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Registered Electrical Engineer — Supply & Distribution (注册电气工程师(供配电), China)

Practice Questions100 questions

2026 Statistics

Key Facts: Registered Electrical Engineer (G&T) Exam

2 Stages

Examination structure (Foundation Exam + Professional Exam)

MOHURD / MOHRSS Registered Engineer System

1 Year

Non-rolling cycle: all subjects per stage must be passed in 1 sitting

National Professional Qualification Examination Regulations

60%

Fixed pass mark as a percentage of total marks across all papers

MOHRSS National Fixed Passing Standards Notice

5% - 12%

Typical national annual pass rate for Professional stage

CPTA Examination Statistics

3 Years

Practising registration renewal cycle with continuing education

MOHURD Electrical Engineer Registration Measures

Lifelong

Statutory legal liability of registered engineer for stamped designs

PRC Construction Law & Quality Regulations

The Registered Electrical Engineer (Generation & Transmission) exam is a national survey-and-design professional qualification. Its professional stage covers power-system analysis, generation, transmission lines, substations, relay protection, insulation coordination, and grounding. This bank is an English-language MCQ study adaptation and does not simulate the official case-analysis work.

Sample Registered Electrical Engineer (G&T) Practice Questions

Try these sample questions to test your Registered Electrical Engineer (G&T) exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1In power system network analysis, how is the diagonal element $Y_{ii}$ (self-admittance) of the nodal admittance matrix $\mathbf{Y}_{bus}$ formulated for a power network without mutual coupling?

A.The sum of the admittances of all branches directly connected to node $i$, including shunt admittances to ground at node $i$

B.The negative sum of the admittances of all branches connected between node $i$ and all other nodes $j$

C.The reciprocal of the sum of the series impedances of all branches terminating at node $i$

D.The sum of the off-diagonal elements in row $i$ divided by the total number of connected nodes

Explanation: Under standard power network matrix theory (DL/T 5218 and power system analysis specifications), the diagonal element $Y_{ii}$ of the nodal admittance matrix $\mathbf{Y}_{bus}$ equals the sum of the admittances of all branches terminating at node $i$, including line charging admittances and shunt capacitor/reactor admittances connected between node $i$ and reference ground ($Y_{ii} = y_{i0} + \sum_{j \neq i} y_{ij}$). The off-diagonal element $Y_{ij}$ equals the negative of the series admittance between nodes $i$ and $j$ ($Y_{ij} = -y_{ij}$).

2In Newton-Raphson power flow calculations, if Node 2 is a PV bus (generator bus) with specified active power generation $P_{G2} = 1.2\text{ p.u.}$ and voltage magnitude $|V_2| = 1.05\text{ p.u.}$, which variables are treated as unknowns in the iterative Jacobian formulation?

A.Voltage phase angle $\theta_2$ and reactive power injection $Q_2$

B.Voltage magnitude $|V_2|$ and voltage phase angle $\theta_2$

C.Active power injection $P_2$ and reactive power injection $Q_2$

D.Voltage magnitude $|V_2|$ and active power injection $P_2$

Explanation: At a PV bus (generator node), the active power $P$ and voltage magnitude $|V|$ are specified known quantities. The state variable to be solved in the iterative power flow algorithm is the voltage phase angle $\theta_2$. Once convergence is achieved, the reactive power injection $Q_2$ is calculated directly from the converged voltage state and checked against generator reactive power limits ($Q_{min} \le Q_2 \le Q_{max}$).

3In the method of symmetrical components, the complex phase operator is defined as $a = e^{j 120^\circ} = -\frac{1}{2} + j\frac{\sqrt{3}}{2}$. Which of the following algebraic relationships is INCORRECT?

A.$1 + a + a^2 = 1$

B.$a^3 = 1$

C.$a^2 = e^{j 240^\circ} = -\frac{1}{2} - j\frac{\sqrt{3}}{2}$

D.$1 - a = \sqrt{3} e^{-j 30^\circ}$

Explanation: The sum of the three balanced symmetrical component unit phasors is identically zero: $1 + a + a^2 = 1 + \left(-\frac{1}{2} + j\frac{\sqrt{3}}{2}\right) + \left(-\frac{1}{2} - j\frac{\sqrt{3}}{2}\right) = 0$. Therefore, stating $1 + a + a^2 = 1$ is mathematically incorrect. All other statements are valid properties: $a^3 = 1$, $a^2 = -0.5 - j0.866$, and $1 - a = 1.5 - j0.866 = \sqrt{3} e^{-j 30^\circ}$.

