Equipment & SystemsQuestion 273 of 521
For the 50 HP, 460 V motor (FLC 65 A), what is the maximum inverse-time circuit breaker per the 250% column of Table 430.52?
a.162.5 A
b.113.75 A
c.195 A
d.520 A
Explanation
Inverse-time breaker: 250% of FLC = 65 x 2.50 = 162.5 A (then next-standard-size rule may apply). 113.75 A is 175% (dual-element fuse), 195 A is 300% (nontime-delay fuse), and 520 A is 800% (instantaneous trip).
Law Reference: 2023 NEC §430.52Practice all 521 questions free — no signup required.
Related questions on this topic
- For the 3 HP, 115 V single-phase motor (FLC 34 A), what minimum branch-circuit conductor ampacity does NEC 430.22 require?
- Using NEC Table 430.250, what is the full-load current of a 50 HP, 460 V, three-phase motor?
- For the 50 HP, 460 V motor (FLC 65 A), what minimum branch-circuit conductor ampacity is required by NEC 430.22?
- For a continuous-duty three-phase motor rated more than 1 HP without integral overload protection, how many overload units are generally required?
- A continuous-duty motor has a nameplate FLA of 62 A and a service factor of 1.15. What is the maximum overload device rating at 125% under NEC 430.32(A)(1)?
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