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That makes perfect sense, but that speed difference is based of the slip. This problem only gives a scenario as if you were handed a nameplate without slip, just speed. So you have to take the RPM at face value and guess at how many poles it has. I don't know why anyone would guess that the...

Are you implying the nameplate speed is for the stator speed and the actual speed is the rotor speed?  I still don't get it, because by this methodology you could buy a motor with a nameplate speed of 1,800 and the damn thing run at 18,000 RPM. 

Yeah this is the method I took. I knew it couldn't be 4.5.  But then when i calculated 4 poles it came out to be 1800 RPM. I can't wrap my mind around the fact a motor will have more RPM's than its nameplate and that be okay. That is the Part I don't understand. I almost picked 4 because its...

Can someone try to put this in terms I can understand? The solution is you should just "assume 4 because the motor is going to run slightly faster than rated speed".  I don't work in the motor fields, but if the motor is designed with 4 poles why wouldn't the name plate say it has 1,800 RPM...

I'm not talking about .85 effective Z values. I'm talking about using the formula of Rcostheta+Xsintheta to obtain a Z. Then use your lengths to get the Z effective. I don't see how that path is ever the "wrong" path and when I "can't" use it. 

All of their voltage drop solutions they just multiple R and X times the distance divided by 1000. They never mention Z or convert. I don't know why they aren't doing it, and am I supposed to be doing it? Every voltage drop problem I solve rather it be single phase or 3 phase involves a Z. 

I used X = .04, and R = .038, then did .038cos(41.4)+.04sin(41.4) = .055. Then (.055*500)/1000 = .027. I*Z = 6.3 V, 600v -6.3 . 

Does anyone else get the same answer they do? I get closer to D. Why aren't they finding the "Z effective Impedance" instead of just using R and X?

Thank you for the kind words and advice. I have a spreadsheet where I track my hours and problems worked for this PE.  By the end of next week I will have over 500 hours studied, and 2,500 problems worked, and haven't even got to take the test due to Covid.  Anyone outside the engineering...

The answer is D. 

I can't find this in NEC or any other literature I was looking in. I could see this being a PE question, but I don't want to take this books "word for it" without verifying it. 

When I solve this problem, I am using Load C and A to get my angle of current. I am giving the PF of the load, so I am taking the conjugant. In this solution It is using 240-angel and 120-angle and I don't know why they are doing that when finding current. 

I want to say I researched this for state of TN or SC, and it was "you can sign up 3 times without question. But after that, you have to prove that you deserve another attempt". When I researched further, that was just stating you signed up for an online class, a masters class, or even just buy...

I was actually wondering this too. 

Can you post a picture or a link of what book you're referring to? I have every reference book (I thought) that Zach recommends, but I don't have that one in my library. 

If you're right, this is another example, of extreme lazy ass garbage literature. Blows my mind people have the audacity to publish work and that be the solution (if yours is right). Don't care for anyone to defend "all people make mistakes" etc.  If my work was as sloppy as this damn book I...

That's how I solved it. 

I'm assuming the "gotcha" here is you don't multiple this current by 1.25% even tho its a continuous motor to size your cable?  

I know for a fact I am super comfortable with any engineering they throw at me. It's going to be nice at least knowing a little SOMETHING (unlike the FE with communications and software blah) but I'm reading the hell out of NEC. I catch myself digging and digging and 20 minutes going by in it...

I would suggest asking your boss. Because if you don't really stamp anything, they aren't going to care. If you DO stamp documents then THEY WILL PAY for you to get your license in Washington.