Electrical Circuits and Network Analysis (DC and AC)
The circuits domain underpins every other ΘΚ topic: the exam expects fast, correct analysis of DC networks and single- and three-phase AC circuits, then reuse of the result (current, power factor, voltage drop) in installation and machine questions. Expect node/mesh analysis, Thevenin equivalents, phasor diagrams, the power triangle and balanced three-phase loads.
Core rules
- Ohm's law, KCL and KVL; superposition for multi-source linear networks; Thevenin/Norton reduction to a source plus impedance - the standard route to supply impedance and fault-level questions.
- AC quantities: v(t) = Vm sin(ωt), Vrms = Vm/√2, form factor 1.11 for a sine, ω = 2πf.
- Complex impedance Z = R + j(XL − XC) with XL = 2πfL and XC = 1/(2πfC); I = V/Z, carrying the angle through every step.
- Power: P = VI cosφ (W), Q = VI sinφ (var), S = VI (VA). Balanced three-phase: P = √3 VL IL cosφ. In star VL = √3 Vph and IL = Iph; in delta IL = √3 Iph and VL = Vph.
- Series resonance at f = 1/(2π√(LC)) gives minimum Z and maximum current; parallel resonance gives maximum Z.
Worked example
A 230 V, 50 Hz supply feeds a coil of R = 8 Ω in series with L = 19.1 mH. XL = 2π × 50 × 0.0191 = 6 Ω, so Z = √(8² + 6²) = 10 Ω and φ = arctan(6/8) = 36.9°. I = 230/10 = 23 A. Then P = I²R = 23² × 8 = 4.23 kW, S = 230 × 23 = 5.29 kVA and Q = √(S² − P²) = 3.17 kvar. To correct to unity power factor, add capacitors absorbing Qc = 3.17 kvar: C = Qc/(2πfV²) = 3170/(314 × 230²) ≈ 191 μF. The supply current then falls to P/V = 18.4 A, although the coil current itself is unchanged.
Traps that cost marks
- Adding rms values arithmetically in a series AC circuit instead of adding phasors or impedances.
- Using peak values in power formulas, or mixing kW and kVA inside an efficiency calculation.
- Forgetting that the neutral carries current in an unbalanced four-wire star load but not in a balanced one.
- Losing track of leading/lagging convention and of the sign of Q in the power triangle.
- Treating a delta load's phase voltage as the line voltage.