This package prepares candidates for 電験二種 (Second-Class Electric Chief Engineer), covering the four primary multiple-choice subjects and the two secondary descriptive subjects, 電力・管理 and 機械・制御. The 20 MCQs are an original supporting-knowledge check, not a simulation of the written secondary papers; use the integrated task to practise open written answers. Confirm current eligibility, subject exemptions, calculator rules, fees, dates and pass criteria with 電気技術者試験センター (JETEC). Use the primary subjects to build calculation fluency and legal/safety awareness, then practise full written answers for 電力・管理 and 機械・制御. The 20 MCQs check supporting knowledge only; they are not a simulation of the descriptive secondary papers. Confirm current rules with 電気技術者試験センター (JETEC).
Credential, scope and assessment structure
The credential is the second-class electric chief engineer qualification in Japan. Its published structure has four primary subjects and two secondary written subjects.
The overview describes knowledge required to serve as chief electrical engineer for business electrical installations below 170 kV. It is not the same credential as the second-class electric construction engineer, and it is not a general electrical-worker licence.
The primary subjects are 理論 (electrical/electronic theory and measurement), 電力 (generation, storage and substation design/operation; transmission and distribution design/operation including indoor wiring; electrical materials), 機械 (machines, power electronics, motor applications, lighting, heating, electrochemistry, electrical machining, automatic control, mechatronics and information transmission/processing), and 法規 (electrical laws limited to safety and electrical facility management).
The primary subjects use multiple-choice mark-sheet answers. The secondary stage is descriptive/written and has two subjects: 電力・管理 and 機械・制御. The secondary is available to same-year primary passers and to those exempt from the primary. The published subject-pass system applies to the primary: a passed subject can be exempted by application in the following two years. A primary pass also exempts the next year's primary if the secondary is failed that year. Confirm current rules, dates and exemptions with JETEC.
Calculators: the published rules allow four arithmetic operations and square root, but prohibit function calculators, programmable/expression-storing calculators and printing calculators; the device must not make sound, and you must bring your own. Treat this as a rule to re-check each year with JETEC.
Preparation priorities
Build the primary stage first, then convert the same technical base into written secondary answers.
The primary stage is the gate. Aim for broad competence rather than one favourite subject. Theory and machinery carry many formula-driven marks; power systems combine theory with network judgement; law/safety rewards careful reading, record-keeping logic and practical protective reasoning. Use a two-pass method: first make a one-page map of formulas and definitions, then solve mixed original problems under time pressure.
For the secondary, do not simply reread primary notes. Practise multi-part written answers that require a calculation, an assumption, a diagram or equivalent explanation, and an operational or safety measure. State units, show intermediate steps and explain why a protective or operating choice is appropriate. Confirm the current examination schedule and subject rules with JETEC.
Electrical theory: phasors, impedance, resonance and transients
Use rms phasors, keep real and reactive power separate, and check whether a circuit is series or parallel before applying a formula.
For a series R–L circuit at 200 V, 50 Hz, R = 6 Ω and X_L = 8 Ω, the impedance magnitude is √(6² + 8²) = 10 Ω. Current is 200/10 = 20 A. Power factor is R/Z = 0.6 lagging. Active power is I²R = 400 × 6 = 2,400 W; reactive power is I²X = 3,200 var; apparent power is 4,000 VA. This consistency check is faster than juggling several formulas.
Resonance occurs when the reactances cancel; for series resonance, f = 1/(2π√(LC)). In an RC charging circuit, τ = RC and the capacitor voltage follows V(1 − e^(−t/τ)); after one time constant it reaches about 63.2% of the supply. Faraday’s law gives average induced emf as NΔΦ/Δt: 100 turns changing by 4 mWb in 0.05 s gives 8 V. Always keep flux units consistent.
Power systems: per-unit values, faults, voltage drop and compensation
Per-unit conversion, three-phase fault current, voltage drop and reactive compensation are the core calculation skills.
On a common voltage base, convert impedance by the MVA-base ratio: a 12 MVA transformer with 9% impedance becomes 9% × (100/12) = 75% on a 100 MVA base. For a three-phase fault, first find the base current I_base = S_base/(√3 V_LL), then I_f = I_base/Z_pu. On a 10 MVA, 11 kV base, I_base ≈ 525 A; with Z = 0.25 pu the fault current is about 2.10 kA. If voltage bases differ, include the voltage-ratio squared term.
Approximate line-to-neutral voltage drop is I(R cosφ + X sinφ); multiply by √3 for line-to-line drop in a balanced three-phase feeder. For power-factor correction, Q_C = P(tanφ₁ − tanφ₂). An 800 kW load corrected from 0.8 to 0.95 lagging needs about 800 × (0.750 − 0.329) ≈ 337 kvar. In grounding and protection, distinguish high-current resistance-grounded faults from small capacitive currents in an ungrounded system; the latter can cause healthy-phase voltage rise and require voltage/directional ground detection rather than plain overcurrent.
Machinery: slip, losses, excitation and back emf
Machine questions reward careful use of speed, loss and excitation relationships.
For an induction motor, n_s = 120f/p and n_r = n_s(1 − s). A 6-pole, 60 Hz motor has n_s = 1,200 rpm; at 3% slip, n_r = 1,164 rpm. In the normal operating region, increasing load increases slip and torque until the breakdown point; starting methods reduce starting current or improve torque but do not change the synchronous-speed formula.
