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Andalusia PAU Technology and Engineering II study materials
Andalusia PAU Technology and Engineering II study resources

Andalusia PAU Technology and Engineering II: Electrical and Electronic Systems: Circuits, Components, Machines and Power

Andalusia PAU Technology and Engineering II · Work through your exam, one topic at a time.

All guide sections

Electrical and Electronic Systems: Circuits, Components, Machines and Power

This domain moves from DC network analysis through single- and three-phase power to the electronics that drive machines. The PAU paper usually mixes a circuit calculation with a machine or power question, so you must be fluent in both DC and AC conventions.

DC circuits

  • Ohm's law and power: V = I·R, P = V·I = I²·R = V²/R.
  • Series: R total adds, current common. Parallel: 1/R total = Σ1/R, voltage common.
  • Kirchhoff: sum of currents at a node is zero; sum of voltage drops around a loop is zero, with consistent signs.
  • Thevenin: any linear two-terminal network reduces to Vth (open-circuit voltage) and Rth (resistance seen with sources killed).

AC and machines

  • RMS values: Vrms = Vpeak/√2 and Irms = Ipeak/√2; mains 230 V is an RMS value.
  • Single phase: P = V·I·cos φ, S = V·I, Q = V·I·sin φ.
  • Three phase: P = √3·V_L·I_L·cos φ. In star V_L = √3·V_phase and I_L = I_phase; in delta I_L = √3·I_phase and V_L = V_phase.
  • Transformer: V₁/V₂ = N₁/N₂ = I₂/I₁; ideal transformers conserve apparent power.
  • Induction motor: synchronous speed ns = 60·f/p (p = pole pairs) and slip s = (ns − n)/ns.

Electronics

  • A diode conducts in forward bias; a bridge rectifier converts AC to pulsating DC.
  • The transistor works as a switch (cut-off/saturation) or as an amplifier; the op-amp inverting gain is −Rf/Rin.
  • Logic gates: AND, OR, NOT, NAND, NOR, XOR; simplify with De Morgan and Karnaugh maps.

Worked example

A three-phase motor is connected to 400 V line voltage, draws 10 A per line at cos φ = 0.85, and has efficiency 0.90. Absorbed power P = √3·400·10·0.85 = 5 889 W ≈ 5.89 kW. Useful shaft power = 0.90 × 5 889 = 5 300 W ≈ 5.30 kW. The apparent power is S = √3·400·10 = 6 928 VA, so Q ≈ 3.65 kvar.

Traps that cost marks

  • Plugging peak voltage into an RMS formula, or the reverse.
  • Confusing line and phase quantities inside a delta connection.
  • Writing P = V·I for a motor without the cos φ factor, or forgetting efficiency.
  • Sign errors in Kirchhoff loops when sources oppose each other.