Electrical Engineering (PEC Curriculum 2023)
The Electrical Engineering domain of the PEC EPE tests whether a candidate can apply circuit theory, power-system analysis, rotating machines, protection, and installation practice to realistic plant problems. Expect numerical items that combine a code limit with a design check, not isolated definitions.
Core rules and concepts
- Work in per-unit on a common base. Convert an impedance by Z_new = Z_old (S_new/S_old)(V_old/V_new)^2; never mix bases across a transformer.
- For balanced three-phase systems, S = √3 V_LL I_L, P = S cos φ, and Q = S sin φ. Power-factor correction adds capacitors with Q_c = P(tan φ1 − tan φ2).
- Transformer fault level is set by source and transformer impedance. On the LV side, I_F ≈ S_rated/(√3 V_LL Z_pu), with Z_pu as a decimal.
- Protection grading needs CT ratio, pickup, time multiplier, and a discrimination margin-typically 0.3 s between consecutive devices. Earth-fault settings must coordinate with neutral earthing.
- Cable and busbar checks cover continuous current after ambient/grouping derating, voltage drop, short-circuit withstand (I²t), and touch voltage.
- Rotating-machine checks use E_b = V − I_aR_a and T = kΦI_a for DC, and slip s = (n_s − n_r)/n_s for induction motors.
Worked example
A 250 kVA, 400 V, 50 Hz transformer has Z = 5%. The LV full-load current is 250,000/(√3 × 400) = 361 A. The bolted LV fault current is 361/0.05 = 7,217 A. A breaker with 10 kA breaking capacity covers the transformer fault level, but the cable’s I²t withstand and the upstream relay’s grading margin must still be checked. If a 90 kW load at 0.80 PF is corrected to 0.95, Q_c = 90(tan 36.9° − tan 18.2°) ≈ 38 kvar.
Traps that cost marks
Using kW in place of kVA, omitting √3 in line quantities, treating 5% transformer impedance as 0.05 Ω, ignoring motor fault contribution, and assuming capacitors reduce active power. Also check whether the question asks for rms symmetrical current or peak asymmetrical current.