PracticeVideosBlogFlashcardsEspañol
All Practice Exams

Free Practice Questions for Tanzania ERB Electrical

Exam-style questions and explanations by OpenExamPrep.

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

Loading practice questions...

Same family resources

Explore More Tanzania ERB Professional Engineers Examination

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

Tanzania ERB Professional Engineers Examination - Civil Engineering

Practice Questions

Tanzania ERB Professional Engineers Examination - Mechanical Engineering

Practice Questions
Exam Review

Key Facts: Tanzania ERB Electrical Exam

100

Practice Questions

OpenExamPrep

50%

Pass Score

Official Guidelines

3.0 hours

Time Limit

Exam Rules

Tanzania ERB Professional Engineers Examination - Electrical Engineering prep course featuring 100 high-quality practice questions and detailed explanations.

Sample Tanzania ERB Electrical Practice Questions

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

1A 50 MVA, 13.8 kV generator has a subtransient reactance of 0.2 pu on its own base. If it is connected to a 13.8/132 kV, 60 MVA transformer, what is the subtransient reactance of the generator in a new system base of 100 MVA, 132 kV?
A.0.167 pu
B.0.2 pu
C.0.043 pu
D.0.4 pu
Explanation: To convert the generator's per-unit reactance to a new system base, the formula X_new_pu = X_old_pu * (S_base_new / S_base_old) * (V_base_old / V_base_new_at_element)^2 is used. The new system base MVA is 100 MVA. The base voltage on the generator side remains 13.8 kV, as it is derived from the system's 132 kV base through the transformer ratio (132 kV * (13.8/132) = 13.8 kV). Therefore, X_gen_new_pu = 0.2 * (100/50) * (13.8/13.8)^2 = 0.2 * 2 * 1 = 0.4 pu.
2What is the primary function of a conservator tank in a power transformer?
A.To provide cooling for the transformer windings
B.To house the primary and secondary windings
C.To filter moisture from the transformer oil
D.To allow for oil expansion and contraction due to temperature variations.
Explanation: The conservator tank, located above the main transformer tank, serves as an expansion vessel for the insulating oil. As the transformer heats up during operation, the oil expands; when it cools down, the oil contracts. The conservator tank accommodates these volume changes, preventing excessive pressure buildup and minimizing exposure of the main tank oil to the atmosphere.
3A long transmission line has negligible resistance and shunt admittance. Its characteristic impedance is Zc, and propagation constant is γ. What is the approximate voltage at the sending end if the receiving end is open-circuited with voltage Vr?
A.Vs = Vr sinh(γL)
B.Vs = Vr e^(γL)
C.Vs = Vr (1 + γL)
D.Vs = Vr cosh(γL)
Explanation: For a long transmission line with negligible resistance and shunt admittance (lossless line), the characteristic impedance is Zc and the propagation constant is γ. For an open-circuited receiving end, the receiving end current Ir is zero. The sending end voltage Vs is given by the hyperbolic cosine relationship: Vs = Vr cosh(γL) + Zc * Ir sinh(γL). Since Ir = 0, the equation simplifies to Vs = Vr cosh(γL).
4A 3-phase, 4-pole, 50 Hz induction motor runs at a speed of 1440 rpm. Calculate the slip of the motor.
A.6%
B.5%
C.3.5%
D.4%
Explanation: First, calculate the synchronous speed (Ns) using the formula Ns = (120 * f) / P, where f is the frequency (50 Hz) and P is the number of poles (4). Ns = (120 * 50) / 4 = 1500 rpm. The slip (s) is then calculated as s = (Ns - Nr) / Ns, where Nr is the rotor speed (1440 rpm). So, s = (1500 - 1440) / 1500 = 60 / 1500 = 0.04. Expressed as a percentage, the slip is 4%.
5Which type of relay is primarily used for the protection of transmission lines against phase faults and ground faults, capable of high-speed operation?
A.Differential relay
B.Overcurrent relay
C.Buchholz relay
D.Distance relay
Explanation: Distance relays are specifically designed for transmission line protection. They measure the impedance between the relay location and the fault point. Since impedance is proportional to distance, the relay operates if the fault is within a predetermined zone, providing fast and selective protection against both phase and ground faults on long lines.
6A 3-phase, 50 Hz, star-connected synchronous generator has a synchronous reactance of 1.0 pu and negligible resistance. It is supplying rated power at unity power factor to an infinite bus. If the excitation voltage is increased by 20% while the prime mover input remains constant, what will be the approximate new power factor?
A.0.92 lagging
B.0.98 leading
C.0.96 leading
D.0.94 lagging
Explanation: For unity power factor operation, the excitation voltage E_f and power angle δ are calculated using P = (E_fV_t/X_s)sinδ and Q = (E_fV_tcosδ - V_t^2)/X_s. For P=1, Q=0, V_t=1, X_s=1, we find E_f = √2 pu and δ = 45°. When E_f increases by 20% to 1.2√2 ≈ 1.697 pu, and P remains constant at 1 pu, the new power angle δ' can be found from sinδ' = P / (E_f'V_t/X_s) = 1 / 1.697, giving δ' ≈ 36.1°. The new reactive power Q' = (E_f'V_tcosδ' - V_t^2)/X_s = (1.697*1*cos(36.1°) - 1^2)/1 ≈ 0.371 pu. The new power factor is cosφ = P / √(P^2 + Q'^2) = 1 / √(1^2 + 0.371^2) ≈ 0.938 lagging.
7A factory has an inductive load of 1.2 MW at a power factor of 0.75 lagging. To improve the power factor to 0.95 lagging, what is the approximate reactive power (kVAr) required from the capacitor bank?
A.525 kVAr
B.800 kVAr
C.960 kVAr
D.664 kVAr
Explanation: First, calculate the initial reactive power (Q1) from P=1.2MW and cos(φ1)=0.75. φ1 = arccos(0.75) ≈ 41.41°, so Q1 = P * tan(φ1) = 1.2 MW * tan(41.41°) ≈ 1.2 MW * 0.8819 = 1.0583 MVAr. Next, calculate the target reactive power (Q2) for cos(φ2)=0.95. φ2 = arccos(0.95) ≈ 18.19°, so Q2 = P * tan(φ2) = 1.2 MW * tan(18.19°) ≈ 1.2 MW * 0.3287 = 0.3944 MVAr. The required reactive power from the capacitor bank is Qc = Q1 - Q2 = 1.0583 - 0.3944 = 0.6639 MVAr, or approximately 664 kVAr.
8Which method is most commonly used for speed control of a DC series motor?
A.Flux weakening control (for shunt motor)
B.Ward-Leonard control (more for precise control of shunt/compound)
C.Armature voltage control
D.Diverter method (field control)
Explanation: A DC series motor's speed is highly dependent on its field flux, which is in series with the armature current. The diverter method involves connecting a variable resistance in parallel with the series field winding. By diverting some armature current away from the field, the field flux is weakened, leading to an increase in motor speed.
9According to common electrical safety practices and standards (e.g., IEC/BS), what is the maximum permissible earth resistance for substation earthing where step and touch potentials must be minimized?
A.Less than 10 Ohms
B.Less than 5 Ohms
C.Less than 2 Ohms
D.Typically less than 1 Ohm
Explanation: Substation earthing is critical for safety, to minimize step and touch potentials, and to provide a low impedance path for fault currents. Industry standards and regulations (e.g., IEC 61936-1) generally recommend an earth resistance of less than 1 Ohm for major substations, especially in areas with significant human activity, to ensure adequate safety and fault current dissipation.
10When two single-phase transformers operate in parallel, which condition is NOT essential for satisfactory parallel operation?
A.Identical kVA ratings
B.Same polarity
C.Same percentage impedance
D.Same voltage ratio
Explanation: For satisfactory parallel operation of transformers, several conditions are essential: identical voltage ratios (to avoid circulating currents), same percentage impedance (to ensure proper load sharing), and same polarity (to avoid short-circuiting windings). While having identical kVA ratings is desirable for optimal utilization and load sharing, transformers with different kVA ratings can operate in parallel, provided the essential conditions are met, albeit with the load sharing proportional to their kVA ratings if their percentage impedances are equal. Thus, identical kVA ratings are NOT strictly essential.

