Electrical and Computer: Power
The power domain of the FE Electrical and Computer exam tests three-phase power, per-unit analysis, transformers, synchronous and induction machines, transmission-line parameters, faults, and protection. Most items reduce to consistent bases and careful accounting of real and reactive power. The exam expects fast conversion between actual quantities and per-unit values, plus recognition of when a machine or transformer is outside its assumed operating condition.
Per-unit and three-phase quantities
- Choose Sbase and Vbase, then compute Zbase = Vbase²/Sbase and Ibase = Sbase/(√3 Vbase) for three-phase systems using line-to-line voltage.
- Per-unit impedance is Zpu = Zactual/Zbase. On a common base, transformer impedances and line impedances can be added directly across voltage levels.
- For a balanced three-phase load, S = √3 VLL IL, P = S cosθ, and Q = S sinθ. Keep lagging power factor associated with positive reactive power in the conventional load convention.
- Wye and delta conversions use ZΔ = 3ZY; line and phase quantities differ by √3 in wye and by √3 in current for delta.
Machines, faults, and protection
- A transformer’s voltage ratio equals its turns ratio; current ratio is inverse. Per-unit impedance does not change through an ideal transformer on a common base.
- Synchronous machine induced voltage depends on field current and speed; the internal emf and synchronous reactance set the terminal voltage under load.
- Induction motor slip is s = (ns − nr)/ns, where synchronous speed is ns = 120f/p. Rotor frequency is sf.
- For a balanced three-phase fault, fault current in per unit is approximately 1/|Zth| with prefault voltage 1.0 pu. Protection coordination uses time-current characteristics and CT ratios.
Worked example
A 13.8 kV, 50 MVA three-phase system has a line reactance of j1.5 Ω. Use 50 MVA and 13.8 kV as bases. Zbase = (13.8 kV)²/50 MVA = 3.809 Ω, so Xpu = 1.5/3.809 = 0.394 pu. Ibase = 50 MVA/(√3 × 13.8 kV) = 2,092 A. With a 1.0 pu prefault voltage and only this reactance, Ipu = 1/0.394 = 2.54 pu, giving a fault current of 2.54 × 2,092 ≈ 5,310 A.
Traps that cost marks
- Using line-to-neutral voltage with a three-phase base formula, or omitting √3 in line current.
- Mixing per-unit bases when a transformer changes voltage level.
- Reversing lagging and leading power-factor signs, or adding real and reactive powers as scalars.
- Forgetting that induction motor rotor frequency is slip times stator frequency.
- Treating a CT ratio as a power ratio, or ignoring relay pickup and coordination margins.