Fault current analysis on the PE Power exam

A balanced three-phase fault problem on the PE Power exam asks for the symmetrical current at a named point, the duty in MVA, or the share one source contributes. The NCEES handbook covers this under Fault Current Analysis in its Transmission and Distribution chapter, which prints the short-circuit MVA relation, and under Per Unit System in its Circuits chapter, which prints the base current, base impedance and change-of-base relations. A balanced three-phase fault uses the positive-sequence network alone. Unbalanced faults are on the symmetrical components page.

What you actually need to own

• Duty in MVA and in kA. S_sc = √3 · V_nom · I_sc, with S_sc in MVA, V_nom the line-to-line voltage in kV and I_sc in kA.
• Source impedance from a stated duty. A utility duty at a bus is a Thévenin impedance behind that bus. In ohms per phase, X_source = V_LN / I_sc. In per unit, X_source = base MVA / S_sc when the base voltage is the bus's nominal voltage.
• Transformer impedance. Nameplate percent impedance is per unit on the transformer's own rating. Its short-circuit capacity is rating / Z_pu. Behind a source stiff enough to ignore, the secondary fault current is full-load current / Z_pu.
• One base before you combine. Per-unit values on a common base add directly along a series path. Values on different bases must be moved first: Z_new = Z_old · (kV_old / kV_new)² · (MVA_new / MVA_old).
• Three routes to the same answer. In ohms: I = V_LN / Z_th, with every impedance referred to the voltage level of the fault. In per unit: I = 1.00 / X_th at nominal prefault voltage, times I_base = base kVA / (√3 · base kV_LL). By the MVA method: each element is rating / Z_pu, elements in series combine as reciprocals, and sources in parallel add.
• A check with round numbers. A 2,000 kVA, 480 V transformer at 5% sits behind a 250 MVA source. On a 2 MVA base the source is 2/250 = 0.008 per unit and the total is 0.058. I_base is 2,406 A, so the fault current is 2,406 / 0.058 = 41.5 kA. The MVA method agrees: 40 MVA in series with 250 MVA is 34.5 MVA.
• Rotating machines are sources. A motor or generator feeds a fault on its own bus with an initial symmetrical current of about its full-load current divided by its subtransient reactance, in per unit on its own rating. For an induction motor the locked-rotor current serves the same purpose, and that contribution dies away within a few cycles. Contributions arriving at one bus add.
• Read the switching state. An open tie or feeder breaker removes its branch whatever the branch impedance is.

Where people lose points

• The wrong voltage. The duty converts to current at the voltage of the faulted bus. On a one-line with two voltage levels, find the fault mark first.
• √3 slips. Leaving √3 out of the base current, or dividing line-to-line volts by Thévenin ohms, inflates the answer by 73%.
• Treating the source as infinite when a duty is given. In the round-number check above, ignoring the 250 MVA source gives 48.1 kA, which is 16% high.
• Quoting the bus duty downstream. Every feeder section between bus and fault adds impedance.
• Leaving out motors, or taking their full-load current at another bus's voltage.
• Ranking buses by MVA. A bus with more fault MVA can have less fault current than a lower-voltage bus.
• Mixing rms and peak in asymmetry questions. Check what a supplied factor returns. An asymmetry factor takes a symmetrical rms current and returns an asymmetrical rms current. A peak factor returns a first-cycle peak and already includes √2. In a high X/R circuit the dc offset is largest in the phase whose fault begins at its voltage zero.

A note on the codes

When a problem says to take a three-phase motor's full-load current from the code, use Table 430.250 in NEC Article 430, not a current worked out from horsepower, efficiency and power factor. The exam supplies its codes, including the 2020 NEC and the 2017 NESC, and scores code answers against those editions. See the PE Power handbook and code books.

How to study it

Draw a one-line with a utility duty, a transformer, a feeder and a motor group. Solve the fault at the bus and at the feeder end in ohms, in per unit and by the MVA method until the three agree. Drill the per-unit system first if base changes slow you down.

Part of the Transmission and Distribution Analysis area of the PE Power exam. → Practice it on CastorPrep