Calculations第 25 / 110 题
A 45-kVA, 208-V, three-phase transformer has 2% impedance. What is the approximate available fault current at its secondary (infinite primary)?
a.6,245 A
b.3,120 A
c.12,500 A
d.9,000 A
解析
Secondary FLA = 45,000 / (1.732 x 208) = 124.9 A. Isc = 124.9 / 0.02 = 6,245 A.
免费刷完整 110 道题库 — 无需注册。
同考点相关题目
- Using the previous circuit (5.06-V drop on a 120-V circuit), what is the approximate percent voltage drop?
- A 208-V three-phase feeder carries 40 A over 150 ft using #6 Cu (CM = 26,240, K = 12.9). What is the approximate voltage drop?
- For a 112.5-kVA, 208-V, three-phase transformer with 2% impedance, what is the approximate available fault current at the secondary terminals (infinite primary)?
- Six current-carrying THHN conductors share one raceway. If the 90 C ampacity of #6 Cu is 75 A, what is the adjusted ampacity after the fill adjustment factor for 4-6 conductors?
- A #3 Cu THHN conductor has a 90 C ampacity of 110 A. In a 50 C ambient (correction factor 0.82), what is the corrected ampacity?
- A #2 Cu THHN (90 C ampacity 130 A) runs with 6 current-carrying conductors (0.80 factor) in a 40 C ambient (0.91 factor). What is the adjusted-and-corrected ampacity?