PracticeVideosBlogFlashcards
All Practice Exams

Free Practice Questions for Taiwan Electrical Engineering Technologist Examination

Exam-style questions and explanations by OpenExamPrep.

✓ No registration✓ No credit card
100+ Questions
100% Free

Loading practice questions...

Same family resources

Explore More Taiwan Professional Engineering Technologist Examinations (專技高考技師)

Continue into nearby exams from the same family. Each card keeps practice questions, study guides, flashcards, videos, and articles in one place.

Taiwan Civil Engineering Technologist Examination (專技高考土木工程技師)

Practice Questions100 questions

Taiwan Mechanical Engineering Technologist Examination (專技高考機械工程技師)

Practice Questions100 questions

Taiwan Structural Engineering Technologist Examination (專技高考結構工程技師)

Practice Questions99 questions
Exam Review

Key Facts: Taiwan Electrical Engineering Technologist Examination Exam

6 papers

Constructed-response subjects, equally averaged.

Rules Articles 9–10 and Appendix 2

12 hours

Six two-hour sessions.

MOEX 2026 timetable

NTD 2,300

Standard paper-based senior-examination fee.

MOEX fee schedule

Review electrical concepts and calculations in English, then practice full written answers using official papers and current Taiwan installation provisions.

Sample Taiwan Electrical Engineering Technologist Examination Practice Questions

Try these sample questions to review concepts for the Taiwan Electrical Engineering Technologist Examination exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1A bipolar junction transistor (BJT) is biased in the forward-active region with a collector DC bias current of I_C = 2.6 mA. Assuming room temperature operation with thermal voltage V_T = 26 mV and common-emitter current gain beta = 100, what are the small-signal transconductance g_m and small-signal base-emitter input resistance r_pi?
A.g_m = 100 mA/V, r_pi = 1.0 kΩ
B.g_m = 10 mA/V, r_pi = 10.0 kΩ
C.g_m = 50 mA/V, r_pi = 2.0 kΩ
D.g_m = 100 mA/V, r_pi = 38.5 Ω
Explanation: Transconductance is calculated as g_m = I_C / V_T = 2.6 mA / 26 mV = 0.1 A/V = 100 mA/V (or 100 mS). The small-signal input resistance looking into the base is r_pi = beta / g_m = 100 / 0.1 A/V = 1000 Ω = 1.0 kΩ.
2A common-source MOSFET amplifier has transconductance parameter k_n'(W/L) = 2.0 mA/V^2, overdrive voltage V_OV = V_GS - V_th = 1.0 V, channel-length modulation coefficient lambda = 0.02 V^-1, and drain load resistance R_D = 5.0 kΩ. What is the small-signal open-circuit voltage gain A_v = v_o / v_i?
A.-10.00 V/V
B.-9.09 V/V
C.-4.55 V/V
D.-18.18 V/V
Explanation: The DC drain current is I_D = 0.5 * k_n'(W/L) * (V_OV)^2 = 0.5 * 2.0 * (1.0)^2 = 1.0 mA. Transconductance is g_m = 2 * I_D / V_OV = 2 * 1.0 mA / 1.0 V = 2.0 mA/V. Small-signal output resistance is r_o = 1 / (lambda * I_D) = 1 / (0.02 * 1.0 mA) = 50 kΩ. The overall gain is A_v = -g_m * (R_D || r_o) = -2.0 mA/V * (5 kΩ || 50 kΩ) = -2.0 * (250 / 55) = -2.0 * 4.545 kΩ = -9.09 V/V.
3An operational amplifier with finite open-loop DC gain A_0 = 1,000 is connected as a non-inverting amplifier with feedback resistors R_1 = 1 kΩ (grounded branch) and R_2 = 9 kΩ (feedback branch). What is the exact closed-loop voltage gain A_CL of this non-inverting amplifier?
A.10.00 V/V
B.9.09 V/V
C.9.90 V/V
D.9.99 V/V
Explanation: The ideal closed-loop gain is A_ideal = 1 + R_2 / R_1 = 1 + 9/1 = 10 V/V, and the feedback factor is beta = R_1 / (R_1 + R_2) = 1 / 10 = 0.1. Accounting for finite open-loop gain A_0, the actual closed-loop gain is A_CL = A_0 / (1 + A_0 * beta) = 1,000 / (1 + 1,000 * 0.1) = 1,000 / (1 + 100) = 1,000 / 101 ≈ 9.901 V/V.
4A second-order unity-gain low-pass Sallen-Key filter uses equal resistors R_1 = R_2 = 10 kΩ and equal capacitors C_1 = C_2 = 15.915 nF. What are its natural pole frequency f_0 = omega_0/(2*pi) and quality factor Q?
A.f_0 = 1.0 kHz, Q = 0.707
B.f_0 = 6.28 kHz, Q = 0.50
C.f_0 = 2.0 kHz, Q = 1.00
D.f_0 = 1.0 kHz, Q = 0.50
Explanation: For this equal-component unity-gain topology, f_0 = 1/(2*pi*R*C) = 1.00 kHz and Q = 0.50. Q = 0.50 gives critical damping with a repeated real pole, not an overdamped response. The natural frequency is distinct from the -3 dB cutoff, which is about 0.644*f_0 for this response.