Transformer efficiency is output/(output + iron loss + copper loss). Copper loss scales with load current squared. For a 100 kVA transformer with 800 W iron loss and 1,600 W full-load copper loss, half-load unity-power-factor output is 50 kW, copper loss is 400 W, and efficiency is 50,000/(50,000 + 800 + 400) ≈ 97.7%. Voltage regulation uses per-phase or line-to-line consistency. A separately excited DC motor has E_b = V − I_aR_a: 200 V, 0.4 Ω and 25 A gives 190 V. Overexcited synchronous generators supply lagging reactive power; underexcited operation absorbs reactive power.
Law, safety and system management
Treat safety as a sequence of verified barriers, not as a memorized slogan.
The chief engineer’s role is supervisory and safety-related: know the installation scope, maintain procedures and records, coordinate inspection and testing, and ensure that work is performed by competent persons under controlled conditions. The law subject is linked to safety-related electrical law and facility management. Because legal details can change, confirm current statutory requirements with JETEC and the current official legal text rather than relying on old summaries.
For safe isolation, identify the energy source, shut down, isolate, lock/tag, verify de-energized, and apply temporary protective grounding where required. Arc-flash prevention follows the same logic: de-energize first; if energized work is unavoidable, require risk assessment, boundaries, qualified persons and suitable PPE. Red flags include unknown or undocumented energy sources, damaged insulation, moisture, unauthorized modifications, repeated overheating and missing test records. Stop, secure the area, escalate and investigate rather than recording the defect and continuing.
Common errors to eliminate
Most lost marks come from mixing bases, phases, units and operating states.
Frequent errors include using line voltage where phase voltage is required; forgetting √3 in balanced three-phase power or voltage drop; mixing line and phase current; converting percent impedance without matching MVA and voltage bases; treating synchronous speed as rotor speed; using full-load copper loss at part load; and adding iron loss and copper loss to input rather than output. In transients, check whether the circuit is charging or discharging before choosing 1 − e^(−t/τ) or e^(−t/τ).
In written answers, another common error is jumping to a number without stating assumptions, units or the protective consequence. In safety questions, avoid shortcuts such as low voltage is safe, test once and work, or tighten it live. Use a verifiable barrier sequence and explain the residual risk. If using older past papers, treat them as practice of concepts and style, not as proof of current rules.
Integrated practice task and self-check rubric
Use this open task to practise the written style that the MCQs cannot simulate.
Scenario: a 6.6 kV industrial bus is fed by a 2 MVA transformer with 6% impedance. A 1.2 MW load operates at 0.85 lagging. A 4-pole, 50 Hz induction motor runs at 4% slip. You plan capacitors to reach 0.95 lagging and later inspect a terminal reported as warm. Tasks: (a) convert the transformer impedance to a 10 MVA base; (b) estimate the three-phase fault current at the bus from the transformer alone; (c) calculate the capacitor kvar; (d) find motor speed; (e) write a safe inspection and repair sequence; (f) state one assumption and one site record you would check.
Model route: (a) Z = 0.06 × (10/2) = 0.30 pu. (b) I_base = 10 MVA/(√3 × 6.6 kV) ≈ 875 A, so I_f ≈ 875/0.30 ≈ 2.92 kA. (c) Q_C = 1.2 MW × [tan(cos⁻¹0.85) − tan(cos⁻¹0.95)] ≈ 1.2 × (0.620 − 0.329) ≈ 0.349 Mvar, or about 349 kvar. (d) n_s = 120 × 50/4 = 1,500 rpm; n_r = 1,500 × 0.96 = 1,440 rpm. (e) De-energize, isolate, lock/tag, verify dead, apply protection as required, repair the connection, investigate the cause and record the finding. (f) Assumptions may include ignoring line resistance or upstream sources; check transformer data, load profile and protection settings.
Self-check rubric, 0–2 per item: base conversion with units; fault calculation with correct √3; compensation using tanφ; motor-speed formula; safety sequence with verification and records; written clarity, assumptions and cause analysis. This is a practice rubric, not an official pass mark.
A practical study sequence
Use short cycles of learning, solving, correcting and writing.
Phase 1: map the four primary subjects and build a formula sheet with units and validity conditions. Phase 2: run daily mixed drills—one theory calculation, one machinery calculation, one power-system calculation and one law/safety scenario. Keep an error log by cause: wrong base, wrong phase quantity, wrong loss model, incomplete safety barrier. Phase 3: solve timed mixed sets and explain why each distractor is wrong. Phase 4: practise secondary-style written answers with full calculations, assumptions and operational recommendations. Phase 5: rehearse the integrated task and reduce your error log. Confirm current dates and rules with JETEC; this sequence is a structure, not a guarantee.
Using the MCQs correctly
The 20 questions check supporting knowledge only; they are not a simulation of the descriptive secondary papers.
Answer without notes first, then write a one-sentence reason for the correct option and for one rejected option. For numerical items, recompute independently and check units. For safety items, identify the barrier or verification step that makes the answer safer. Because the secondary examination is written/descriptive, the MCQ sample cannot replace open-task practice: use the integrated task above to write full answers with diagrams or equivalent explanations, calculations, assumptions and management measures. This is a preview, not full syllabus coverage.
References and further reading
Use these references to explore the concepts and check the latest information from the relevant organizations.