4A single line-to-ground (LG) solid fault occurs on Phase A of an unloaded 220 kV power system with pre-fault phase voltage $E = 1.0\text{ p.u.}$. The system sequence impedances viewed from the fault point are $Z_1 = j0.12\text{ p.u.}$, $Z_2 = j0.12\text{ p.u.}$, and $Z_0 = j0.24\text{ p.u.}$. What is the magnitude of the total fault current $I_A$ in per-unit?

A.$6.25\text{ p.u.}$

B.$2.08\text{ p.u.}$

C.$4.17\text{ p.u.}$

D.$8.33\text{ p.u.}$

Explanation: For a single line-to-ground (LG) fault, the positive, negative, and zero sequence networks are connected in series. The symmetrical sequence currents are: $I_{A1} = I_{A2} = I_{A0} = \frac{E}{Z_1 + Z_2 + Z_0} = \frac{1.0}{j0.12 + j0.12 + j0.24} = \frac{1.0}{j0.48} = -j2.0833\text{ p.u.}$. The total fault current in the faulted phase A is $I_A = I_{A1} + I_{A2} + I_{A0} = 3 I_{A1} = 3 \times 2.0833 = 6.25\text{ p.u.}$.

5A bolted line-to-line (LL) fault occurs between Phase B and Phase C in a transmission network where $Z_1 = Z_2 = j0.15\text{ p.u.}$ and $E = 1.0\text{ p.u.}$. What is the magnitude of the fault current flowing between the faulted phases ($|I_B| = |I_C|$)?

A.$5.77\text{ p.u.}$

B.$3.33\text{ p.u.}$

C.$6.67\text{ p.u.}$

D.$2.89\text{ p.u.}$

Explanation: For a bolted line-to-line (B-C) fault, the boundary conditions give $I_{A0} = 0$, $I_{A1} = -I_{A2} = \frac{E}{Z_1 + Z_2} = \frac{1.0}{j0.15 + j0.15} = -j3.333\text{ p.u.}$. The phase current is $I_B = a^2 I_{A1} + a I_{A2} = (a^2 - a) I_{A1} = -j\sqrt{3} I_{A1} = -j\sqrt{3} (-j3.333) = -\sqrt{3} \times 3.333 = -5.7735\text{ p.u.}$. The magnitude is $|I_B| = \sqrt{3} \times 3.333 = 5.77\text{ p.u.}$, which is $\frac{\sqrt{3}}{2} = 86.6\%$ of the three-phase fault current ($6.67\text{ p.u.}$).

6In symmetrical component fault analysis, how are the positive-, negative-, and zero-sequence networks interconnected at the fault point for a bolted double line-to-ground (LLG) fault?

A.All three sequence networks ($Z_1, Z_2, Z_0$) are connected in parallel at the fault port

B.All three sequence networks ($Z_1, Z_2, Z_0$) are connected in series at the fault port

C.The positive and negative sequence networks are connected in series, and the zero sequence network is open-circuited

D.The positive and zero sequence networks are connected in parallel, and the negative sequence network is open-circuited

Explanation: Under standard symmetrical component boundary conditions for a bolted double line-to-ground (B-C-G) fault ($V_B = V_C = 0, I_A = 0$), the sequence voltages are equal: $V_{A1} = V_{A2} = V_{A0}$. Consequently, the positive-, negative-, and zero-sequence networks are connected in parallel at the fault port. The sequence current is $I_{A1} = \frac{E}{Z_1 + \frac{Z_2 Z_0}{Z_2 + Z_0}}$.

7For a salient-pole synchronous generator, which relationship accurately ranks its fundamental sequence reactance parameters during power system fault calculations?

A.$X_d > X'_d > X''_d \approx X_2 > X_0$

B.$X_0 > X_2 > X_d > X'_d > X''_d$

C.$X''_d > X'_d > X_d > X_2 > X_0$

D.$X_d = X_2 = X_0 > X'_d > X''_d$

Explanation: In synchronous machine theory, the steady-state direct-axis synchronous reactance $X_d$ is the largest (typically $1.0 - 2.2\text{ p.u.}$). When a sudden fault occurs, the damper windings and rotor circuits trap magnetic flux, yielding the transient reactance $X'_d$ ($0.2 - 0.4\text{ p.u.}$) and subtransient reactance $X''_d$ ($0.12 - 0.25\text{ p.u.}$). The negative-sequence reactance $X_2 \approx \frac{X''_d + X''_q}{2} \approx X''_d$. The zero-sequence reactance $X_0$ ($0.03 - 0.10\text{ p.u.}$) is the smallest because spatial third-harmonic air-gap MMFs cancel out. Thus, $X_d > X'_d > X''_d \approx X_2 > X_0$.