5An operational amplifier has a specified slew rate of SR = 2.0 V/μs. If the amplifier produces an undistorted sinusoidal output voltage v_o(t) = V_p * sin(2*pi*f*t) with peak amplitude V_p = 10.0 V, what is the maximum allowable signal frequency f_max (full-power bandwidth)?
A.31.83 kHz
B.100.0 kHz
C.20.0 kHz
D.15.92 kHz
Explanation: The maximum rate of change of a sinusoidal signal is (d v_o / dt)_max = 2 * pi * f * V_p. Slew rate limitation requires 2 * pi * f_max * V_p <= SR. Substituting the given values: f_max = SR / (2 * pi * V_p) = (2.0 * 10^6 V/s) / (2 * pi * 10.0 V) = 100,000 / pi ≈ 31.83 kHz.
6A differential instrumentation amplifier exhibits a differential-mode voltage gain of A_d = 80 dB and a common-mode voltage gain of A_cm = -40 dB. What is the Common-Mode Rejection Ratio (CMRR) in decibels?
A.40 dB
B.120 dB
C.80 dB
D.60 dB
Explanation: CMRR in decibels is defined as CMRR_dB = 20 * log10(|A_d / A_cm|) = 20 * log10(|A_d|) - 20 * log10(|A_cm|). Substituting the given values: CMRR_dB = 80 dB - (-40 dB) = 120 dB (in linear terms, A_d = 10,000 and A_cm = 0.01, so CMRR = 1,000,000 = 120 dB).
7An open-loop amplifier has a forward voltage gain A = 10,000 V/V and an input resistance R_in = 20 kΩ. Negative feedback is applied using a series-shunt (voltage-mixing, voltage-sensing) topology with feedback factor beta = 0.009. What are the closed-loop voltage gain A_f and closed-loop input resistance R_in,f?
A.A_f = 111.11 V/V, R_in,f = 219.8 Ω
B.A_f = 109.89 V/V, R_in,f = 20 kΩ
C.A_f = 109.89 V/V, R_in,f = 1.82 MΩ
D.A_f = 1000.0 V/V, R_in,f = 200 kΩ
Explanation: The amount of feedback is 1 + A * beta = 1 + 10,000 * 0.009 = 1 + 90 = 91. The closed-loop gain is A_f = A / (1 + A * beta) = 10,000 / 91 ≈ 109.89 V/V. For a series input connection (series-shunt feedback), the input resistance is multiplied by the feedback amount: R_in,f = R_in * (1 + A * beta) = 20 kΩ * 91 = 1820 kΩ = 1.82 MΩ.
8A step-down (buck) DC-DC converter operates in continuous conduction mode (CCM) with DC input voltage V_in = 48.0 V, switching frequency f_s = 100 kHz, and duty ratio D = 0.375. Assuming ideal components with zero voltage drops across the switch and diode, what is the average output DC voltage V_o?
A.12.0 V
B.24.0 V
C.30.0 V
D.18.0 V
Explanation: In an ideal buck converter operating in continuous conduction mode, the DC conversion ratio is V_o / V_in = D. Thus, V_o = D * V_in = 0.375 * 48.0 V = 18.0 V.
9A step-up (boost) DC-DC converter operates in continuous conduction mode (CCM) with DC input voltage V_in = 24.0 V, regulated output voltage V_o = 60.0 V, switching frequency f_s = 50 kHz, and load resistance R = 30.0 Ω. What is the critical minimum inductance L_min required to maintain continuous conduction mode at this load?
A.28.8 μH
B.57.6 μH
C.14.4 μH
D.72.0 μH
Explanation: In CCM, the duty ratio is D = 1 - (V_in / V_o) = 1 - 24/60 = 0.60. Average output current is I_o = V_o / R = 60 / 30 = 2.0 A. Average inductor current is I_L = I_o / (1 - D) = 2.0 / 0.40 = 5.0 A. At the CCM/DCM boundary, peak-to-peak inductor ripple is Delta_i_L = 2 * I_L = 10.0 A. Since Delta_i_L = (V_in * D) / (L * f_s), the critical inductance is L_min = (V_in * D) / (Delta_i_L * f_s) = (24 * 0.60) / (10.0 * 50,000) = 14.4 / 500,000 = 28.8 * 10^-6 H = 28.8 μH. (Direct formula: L_min = D * (1 - D)^2 * R / (2 * f_s) = 0.60 * 0.16 * 30 / 100,000 = 28.8 μH).
10An ideal conventional inverting buck-boost DC-DC converter operates in continuous conduction mode with input voltage V_in = 12.0 V and duty cycle D = 0.60. What is the steady-state output voltage relative to the common input ground?
A.+18.0 V (non-inverted polarity)
B.-18.0 V (inverted polarity)
C.-8.0 V (inverted polarity)
D.+30.0 V (non-inverted polarity)
Explanation: For an inverting buck-boost converter in CCM, the steady-state output voltage is V_o = -V_in * (D / (1 - D)) = -12.0 V * (0.60 / 0.40) = -12.0 * 1.5 = -18.0 V. The output voltage magnitude is 18.0 V, and its polarity is negative (inverted) relative to the common input ground.

