Electrical and electronic systems: circuits, machines, power electronics and digital electronics
This domain covers DC and AC circuit analysis, transformers and rotating machines, power converters and Boolean logic. The exam usually mixes a quantitative circuit problem with a simplification of a logic function.
Direct current
- Ohm's law, Kirchhoff's laws (meshes and nodes), voltage divider, and Thévenin (V_th = open-circuit voltage, R_th = resistance seen with sources set to zero).
- Power P = V·I = I²·R.
Alternating current
- V_rms = V_peak/√2; Z = √(R² + (X_L − X_C)²); X_L = 2πfL; X_C = 1/(2πfC).
- P = V·I·cos φ; Q = V·I·sin φ; S = V·I. Power-factor correction with a capacitor: Q_C = P(tan φ₁ − tan φ₂).
- Three-phase: P = √3·V_L·I_L·cos φ.
Machines and power electronics
- Transformer: V₁/V₂ = N₁/N₂ = I₂/I₁; apparent power S = V·I is conserved.
- Induction motor: n = (1 − s)·n₁ with n₁ = 60f/p (p = pole pairs).
- Diode rectifier (half-wave, bridge), thyristor phase control, chopper (DC–DC), inverter (DC–AC).
Digital electronics
- Boolean algebra, De Morgan, Karnaugh maps; gates AND, OR, NOT, NAND, NOR, XOR. Example: A'B + AB + AB' simplifies to A + B.
Worked example
A 10 kW load with cos φ = 0.7 (inductive). S = 10/0.7 = 14.29 kVA and Q = 10·tan 45.6° = 10.2 kvar. To reach cos φ' = 0.95: Q' = 10·0.329 = 3.29 kvar, so Q_C = 10.2 − 3.29 = 6.91 kvar. At 230 V and 50 Hz: C = Q_C/(2πf·V²) = 6910/(314·230²) = 4.16·10⁻⁴ F ≈ 416 µF.
Marks lost
- Using peak instead of rms values.
- Forgetting cos φ, or reporting kVA as if it were kW.
- Adding impedances arithmetically instead of vectorially.
- Inverting the transformer current ratio.
- Wrong De Morgan application, or failing to give the minimum simplified expression.