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

Available fault current is calculated as AFC = V / Ztotal, where Z is the total impedance of the fault current path from the source to the point of calculation.

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

Full-load current = 75,000 / 240 = 312.5A. AFC = FLA / Z% = 312.5 / 0.05 = 6,250A.

Question 3

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

Accepted Answer

Fault current decreases

Answer

Longer conductor runs add impedance to the fault path, which reduces the available fault current at the downstream end.

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 240.86 covers series ratings, allowing a tested combination of devices where the upstream device limits fault current to protect a downstream device with a lower interrupting rating.

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

I squared t (current squared times time) represents the total thermal energy let-through during a fault, which determines the damage potential to conductors and equipment.

TN LLE - Tennessee Limited Licensed Electrician Practice Test

TN LLE - Tennessee Limited Licensed Electrician Practice Test

TN LLE Fault Current and OCPD Ratings 3