Power Systems and Electrical Machines: practice problems and exam questions

Three-phase systems, transformers, AC power and power factor correction: the power engineering that follows the circuits courses.

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Cram pack (PDF)Formula sheets, method steps and practice problems with full solutions, ready to print.

Topics in this course

  1. Three-phase powerbalanced Y and Δ voltages and currents · three-phase power · power factor correctioncalculations
  2. Transformersturns ratio · currents and power · efficiency and lossescalculations
  3. AC power & power factorP and Q of a load · line current · power factor correctioncalculations
  4. AC impedance & phasorsseries RL · series RLC · capacitor voltage and resonancecalculations
  5. DC motorsback-EMF, current and torque · speed–torque and efficiency · starting and speed controlcalculations

Exam tips for Power Systems and Electrical Machines

  • Always check whether a voltage is line-to-line or per phase.
  • Total three-phase power is √3·V_L·I_L·cos φ, or three times the per-phase power.
  • An ideal transformer keeps the power: voltage goes down by the turns ratio, current goes up by it.

Worked examples

Three-phase powerEasy
ABCVL​ = 400 VIL​ = ?Z = 24 + j7 Ωper phase

A balanced Δ-connected load has an impedance of 24 + j7 Ω per phase. It is supplied from a balanced three-phase source with a line-to-line voltage of 400 V. Find the magnitude of the line current IL.

VL = 400 VZ = 24 + j7 Ω per phase

Find IL (A)

Show hints
  1. In a Δ connection each impedance is connected between two lines.
  2. So each impedance sees the full line-to-line voltage, and IL = √3·Iph.
  3. Iph = 400/25 = 16 A.
Show the answer and the step-by-step solution

Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ.

Impedance magnitude

|Z| = √(24² + 7²) = 25 Ω

Phase current

Each impedance is connected between two lines and sees VL.

Iph = VL/|Z| = 400/25 = 16 A

Line current

IL = √3·Iph = √3·16 = 27.71 A

Answer

IL = 27.71 A
TransformersMedium
4 ΩN1​N2​256 Ω

An amplifier with an output resistance of 256 Ω drives a 4 Ω loudspeaker through an ideal transformer. For maximum power transfer the loudspeaker must look like 256 Ω from the amplifier. Find the turns ratio a = N₁/N₂.

Rs = 256 ΩRL = 4 Ω

Find a

Show hints
  1. Seen from the primary, the load looks like a²·RL.
  2. Set a²·RL equal to the amplifier’s output resistance.
  3. a² = 256/4.
Show the answer and the step-by-step solution

Ideal transformer: the voltages are in the ratio of the turns and the currents in the inverse ratio: V₁/V₂ = N₁/N₂ = a and I₁/I₂ = N₂/N₁ = 1/a, so the power in equals the power out, V₁·I₁ = V₂·I₂. A load impedance Z on the secondary looks like a²·Z from the primary.

Match the referred load

a²·4 = 256
a = √(256/4) = √64 = 8

Answer

a = 8

How practising works

  • “Practise this course” mixes problems from all its topics, like a real exam.
  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Mock exams are timed and have no hints, and you get your score and a full review at the end.

Calculators for this course

RLC impedance calculatorPower factor correction calculatorThree-phase power calculatorTransformer turns ratio calculatorVoltage drop calculator

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Practise this course for freePractise this courseTake a mock exam5 free problems every day. No card needed.