8A step-up transformer has a delta primary winding connected to a generator and an ungrounded wye (Y) secondary winding connected to a 110 kV transmission line. Viewed from the 110 kV transmission line, what is the zero-sequence impedance of this transformer?

A.Infinite (open-circuit)

B.Equal to the transformer leakage reactance $X_T$

C.Equal to the zero-sequence magnetizing impedance $X_{m0}$

D.Zero (short-circuit to ground)

Explanation: According to transformer zero-sequence equivalent network rules (GB/T 14285 and Power Engineering Electrical Design Manual), zero-sequence currents can only flow into a wye winding if its neutral point is grounded. Since the 110 kV secondary wye neutral is ungrounded (floating), no zero-sequence current can enter from the line, making the zero-sequence impedance viewed from the 110 kV side infinite (open circuit).

9A 500 kV substation bus is fed by an equivalent system with positive-sequence reactance $X_1 = 0.05\text{ p.u.}$ on a base of $S_B = 1000\text{ MVA}, U_B = 500\text{ kV}$. What is the three-phase initial symmetrical short-circuit current $I''_k$ at this bus?

A.$23.09\text{ kA}$

B.$40.00\text{ kA}$

C.$11.55\text{ kA}$

D.$20.00\text{ kA}$

Explanation: 1) Calculate the base current: $I_B = \frac{S_B}{\sqrt{3} U_B} = \frac{1000\text{ MVA}}{\sqrt{3} \times 500\text{ kV}} = 1.1547\text{ kA}$. 2) Calculate the per-unit short-circuit current: $I''_k(\text{p.u.}) = \frac{E}{X_1} = \frac{1.0}{0.05} = 20.0\text{ p.u.}$. 3) Convert to actual kiloamperes: $I''_k = 20.0 \times 1.1547\text{ kA} = 23.094\text{ kA}$. Alternatively, short-circuit capacity $S''_k = \frac{S_B}{X_1} = \frac{1000}{0.05} = 20,000\text{ MVA}$, so $I''_k = \frac{20000}{\sqrt{3} \times 500} = 23.09\text{ kA}$.

10A synchronous generator is connected to an infinite bus through a transmission line. The internal EMF is $E' = 1.15\text{ p.u.}$, the infinite bus voltage is $U = 1.0\text{ p.u.}$, and total transfer reactance is $X_{\Sigma} = 0.50\text{ p.u.}$. Neglecting armature and line resistances, what is the maximum electromagnetic power $P_{max}$ that can be transferred under steady-state conditions?

A.$2.30\text{ p.u.}$

B.$1.15\text{ p.u.}$

C.$0.575\text{ p.u.}$

D.$4.60\text{ p.u.}$

Explanation: The steady-state power-angle characteristic is given by $P_e = \frac{E' U}{X_{\Sigma}} \sin\delta$. The maximum transferable power occurs at a power angle of $\delta = 90^\circ$, giving $P_{max} = \frac{E' U}{X_{\Sigma}} = \frac{1.15 \times 1.0}{0.50} = 2.30\text{ p.u.}$.

About the Registered Electrical Engineer (G&T) Exam

The China Registered Electrical Engineer Examination — Generation & Transmission (全国勘察设计注册电气工程师(发输变电)执业资格考试) is the national survey-and-design registered-engineer qualification for generation, transmission-line, substation, and power-system design. It assesses power-system analysis, equipment and substation arrangements, transmission design, protection, insulation coordination, and grounding. This bank is an English-language MCQ study adaptation, not an official translation or format simulation and not a substitute for the open-book professional case paper.

Assessment

Two-stage qualification: Stage 1 Foundation Exam (General Engineering Foundation 120 MCQs, 4.0 h + Professional Electrical Foundation 60 MCQs, 4.0 h; total 240 marks); Stage 2 Professional Exam (Day 1: Professional Knowledge AM 40 single + 30 multi, PM 40 single + 30 multi, 6.0 h total, 200 marks; Day 2: Professional Case Analysis AM 25 calculation cases, PM 25 calculation cases, 6.0 h total, 100 marks). Non-rolling annual qualification requirement.