About the Taiwan Electrical Engineering Technologist Examination Exam

Independent English-language MCQ study adaptation for 電機工程技師. Ordinary official papers are Chinese constructed-response examinations. These questions review selected principles and calculations, not an official translation, format simulation, or substitute for complete written derivations and design work. Article 12's special reciprocal route has separate language and assessment provisions.

Exam sponsor: Ministry of Examination (考選部 / MOEX). The requirements and fees below concern the certification or admission exam, separate from our free practice resources.

Assessment

2026 day 1: 電子學（包括電力電子學）, 電力系統, 電路學. Day 2: 工程數學（包括線性代數、微分方程、複變函數與機率）, 電機機械, 工業配電. Cases and designs are answered within written papers. The ordinary exam has no separate mandatory assignment, oral, interview, or hands-on practical stage.

Time Limit

12 testing hours: six 2-hour papers on November 14–15, 2026.

Passing Score

Article 10: mean at least 60 and no zero; if fewer than 16% qualify, the top-16% mechanism requires mean at least 50 and no zero, with cutoff ties included.

Exam / Certification Fees

NTD 2,300 standard registration fee.

Exam sponsor website

Fees, eligibility, and exam policies can change. Confirm them with the exam sponsor before applying or paying.

Official sources

Our practice resources: topics covered

We aim to reflect publicly available exam outlines and topic information in our study resources. Coverage, format, and difficulty may differ from the actual exam, and we cannot guarantee that every detail is accurate or current. Confirm exam requirements, fees, and policies with the official exam sponsor.

17 study questions; 1/6 subject weight

Electronics (電子學，包括電力電子學)

Devices, amplifiers, filters, switching converters, and rectifiers.

17 study questions; 1/6 subject weight

Electric Circuits (電路學)

Network analysis, AC power, transients, and equivalent circuits.

16 study questions; 1/6 subject weight

Engineering Mathematics (工程數學，包括線性代數、微分方程、複變函數與機率)

Linear systems, differential equations, transforms, complex analysis, and probability.

17 study questions; 1/6 subject weight

Electric Machinery (電機機械)

Magnetic circuits, transformers, and rotating machines.

17 study questions; 1/6 subject weight

Power Systems (電力系統)

Lines, power flow, faults, protection, stability, and operation.

16 study questions; 1/6 subject weight

Industrial Power Distribution (工業配電)

Installation rules, grounding, protection, capacity planning, and power quality.

Preparing for the Taiwan Electrical Engineering Technologist Examination Exam

What You Need to Know

  • Passing score: Article 10: mean at least 60 and no zero; if fewer than 16% qualify, the top-16% mechanism requires mean at least 50 and no zero, with cutoff ties included.
  • Assessment: 2026 day 1: 電子學（包括電力電子學）, 電力系統, 電路學. Day 2: 工程數學（包括線性代數、微分方程、複變函數與機率）, 電機機械, 工業配電. Cases and designs are answered within written papers. The ordinary exam has no separate mandatory assignment, oral, interview, or hands-on practical stage.
  • Time limit: 12 testing hours: six 2-hour papers on November 14–15, 2026.
  • Exam / certification fees: NTD 2,300 standard registration fee. Official sources

Using Our Practice Resources

  • Work through all 100 available questions
  • Review every answer and explanation
  • Track weak areas and revisit them
  • Use our AI tutor for tough concepts

Taiwan Electrical Engineering Technologist Examination: Suggested Study Strategy

1State phasor, per-unit, current-direction, and sign conventions.
2Check units and assumptions before calculating.
3Use current Taiwan electrical provisions rather than old article numbers or foreign-code defaults.
4Practice complete written circuit, machine, and power-system solutions.

Frequently Asked Questions

Who can apply?

Appendix 1 includes recognized junior-college-or-higher electrical graduates and eligible graduates with seven listed subjects and 20 credits, including specified core subjects. MOEX assesses credentials and other statutory routes.

Is the pass rate fixed at 16%?

No. Article 10's top-16% mechanism supplements the 60-average rule, subject to a 50 minimum and no zero subject. It is not an observed pass rate or an upper quota.

Is this an official English exam simulation?

No. This is independent English MCQ study practice, not an official translation or format simulation. Practice extended written responses and follow official paper instructions; Article 12's reciprocal route does not make every ordinary paper available in English.

Does passing grant a practice license?

Passing leads to examination certification. A technologist certificate and practice license involve separate steps under 技師法.