Electrical Power System Fundamentals
What the exam asks
Power System Fundamentals questions check whether you can move between the physical system and its numbers: three-phase relationships, percent impedance, transformer ratings, power factor and fault levels. Expect short calculations rather than derivations, plus qualitative items on transformer connections, grounding and power quality.
Core rules
- Three-phase math. S (kVA) = √3 × V_LL × I_L, and P (kW) = √3 × V_LL × I_L × PF. Line current pairs with line-to-line voltage; when you work in phase quantities the √3 disappears.
- Percent impedance lives on the device's own base. I_sc ≈ I_FLA ÷ (%Z/100), assuming an infinite source and negligible resistance.
- Winding connections change behaviour. Delta windings block zero-sequence (triplen) current; a Dyn transformer gives a 30° shift and a low-impedance path for triplen harmonics.
- Power-factor correction. kVAR_c = P (tan θ1 − tan θ2).
- Sequence components. Positive sequence is balanced load; negative sequence is unbalance and rotor heating; zero sequence is ground faults.
Worked example
A 1500 kVA, 13.8 kV / 480–277 V transformer has 5.75% impedance. Find the secondary full-load current and the bolted three-phase fault current.
I_FLA = 1,500,000 VA ÷ (√3 × 480 V) = 1804 A. I_sc = 1804 ÷ 0.0575 = 31.4 kA rms symmetrical.
Interpretation: this is the fault level at the secondary terminals with an infinite primary. For equipment rating, add decaying motor contribution (roughly 3–6 × motor FLA during the first cycles) and apply the nameplate %Z tolerance (often ±7.5%), giving about 33.9 kA worst case. Check peak (crest) current as well - it can exceed twice the rms symmetrical value.
Traps that cost marks
- Putting line-to-neutral voltage into a √3 formula, or dropping √3 when converting kVA to amps.
- Using kW where the question supplies kVA.
- Quoting nameplate %Z as the worst case while ignoring tolerance and motor contribution.
- Converting per-unit values between bases without squaring the voltage ratio.