HK EMSD Grade C Electrical Worker - independent technical foundations study materials
HK EMSD Grade C Electrical Worker - independent technical foundations study resources

HK EMSD Grade C Electrical Worker - independent technical foundations: Electrical circuits, machines and digital logic

HK EMSD Grade C Electrical Worker - independent technical foundations · Work through your exam, one topic at a time.

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Electrical circuits, machines and digital logic

This domain covers DC and AC circuit analysis, transformers and rotating machines, power converters and Boolean logic. This foundations field 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 is found by deactivating independent sources appropriately while retaining dependent-source relations; a test source may be necessary).
  • Power P = V·I = I²·R.

Alternating current

  • For sinusoidal signals, V_rms = V_peak/√2. For a series RLC circuit, |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).
  • For a balanced sinusoidal three-phase load, P = √3·V_L·I_L·cos φ using line RMS values.

Machines and power electronics

  • Ideal-transformer magnitude relations: V₁/V₂ = N₁/N₂ = I₂/I₁; input and output apparent powers match in this idealization. Real devices have losses and operating limits.
  • Induction motor: n = (1 − s)·n₁ with n₁ = 60 f/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

In a stipulated single-phase sinusoidal model, a 10 kW inductive load has cos φ =0.7. The following is an ideal calculation, not capacitor installation advice. 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.