Electrical Science and Principles
Electrical Science and Principles underpins every design and fault-finding question in the paper. The exam expects fluent, error-free manipulation of Ohm's law, power relationships, AC impedance and three-phase quantities, plus a working grasp of magnetism, transformers and motors.
Core relationships
- Ohm's law: V = IR; power: P = VI = I²R = V²/R.
- Series: resistances add, current is common. Parallel: 1/R = 1/R1 + 1/R2, voltage is common.
- AC: Vrms = Vpeak / √2. Inductive reactance XL = 2πfL, capacitive reactance XC = 1/(2πfC), impedance Z = √(R² + (XL − XC)²), and current lags or leads by the phase angle φ where cos φ is the power factor.
- Single-phase power: P = VI cos φ; reactive power Q = VI sin φ in VAr; apparent power S = VI in VA.
- Three-phase: P = √3 × VL × IL × cos φ for a balanced load, with VL = √3 × Vphase in star.
Magnetism, transformers and motors
A changing flux links a coil to induce an emf (Faraday); the transformer emf equation E = 4.44 f N Φ links turns, flux and voltage, giving the turns ratio V1/V2 = N1/N2. In a three-phase induction motor the stator field rotates at synchronous speed Ns = 120f / p, and the rotor runs slightly slower; the fractional slip is (Ns − Nr)/Ns.
Worked example
A 15 kW, 380 V, three-phase motor has efficiency 90% and power factor 0.82. Input power = 15 / 0.90 = 16.67 kW. Line current = 16 667 / (√3 × 380 × 0.82) = 16 667 / 539.6 = 30.9 A. If the power factor were corrected to 0.95, the capacitor bank would need Qc = P (tan φ1 − tan φ2) = 16.67 (0.698 − 0.329) = 6.2 kVAr, and the current would fall to about 26.7 A.
Traps that cost marks
Using 380 V in a single-phase formula; forgetting efficiency when converting output kW to input current; adding power factors arithmetically instead of resolving kW and kVAr vectorially; confusing kW, kVA and kVAr in the answer; and quoting peak voltage where the question asks for RMS.