Electrical Engineering Fundamentals: practice problems and exam questions

The electrical course that mechanical, civil and chemical engineering students take: DC circuits, equivalent circuits, transients and AC power. The problems are shorter than in an electrical engineering degree, but they cover a lot of ground.

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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. Series & parallel resistors3 resistors · 4 resistors · ladder of 6 resistorscalculations
  2. Nodal analysis1 unknown node · 2 unknown nodes · 2 nodes, 3 sourcescalculations
  3. Thévenin & Nortondivider network · with series resistor and load · with current source and loadcalculations
  4. RC & RL transientsRC charging · RL circuit · capacitor with initial chargecalculations
  5. AC impedance & phasorsseries RL · series RLC · capacitor voltage and resonancecalculations
  6. AC power & power factorP and Q of a load · line current · power factor correctioncalculations
  7. Three-phase powerbalanced Y and Δ voltages and currents · three-phase power · power factor correctioncalculations
  8. Transformersturns ratio · currents and power · efficiency and lossescalculations

Exam tips for Electrical Engineering Fundamentals

  • Simplify series and parallel resistors first; many exam circuits shrink to one loop.
  • A capacitor acts as an open circuit and an inductor as a short circuit once a DC circuit has settled.
  • For AC power, write down P, Q and S as a triangle: S² = P² + Q².

Worked examples

Series & parallel resistorsEasy
R1​ = 5 ΩIR1​+−V1​18 VR2​6 ΩAR3​12 Ω

Find the current IR1 delivered by the source, which flows through R1.

V1 = 18 VR1 = 5 ΩR2 = 6 ΩR3 = 12 Ω

Find IR1 (A)

Show hints
  1. Reduce the network step by step. In series, resistances add. Two resistors in parallel combine as R₁R₂/(R₁ + R₂).
  2. R₂ and R₃ both join node A to the bottom wire, so they are in parallel. That pair is in series with R₁.
  3. Req = 9 Ω.
Show the answer and the step-by-step solution

Series and parallel resistors: in series the same current flows through each resistor and R = R₁ + R₂ + …. In parallel each resistor has the same voltage and 1/R = 1/R₁ + 1/R₂ + …. A ladder network is reduced one pair at a time until a single equivalent resistance remains.

R1​ = 5 Ω+−V1​18 VR2​6 ΩAR3​12 Ω

Combine resistors from the far end

Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.

R2 ∥ R3 = 6·12/(6+12) = 4 Ω
Req = R1 + R2 ∥ R3 = 5 + 4 = 9 Ω

Current from the source

The source sees only Req, so Ohm’s law gives the current it delivers. That current flows through R₁.

IR1 = V1/Req = 18/9 = 2 A

Answer

IR1 = 2 A
Nodal analysisMedium
−+Vs​ = 3 VIx​Is​1.5 AAR1​6 ΩR2​3 ΩB

The voltage source Vₛ between A and B has its + terminal at B, so it makes VB − VA = Vₛ. Use nodal analysis with a supernode to find the current Ix through the voltage source, from A to B as shown by the arrow.

Is = 1.5 AVs = 3 VR1 = 6 ΩR2 = 3 Ω

Find Ix (A)

Show hints
  1. A voltage source between two non-ground nodes fixes their difference: VB − VA = Vs.
  2. Write one KCL equation for the supernode around A, B and the source: VA/R₁ + VB/R₂ = Is.
  3. VA = 1 V.
Show the answer and the step-by-step solution

Nodal analysis: choose a reference node, label the unknown node voltages, write Kirchhoff’s current law at each unknown node using Ohm’s law (I = ΔV/R), and solve the resulting equations.

Form a supernode

The voltage source between A and B carries an unknown current, so KCL cannot be written at A or B separately. Enclose A, B and the source in one supernode: the current entering it equals the current leaving it.

VB = VA + Vs = VA + 3
VA/R1 + VB/R2 = Is

Solve

VA/6 + (VA + 3)/3 = 1.5
VA = (1.5 − 3/3)/(1/6 + 1/3) = 1 V
VB = 1 + 3 = 4 V

Current through the voltage source

Everything that passes through the source from A to B leaves node B through R₂.

Ix = VB/R2 = 4/3 = 1.333 A

Answer

Ix = 1.333 A

A negative result means the real current flows against the arrow.

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

Series and parallel resistor calculatorVoltage divider calculatorRC time constant calculatorRLC impedance calculatorPower factor correction calculatorThree-phase power calculatorTransformer turns ratio calculatorVoltage drop calculatorThévenin and Norton equivalent calculatorOhm’s law calculator (voltage, current, resistance and power)Resistor colour code calculator (4, 5 and 6 bands)Capacitor calculator (series, parallel, charge and energy)

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FE Electrical and Computer exam practiceCircuit Analysis 2: AC circuitsCircuit Analysis 1Power Systems and Electrical MachinesPhysics 2: Electricity and MagnetismElectronics 1Power Electronics and DrivesMeasurement and Instrumentation
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