Complex Imaginary Power PE Exam Set 4 Question 67 - Sizing conductor for 150 HP motor on continuous duty

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Well-known member
Feb 7, 2020
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Hi guys,

I have a question regarding sizing a conductor for a 150 HP motor, based from the Complex Imaginary PE Power Set 4 exam.

This is question 67 from Set 4 of the Complex Imaginary PE Power Set 4 exam:

"What is the minimum required size for the single-insulated, 60-degree C, TW, copper conductors feeding a 460 V, 3-phase, AC motor rated 150 HP operating on continuous duty?"

The thing that confuses me is what I don't see in the provided solution.

The given solution says to refer to NEC Table 430.250 to find the full load current of a 3-phase, 460 C, 150 HP motor, which is 180 A.

However, it does not then multiply this FLC by 125%, which I would expect for a continuous duty motor. The provided solution simply uses this 180-A full-load current to size the conductors.

Is there some sort of exception that I'm not seeing in the NEC regarding higher-HP motors not needing the 125% multiplier for continuous duty? I thought that all continuous-duty motors need the 125% multiplier to their full-load current, and I never saw any exceptions regarding higher HP on motors. Is it possible that this given solution is incorrect?

I appreciate anyone who spends their time helping me on this!

This problem also bugs me since I think that the solution was incorrect.

At the first place, the conductor ampacity table  should be referred to T310.15(B)(17) since the problem says "single insulated", and yes the FLC should be multiplied by 1.25 per Article 430.22.

180x1.25=225A (#2/0 AWG)

I would still answer letter A(#4/0AWG) since this is the  "minimum" size among the the other choices. 

What will be the correct answer then #2/0 or #4/0?

Technically it should be #2/0.