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Create a 3x3 matrix M = [1/3, -1/3, 0; 0, 1/3, -1/3; -1/3, 0, 1/3]. Create a column vector of line currents I = [IA; IB; IC]. Multiply M*I to get vector of phase currents [IAB; IBC; ICA]. The derivation of M was a little painful; it's the pseudoinverse of the non-invertible transformation...

I understand. The sending end reactive power is larger than the receiving end reactive power by I^2X. I fixed the pdf to reflect that.

The study guide equation in the original post is incorrect as the cosine term is not outside the fraction. See the derivation of equations for both active and reactive power flow from the sending end in the attached pdf. (Corrected per Zach's comment below)

First, both the transformer and generator need to be on the same base for their pu values to be combined. The solution chooses to change the reactance of the transformer to the same base as the reactance of the generator, so it goes from 0.04 to 0.0533 after that. Next, the last sentence is...

This question makes no sense to me. Voltage regulation of a generator is always determined by the terminal voltage magnitude change when current goes from no load to full load with constant excitation. As far as I know, it has nothing to do with internal voltage, which remains the same...

There isn't another CT, only taps to adjust the way the two currents oppose each other in the restraint coils. Think about it this way: the operating coil shouldn't see much current under full load, so the taps selected need to balance the currents through the restraint coils. With one current...

The H is a way of indicating each group of four digits is in hexadecimal.

List out the address range spaces 15 to 0, then start putting a 1 or 0 as needed for each space. For CS4 to go low, A15 = 1, A14 = 0, A13 = 0, so those are defined as such. This leaves A12 to A0 as don't care bits. When coupled with 1 0 0 for bits 15-12, if A12 through A0 are all 0 to find the...

For Relay A, the primary impedances for the required distance settings are: the Z1 setting is 80% of impedance 1-2, or 0.8(4 + j20) ohms = 3.2 + j16 ohms. the Z2 setting is 120% of impedance 1-2, or 1.2(4+l20) ohms = 4.8 + j24 ohms. the Z3 setting is 100% of impedance 1-2 plus 120% impedance...

You are correct. Only a qualified person can pass the limited approach boundary.

That is the primary current with rated primary voltage applied and the secondary shorted. The current you need to use for I^2R loss is the primary current when the secondary is at full load, not shorted, i.e., primary full load current.

The question asks for the value of the current-limiting resistor to limit the peak charging current to 10 A, so you use the peak voltage that comes out of the rectifier, not the average voltage.

Sometimes rolling the phases reduces the current spikes that can be caused by a wye/delta open transition so that nuisance breaker trips don't occur, but that doesn't always help. Since the purpose of wye/delta starting is to reduce inrush currents, having large current spikes when open...

The solution manual and all of the responses above are wrong: a transformer's impedance in ohms does not change when the applied voltage changes. It is a function of core and winding construction only. The short-circuit test is used on a transformer to determine this value, and nominal voltage...