A master electrician is performing a short-circuit current calculation for a commercial service. The utility transformer is rated 1000 kVA, 480Y/277V, with an impedance of 5%. The service conductors have a total impedance (including busway) of 0.0005 ohms. What is the approximate available symmetrical short-circuit current at the service entrance?
- A12,000 A
- B24,000 A
- C36,000 A
- D30,000 A
Show answer & explanationAnswer & explanation
Correct answer: B. 24,000 A
1. Calculate full-load current (FLA) of the transformer: FLA = kVA / (V_LL * sqrt(3)) = 1000 kVA / (0.480 kV * 1.732) = 1000 / 0.831 = 1203.36 A. 2. Calculate transformer impedance in ohms: Z_T = (%Z / 100) * (V_LL / (FLA * sqrt(3))) = (5/100) * (480V / (1203.36A * 1.732)) = 0.05 * (480 / 2084.2) = 0.05 * 0.23 = 0.0115 ohms (Per unit impedance * (V^2/kVA)). 3. A simpler way for transformer impedance in ohms: Z_T = (%Z * V_LL^2) / (kVA * 1000) = (5 * 480^2) / (1000 * 1000) = (5 * 230400) / 1000000 = 1152000 / 1000000 = 1.152 ohms (This is wrong, need to use base impedance calc). Let's use the per-unit method for total impedance: Base kVA = 1000 kVA, Base V = 480V (Line-to-Line). Base Current (I_base) = 1000 kVA / (sqrt(3) * 0.480 kV) = 1202.8 A. Base Impedance (Z_base) = V_LL^2 / (kVA * 1000) = 480^2 / (1000 * 1000) = 0.2304 ohms. Transformer impedance in ohms = Z_pu * Z_base = (0.05) * 0.2304 = 0.01152 ohms. Total Impedance (Z_total) = Z_T + Z_conductors = 0.01152 ohms + 0.0005 ohms = 0.01202 ohms. Symmetrical Short-Circuit Current (I_sc) = V_LL / (sqrt(3) * Z_total) = 480V / (1.732 * 0.01202 ohms) = 480 / 0.0208 = 23076 A. Closest answer is 24,000 A.
Why the other options are wrong
- A. This value is too low, likely from calculation errors or incorrect impedance assumptions.
- C. This value is significantly too high, indicating a major error in impedance calculation.
- D. This value is too high, likely from calculation errors or incorrect impedance assumptions.
Short-Circuit Current Calculation
Short-circuit current is the maximum current that can flow during a fault condition. Calculating it is essential for properly sizing overcurrent protective devices (OCPDs) and ensuring equipment withstand ratings.
- Calculated using the impedance of the utility source, transformer, and conductors.
- Expressed in symmetrical and asymmetrical components.
- Typically uses per-unit or ohmic methods for calculation.
- Crucial for arc-flash and equipment damage prevention.
Memory trick: Short Circuit: Impedance Limits the Blast.