Time Limit

Foundation Exam: 8.0 hours (4.0 h morning + 4.0 h afternoon); Professional Exam: 12.0 hours over 2 days (3.0 h per session across Professional Knowledge AM/PM and Professional Case AM/PM)

Passing Score

Foundation: 132/240 points (55%); Professional Knowledge: 120/200 points (60%); Professional Case Analysis: 60/100 points (60%)

Exam Fee

Set by the provincial examination authority; consult the current registration notice (Ministry of Housing and Urban-Rural Development (MOHURD) & Ministry of Human Resources and Social Security (MOHRSS))

Registered Electrical Engineer (G&T) Exam Content Outline

18%

Power System Analysis & Grid Stability (电力系统分析与稳定控制)

Per-unit system calculations, complex power flows, nodal admittance matrix Y_bus, symmetrical components method for unsymmetrical faults (single line-to-ground LG, line-to-line LL, double line-to-ground LLG), sequence impedance networks (Z1, Z2, Z0), power-angle curve P = (E' U / X') sin(delta), transient stability equal-area criterion, swing equation M (d²delta/dt²) = Pm - Pe, frequency and voltage regulation (AGC / AVR), and power system security guidelines (GB 38755).

16%

Power Plants & Generating Electrical Systems (发电厂与发电机电气主系统)

Synchronous generator parameters (Xd, X'd, X''d), excitation control modes and P-Q capability curve thermal/stability limits, electrical main wiring schemes (single bus, double bus with bypass, breaker-and-a-half 3/2 configuration, bridge scheme), generator-transformer unit connections, and thermal/hydro/nuclear plant auxiliary power distribution (6kV/10kV station service systems, fast bus transfer, motor starting voltage drop).

16%

High-Voltage Overhead Transmission Lines & HVDC (高压输电线路与直流输电)

Transmission line electrical parameters (R, XL, BC), bundle conductor sub-spacing and surface electric field gradient for corona control, mechanical sag and tension state equation (parabolic and catenary models), insulator string creepage ratio (爬电比距) and potential distribution, HVDC transmission converter transformers, 12-pulse converter valve bridges, smoothing reactors, and DC filter configurations.

18%

Substations & High-Voltage Electrical Apparatus (变电站设计与高压电气设备选择)

110kV/220kV/500kV/750kV/1000kV UHV AC substation layouts, SF6 Gas Insulated Switchgear (GIS) design and very fast transient overvoltage (VFTO), power transformer vector groups (YN,d11; YN,yn0,d11), auto-transformer neutral grounding and tertiary stabilizing delta winding, shunt reactor compensation for Ferranti voltage rise, and instrument transformers (CT 10P/5P accuracy classes, 10% saturation curve, CVT ferroresonance suppression).

18%

Relay Protection & Power System Automation (继电保护与电力自动化)

GB/T 14285 and DL/T 584 protection engineering standards, transmission line distance protection (quadrilateral and mho characteristics, Zone I reach 80-85% ZL, Zone II/III coordination, infeed coefficient), current differential protection, transformer longitudinal differential protection (2nd harmonic restraint for magnetizing inrush), generator stator 100% earth-fault and loss-of-excitation protection, breaker failure protection, and single/three-phase automatic reclosing schemes.

14%

Overvoltage Protection, Insulation Coordination & Grounding (过电压保护、绝缘配合与接地设计)

DL/T 620 / GB/T 50064 internal overvoltages (line switching, capacitive de-energization, linear/ferroresonance), lightning overvoltages (direct strikes, backflashover, shielding failure, shielding angle alpha), metal-oxide surge arrester (MOA continuous operating voltage Uc, rated voltage Ur, residual voltage Ures), substation grounding grid design under DL/T 621 / GB/T 50065 (touch voltage Ut, step voltage Us, seasonal soil resistivity, grounding grid impedance R <= 2000/I).

How to Pass the Registered Electrical Engineer (G&T) Exam

What You Need to Know

Passing score: Foundation: 132/240 points (55%); Professional Knowledge: 120/200 points (60%); Professional Case Analysis: 60/100 points (60%)
Assessment: Two-stage qualification: Stage 1 Foundation Exam (General Engineering Foundation 120 MCQs, 4.0 h + Professional Electrical Foundation 60 MCQs, 4.0 h; total 240 marks); Stage 2 Professional Exam (Day 1: Professional Knowledge AM 40 single + 30 multi, PM 40 single + 30 multi, 6.0 h total, 200 marks; Day 2: Professional Case Analysis AM 25 calculation cases, PM 25 calculation cases, 6.0 h total, 100 marks). Non-rolling annual qualification requirement.
Time limit: Foundation Exam: 8.0 hours (4.0 h morning + 4.0 h afternoon); Professional Exam: 12.0 hours over 2 days (3.0 h per session across Professional Knowledge AM/PM and Professional Case AM/PM)
Exam fee: Set by the provincial examination authority; consult the current registration notice

