Power System Fundamentals and Circuit Analysis
Three-phase relationships
The single most valuable set of formulas: S = √3 × V_LL × I_L for three-phase, and S = V × I for single-phase, with P = S × power factor. Impedance base conversions follow Z_base = V_LL² / S and I_base = S / (√3 × V_LL). A persistent trap is substituting line-to-neutral voltage into the √3 formula, or forgetting √3 entirely on a three-phase fault calculation.
Per-unit and percent impedance
Percent impedance is impedance on the equipment's own base. To combine a transformer and a source, convert everything to a common kVA base: Z_new = Z_old × (kVA_new / kVA_old). Fault current from a transformer is approximately I_FL / Z_pu - valid only when the source is stiff and motor contribution is ignored.
Worked example
A 1500 kVA, 480 V transformer with 5.75% impedance. Full-load current = 1,500,000 / (1.732 × 480) = 1804 A. Available short-circuit current ≈ 1804 / 0.0575 = 31,400 A. That number then drives breaker interrupting rating and arc flash study inputs.
Sequence components and power factor
Unbalanced faults are handled with symmetrical components (positive, negative, zero sequence); zero-sequence current needs a return path, which is why delta-wye grounding configuration matters. Capacitor correction raises power factor and reduces line current but does not reduce kW load. Watch for questions asking for the new line current after correction - use kW / (√3 × V × new PF).
Traps to expect
Mixing kW and kVA, treating an impedance in ohms as if it were per-unit, and ignoring that parallel impedances divide while series impedances add.