqrt 3. Then I assume you know the next step?

When in doubt, always review the exam specification: I assume you are taking Electrical and Electronics Circuit Analysis 25% 1. Passive components 2. DC circuits 3. Sinusoidal analysis 4. Transient analysis 5. Power and energy calculations 6. Battery characteristics and...

Maybe think of it this way. They chose the line-neutral voltage of 120 at 0 and that set the current relative that at -31 degrees as the power factor of the load is voltage across it to current through it. So the current has to lag the voltage which is at 0 degrees by 31. Now just think of it...

You're going to have to provide a few more details for those of us that don't have CI exams? Conductors ? distance? etc

In short yes, since voltage is assumed constant, the only way one could go from 25% kVA to 50% of some kVA is to double current. Same as if you go from 50 % kVA to 100% kVA current needs to double. If you have a transformer rated 25kVA and is is operating at 50% load and at a later time it is...

I don't think Wildi says that...in fact for the D-Y transformer you mentioned, the y connected seconday would only lead if it was a step up transformer (see above high side leads low side). If it is step down then not true. What Wildi states in my copy (6th ed) under polarity marking of 3-phase...

I only have a copy of Stevenson handy (predecessor of Grainger/Stevenson)... It states " The American standard of designating terminals H1 and X1 on Y-Delta transfomers requires that the positive-sequence voltage drop from H1 to neutral lead p-sequence voltage drop from X1 to neutral by 30...

I think ANSI standard for transformer connection is to have high side lead low side. I think this is reference but you should check. Look harder in Grainger/Stevenson as I think it mentions it. I'll look tonight if you don't find. IEEE Standard Terminal Markings and Connections for...

I have what I think is latest version copyright 2011 with a light blue cover. I believe the earlier version has the same problems/solutions and different cover. There may be some errata as well. If you have earlier version, be aware of the following. If you go to NCEES (they make the book)...