Electrical and electronic systems: DC, AC, electromagnetism and components
The exam combines circuit calculation with component behaviour: resistive networks, series RLC in alternating current, transformer ratios, and diode/transistor/op-amp circuits used as switches and amplifiers.
Direct current
- Ohm's law, series and parallel equivalents, voltage divider Vout = V·R2/(R1+R2), Kirchhoff's current and voltage laws, Thevenin equivalent for a two-terminal network.
- Power P = V·I = I²R = V²/R; a resistor is rated by the power it can dissipate.
Alternating current
- RMS values: Vrms = Vp/√2; mains 230 V is an RMS value, so Vp ≈ 325 V.
- XL = 2πfL, XC = 1/(2πfC), Z = √(R² + (XL − XC)²), cos φ = R/Z.
- P = V·I·cos φ (W), Q = V·I·sin φ (var), S = V·I (VA); resonance at f0 = 1/(2π√(LC)), where Z = R and cos φ = 1.
Electromagnetism
Faraday–Lenz: e = −N·ΔΦ/Δt; the minus sign is the opposition of the induced effect. Ideal transformer: V1/V2 = N1/N2 = I2/I1, with S1 = S2; real losses are copper (I²R) and iron (hysteresis, eddy currents).
Electronics
- Silicon diode: 0.7 V forward drop; half-wave and bridge rectifiers; Zener diode regulating in reverse breakdown.
- NPN transistor as a switch: VBE ≈ 0.7 V, saturation when IB > IC/β.
- Op-amp: inverting gain −R2/R1, non-inverting 1 + R2/R1, output limited by the supply rails (±Vcc).
Worked example
Series RLC with R = 30 Ω, XL = 80 Ω, XC = 40 Ω at 230 V, 50 Hz. Z = √(30² + 40²) = 50 Ω; I = 230/50 = 4.6 A; cos φ = 30/50 = 0.6 inductive; P = 230 × 4.6 × 0.6 = 635 W; Q = 230 × 4.6 × 0.8 = 846 var; S = 1058 VA. Check: √(635² + 846²) ≈ 1058 VA.
Traps that cost marks
- Using peak voltage where RMS is required, or vice versa.
- Adding XL and XC instead of subtracting them.
- Forgetting the 0.7 V diode drop in series paths.
- Expecting an op-amp output beyond its supply rails.
- Believing power factor correction changes the active power consumed.