A conductor starts at 55 amperes in the permitted insulation column, and the supplied adjustment factor is 70 percent. What adjusted ampacity results before terminal limits?
- A55 amperes
- B16.5 amperes
- 38.5 amperes
- D78.6 amperes
Why C is correct
Multiply the starting ampacity by the supplied factor: 55 amperes times 0.70 equals 38.5 amperes. Terminal limitations would be checked afterward if the problem supplied them.
