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Ha, thank you! I had blinders on staring at this too long.

(I think this problem has a typo; 0.8 lagging power factor should be 0.85 lagging.) When the solution starts solving for V_S, the sending-end voltage, 5.11 degrees is subtracted from the current angle. Is that the angle from V_R, the receiving-end voltage? If so, how do you solve for that angle?

To cancel out the "j" going from the second to third line, a "-1" multiplier would appear on either side of the equation. Should the third line actually be Q = - (VA-VB)^2/X?

Ok, I figured out my mistake. The rated percentages should also be squared with the current. I was using: P_50% = 0.5 * I^2 * R + P_NoLoad When it's supposed to be: P_50% = (0.5 * I)^2 * R + P_NoLoad So then I^2*R should be: I^2*R = (P_50% - P_NoLoad) / 0.5^2 = (2370 - 460)/0.25 = 7640 W...

(Practice exam book printed in 2014) Can someone tell me where the multiplication factor of 4 comes from when calculating the 100% rated load loss? And where that line of calculation even comes from? Also, with the given information, are the I^2*R losses calculated the following way...