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Registered Electrical Engineer (G&T) Study Tips from Top Performers

1Master symmetrical component sequence networks: Remember that for a single line-to-ground (LG) fault, the sequence networks are connected in series ($I_1 = I_2 = I_0 = E / (Z_1 + Z_2 + Z_0 + 3Z_f)$); for a line-to-line (LL) fault, positive and negative sequence networks connect in parallel opposing ($I_1 = -I_2, I_0 = 0$); and for a double line-to-ground (LLG) fault, all three sequence networks connect in parallel.

2Understand power-angle curves and equal area criteria: Steady-state power limit is $P_{max} = E' U / X'$. In transient stability, accelerating area $A_{acc} = \int_{\delta_0}^{\delta_c} (P_m - P_{e,fault}) d\delta$ must be less than or equal to the maximum available decelerating area $A_{dec} = \int_{\delta_c}^{\delta_{max}} (P_{e,post} - P_m) d\delta$ to maintain synchronism.

3Know 3/2 (breaker-and-a-half) arrangement operational rules: In a standard bay with two main buses and three circuit breakers feeding two outgoing circuits, isolating a bus requires tripping only the adjacent bus breaker without interrupting line power flow. A breaker failure trips only the adjacent breaker and the associated bus differential or line trip.

4Memorize transmission line parameter formulas: Inductive reactance $X_L = 0.1445 \log_{10}(D_{eq}/r_{eq}) + 0.0157 \ \Omega/\text{km}$, capacitive susceptance $B_C = \frac{7.58}{\log_{10}(D_{eq}/r_{eq})} \times 10^{-6} \ \text{S/km}$, where $r_{eq} = \sqrt[n]{n r R^{n-1}}$ is the equivalent bundled radius.

5Apply transformer and distance protection rules: Zone I distance protection reach is standardly $Z_I = (0.80 \sim 0.85) Z_L$ without intentional time delay; Zone II coordinates with the adjacent line Zone I ($Z_{II} = 0.8 (Z_L + Z_{I,adj})$); transformer differential protection uses 2nd harmonic current restraint (typically $I_{2}/I_{1} \ge 15\% - 20\%$) to prevent tripping on magnetizing inrush.

6Calculate grounding grid safety potentials under DL/T 621 / GB/T 50065: Permissible touch potential is $U_t = \frac{174 + 0.17 \rho_s}{\sqrt{t}}$ and permissible step potential is $U_s = \frac{174 + 0.7 \rho_s}{\sqrt{t}}$, where $\rho_s$ is the surface crushed gravel layer resistivity and $t$ is the fault clearance duration.

Frequently Asked Questions

What is the China Registered Electrical Engineer (Generation & Transmission) Examination?

It is the national survey-and-design professional qualification examination for generation-and-transmission electrical engineering, administered under the MOHURD and MOHRSS framework.

How is the Professional stage structured and scored?

The Professional stage spans two consecutive days (12 hours total). Day 1 covers Professional Knowledge (专业知识), numerical concepts and code rules in two 3-hour sessions with 40 single-choice and 30 multi-choice questions each (200 points total, pass mark 120). Day 2 covers Professional Case Analysis (专业案例), comprising two 3-hour sessions with 25 calculation cases each (100 points total, pass mark 60). In the Case exam, candidates must record full step-by-step mathematical calculations and standard citations on the test booklet in addition to bubbling the machine answer sheet.

What is the passing rule under non-rolling management (非滚动管理)?

Unlike multi-year rolling exams, the Registered Electrical Engineer examination requires candidates to pass ALL components of a stage in a single examination year. If a candidate passes Professional Knowledge but fails Professional Case Analysis (or vice versa), the passing score is forfeited, and both subjects must be retaken in a subsequent year.

What reference materials are permitted during the Professional exam?

The Professional stage is strictly open-book. Candidates are permitted to bring officially published national codes, electric power industry standards (DL/T), and recognized design handbooks (such as the Power Engineering Electrical Design Manual / 电力工程电气设计手册). Electronic devices, handwritten notebooks, and unapproved materials are prohibited.

What is the lifelong statutory liability of a Registered Electrical Engineer in China?

Under the PRC Construction Law and Regulations on the Administration of Quality of Construction Works, registered engineers bear lifelong statutory liability (质量终身责任制) for the safety, stability, and code compliance of their stamped engineering designs throughout the design working life of the power installation.