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If the motor doesn't start then yes use the percentages in 430.32.C

Sizing conductors for motors is 430.22 for a single motor. continuous duty is 125% of the FLA in the Motor tables. Several motors is 430.24. Example: (4) motors, (1) is 25HP, the other three are 5HP. Size 125% of the FLA of the 25HP motor and then add the FLA 's of the 5HP. For conductor sizing...

The overload is can not go above a certain amperage. So, using NEC 430.32, at 68A with the whatever percentage used for service factor/temp/all other motors, it will be 60A breaker. "This device shall be selected to trip or shall be rated AT NO MORE THAN the following percent of the motor...

Go Pack.

Turns ratio here. If only given voltages, then use the voltage ratios. When using voltage ratio is this senerio, need to make sure you do the ratio of either Vphase or VLL  on each side.

Article 110 is basically rules for installing electrical equipment. I use it a lot in my field when designing. I think as long as you index it, then it will be easy to flip to when asked about installations. 

Edit: No need for references for power stability. I didn't realize Grainger had it on page 695.

Anyone care to share any websites with some information on Lightning arresters, BIL/SIL, batteries, and power stability studies? Thanks

I think they did miss the 1.25 factor. I was just making sure I wasn't missing something. Thanks. I believe this part of the NEC hasn't changed from 2011 to 2014. Thanks for checking.

See attached.

Question: "Refer to 2014 edition of the NEC. What is the minimum required size for single-insulated, 60degreeC, TW copper conductors feeding a 460V 3-ph AC motor rated 150hp operating on continuous duty? A. 4/0 AWG B. 300 kcmil C. 400 Kcmil D. 500 Kcmil " I got B, 300 kcmil. Per NEC...

I just did this problem. I solved it using the turns ratio. I found the current going to the low side first. Turns ratio for the system is (5+2):5. I2 going out onto the 200kv/1.732 line is 4330A. I2 coming from the 230kv/1.732 is 3090A. The Ic is I2 - I1 = 1240A. Use Ic to find S in the...

In this problem, it is stated that the load draws 47A with the pf= 0.82 lagging. I took that as taking the pf at the load.

I had this exact question. The power factor is only to find the angle BETWEEN the Voltage and the current, not Current angle itself. Find the angle, arcos(.82). So the current is lagging the V by angle 34.9. We are given the angle for Voltage of 14, so the current angle will be 14-34.9, which is...

Can you cite the code section? for conductors I am referring to NEC 240.4(B), which states that the OCPD can be higher standard rating.

It should be 200A. their amps that they got does not match the NEC tables.

I had the same issue as well on this practice exam. I figured the same as you.

Yes, since the Generator is on the high side, then the voltages are equal so the MVA is the only thing that changes.

Yes, that is why I chose 200A. CI chose the 180A.

Problem: "Of the rated current listed below, which is the largest allowable for an overcurrent device to protect a 300kcmil, TW, 60degree Aluminum conductor feeding a 277V load. Assume conductor is in a steel conduit. A. 150 A B. 180 A C. 200 A D. 225 A " NEC says 300kcmil, TW at 60...