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First you must know what the sc test is: basically short the sec, increase v prim until you get rated i on sec (and prim) current sec and prim = 1 pu, Z = 0.10 pu v pu = i x Z = 1 x 0.1 = 0.1 pu v = 0.1 x vbase = 0.1 x 100 = 10 kv

A good basic reference https://www.instrumart.com/assets/Megger-Guide-to-Insulation-Testing.pdf

I get 10k either way   Sorry,  in the S equation should be the Vdrop not 500kv, but 707 kv

Losses are I^2 R so 2 times the I therefore 2^2 = 4 times the losses 460 are no load losses, independent of load 50% I2R losses = 2370 - 460 = 1910 I2R at 100% = 2^2 x 1910 = 7640

Bear in mind it's limited by the rated winding current Buck: 25 kva / 120 x 360 = 75 kva if used as boost: 25 kva / 120 x 600 = 125 kva easy way AT kva = Vout/Vbuck or boost x xfmr kva 360/120 x 25 = 75 kva or 600/120 x 25 = 125 kva either way windings rated current, V and ratio applies 4:1...

The wording is important FAULT power...NOT total input power obviously the xfmr has losses but input P = fault + losses output = fault therefore the FAULT P component is the same on both sides but total in vs out is not another take: inf bus Z source = 0, ie, no Vdrop input P=1/pu Z, no...

Most texts on transients have a chapter on lightning, most common cause of outages This is the one I used class was taught by a guy taught by the author for the price a must have imho...

Most texts on transients have a chapter on lightning, most common cause of outages

Lethality has 2 factors Duration/time and i magnitude i  = 175/(sqrt t) i mA, t in sec, , 175 is avg wt in lbs look up Dalziel (invented the gfci), an EE prof at ucla iirc a static charge is uSec so a higher i can be tolerated, TC ~ RC, typ body R 600-1200 Ohm, very little C

NGR's are required by law in mining to limit frame V to <40 An add on ? what is the N-G V of the ungrounded phases during a gnd fault? hint: what is the N V elevated to?

My approach

B during + swing 2 x sqrt2/Pi x 240/20 x 10 = 108 during - swing RL shunted so RL sees no i so V = 0 avg = 54