Master Electrician Exam (NEC)General Electrical KnowledgeHard
A master electrician is performing a short-circuit fault current calculation for a new 480V, 3-phase commercial service. The utility provides a maximum available fault current of 25,000A at the service entrance. The service conductors are 500 kcmil copper, 100 feet long, and the transformer impedance is negligible. What is the approximate impedance of the service conductors per phase, considering a conductor resistance of 0.025 ohms per 1000 feet?
- A0.025 ohms
- B0.0025 ohms
- C0.005 ohms
- D0.05 ohms
Show answer & explanationAnswer & explanation
Correct answer: B. 0.0025 ohms
The resistance of the conductor is given as 0.025 ohms per 1000 feet. For a 100-foot run, the resistance is (100 feet / 1000 feet) * 0.025 ohms = 0.1 * 0.025 ohms = 0.0025 ohms. Since the transformer impedance is negligible and we're looking for the conductor impedance per phase, this resistance value approximates the impedance.
Why the other options are wrong
- A. This is the resistance per 1000 feet, not for 100 feet.
- C. This might result from an incorrect calculation, possibly multiplying by 2 (for both ways) when the resistance is already per length.
- D. This is an incorrect calculation, possibly a misinterpretation of the unit or length.
Conductor Impedance Calculation
Conductor impedance, a key factor in short-circuit calculations, is primarily determined by its resistance and reactance, which vary with conductor material, size, and length.
- Impedance (Z) = √(R² + X²), where R is resistance and X is reactance.
- Resistance (R) varies with conductor material, size, temperature, and length.
- Reactance (X) varies with conductor size, spacing, and frequency.
- For short runs, resistance often dominates impedance, especially in lower voltage systems.
Memory trick: Fault current needs to 'Z'ee the path, where 'Z' is impedance.