About the Tanzania ERB Electrical Exam

The ERB Tanzania electrical engineering professional registration exam checks competence in power engineering, electrical circuits, control telemetry, local safety codes, and ethics.

Exam sponsor: Engineers Registration Board (ERB) Tanzania. The requirements and fees below concern the certification or admission exam, separate from our free practice resources.

Assessment

Multiple-choice computerised exam administered by Engineers Registration Board (ERB) Tanzania.

Time Limit

3.0 hours

Passing Score

50%

Exam / Certification Fees

150,000 TZS

Exam sponsor website

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

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.

25%

Power Systems And Electrical Machines

Practice questions covering the domain: power systems and electrical machines.

25%

Electronics And Telecommunications

Practice questions covering the domain: electronics and telecommunications.

25%

Control Systems And Instrumentation

Practice questions covering the domain: control systems and instrumentation.

25%

Electrical Safety Codes And Erb Ethics

Practice questions covering the domain: electrical safety codes and erb ethics.

Preparing for the Tanzania ERB Electrical Exam

What You Need to Know

  • Passing score: 50%
  • Assessment: Multiple-choice computerised exam administered by Engineers Registration Board (ERB) Tanzania.
  • Time limit: 3.0 hours
  • Exam / certification fees: 150,000 TZS 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

Tanzania ERB Electrical: Suggested Study Strategy

1Carefully study all regulatory and legislative requirements.
2Practice sample calculations and review real-world scenario items.
3Review the explanations for all incorrect practice questions to build core conceptual clarity.

Frequently Asked Questions

What is the passing score for Tanzania ERB Electrical?

The passing score is 50%.

Who administers the Tanzania ERB Electrical exam?

The exam is administered by the Engineers Registration Board (ERB) Tanzania.