TN LLE Fault Current and OCPD Ratings 3

Question 1

What is the formula for calculating single-phase available fault current at the end of a conductor run?

Accepted Answer

AFC = V / Z (where Z is total impedance)

Answer

AFC = V x I / R

Answer

AFC = kVA x 1000 / V

Answer

AFC = V squared / P

Question 2

A transformer nameplate shows 5% impedance and 75 kVA rating at 240V secondary. What is the maximum available fault current at the transformer terminals?

Accepted Answer

6,250 amperes

Answer

3,125 amperes

Answer

12,500 amperes

Answer

1,500 amperes

Question 3

What is the effect of longer conductor runs on available fault current?

Accepted Answer

Fault current decreases

Answer

Fault current increases

Answer

No effect on fault current

Answer

Fault current doubles

Question 4

What NEC section covers series-rated systems where an upstream device protects a downstream device with a lower AIC rating?

Accepted Answer

NEC 240.86

Answer

NEC 240.85

Answer

NEC 240.83

Answer

NEC 110.22

Question 5

What is I squared t and why is it important in overcurrent protection?

Accepted Answer

Current squared times time — measures total energy let-through during a fault

Answer

Current times two times time — measures trip speed

Answer

Impedance squared times temperature — measures conductor heating

Answer

Inrush current times trip time — measures starting energy

Question 6

Per NEC 240.87, what additional protection is required for circuit breakers rated 1200A or higher?

Accepted Answer

An arc energy reduction method such as zone-selective interlocking, differential relaying, or energy-reducing maintenance switch

Answer

No additional protection required

Answer

A secondary fuse must be installed

Answer

Automatic reclosing capability