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Exam Review

Key Facts: GER Category 2 Exam

1 kV

Maximum Voltage

GER Regulations

500 kVA

Maximum Capacity

GER Regulations

$15,000 JMD

Examination Fee

GER Portal

70%

Estimated Passing Score

GER Board

3 Years

License Validity

GER Portal

5 Years

Experience Prerequisite

GER Portal

The GER Category 2 Registered Electrician Examination in Jamaica licenses professionals to work on installations up to 1 kV and 500 kVA capacity. It requires passing both a written theory test (covering safety, electrical theory, and the JS316 code) and a practical assessment. Preparation requires at least 80-120 hours of study.

Sample GER Category 2 Practice Questions

Try these sample questions to review concepts for the GER Category 2 exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1A single-phase 240V circuit supplies a resistive heating load of 4.8 kW. What is the current flowing in this circuit?

A.20 A

B.10 A

C.40 A

D.15 A

Explanation: Using Ohm's Law and the power formula (P = V * I), the current (I) is calculated as I = P / V. Here, I = 4800 Watts / 240 Volts = 20 Amperes.

2A single-phase 240V load draws a current of 30 A with a power factor of 0.8 lagging. What is the active power (real power) in kW of this load?

A.7.20 kW

B.5.76 kW

C.9.00 kW

D.4.32 kW

Explanation: Active power (P) in kW is calculated using P = V * I * PF / 1000. Therefore, P = 240V * 30A * 0.8 / 1000 = 5.76 kW. Apparent power is 7.20 kVA.

3What is the total resistance when three resistors of 10 ohms, 20 ohms, and 30 ohms are connected in parallel?

A.5.45 ohms

B.60.00 ohms

C.15.00 ohms

D.3.33 ohms

Explanation: For resistors in parallel, 1/Rt = 1/R1 + 1/R2 + 1/R3. So, 1/Rt = 1/10 + 1/20 + 1/30 = 0.1 + 0.05 + 0.0333 = 0.1833. Rt = 1 / 0.1833 = 5.45 ohms.

4A copper conductor has a resistance of 0.5 ohms. If its length is doubled and its cross-sectional area is halved, what will be the new resistance?

A.2.0 ohms

B.1.0 ohms

C.0.5 ohms

D.0.25 ohms

Explanation: Resistance is directly proportional to length (L) and inversely proportional to cross-sectional area (A): R = p * L / A. Doubling length doubles resistance (0.5 * 2 = 1.0 ohm), and halving area doubles it again (1.0 * 2 = 2.0 ohms).

5A 240V single-phase branch circuit is 50 meters long and carries a current of 16 A. If the conductor has a resistance of 2.5 ohms per kilometer, what is the voltage drop in volts?

A.4.0 V

B.2.0 V

C.8.0 V

D.1.6 V

Explanation: Voltage drop in a single-phase circuit is calculated as VD = 2 * L * I * R_per_meter. Here, length L = 50m (one-way, total wire run is 2 * 50 = 100m). R = 2.5 ohms/1000m = 0.0025 ohms/m. VD = 2 * 50m * 16A * 0.0025 ohms/m = 4.0 V.

6What is the primary function of a capacitor in an AC electrical system?

A.To introduce inductive reactance to limit current

B.To store electrical energy in a magnetic field

C.To provide power factor correction by supplying leading reactive power

D.To step up or step down voltage levels

Explanation: Capacitors store energy in an electric field and produce capacitive reactance, which opposes inductive reactance. In AC systems, they supply leading reactive power (kVAR) to offset the lagging reactive power drawn by inductive loads like motors, correcting the power factor.

7A single-phase transformer has a primary voltage of 2400V and a secondary voltage of 240V. If the secondary current is 50 A, what is the primary current (assuming an ideal transformer)?

A.5.0 A

B.50.0 A

C.500.0 A

D.2.5 A

Explanation: The turns ratio (Np/Ns) equals the voltage ratio (Vp/Vs) and is inversely proportional to the current ratio (Is/Ip). Turns ratio = 2400 / 240 = 10. Thus, the primary current is Is / 10 = 50 A / 10 = 5.0 A.

8What is the active power (kW) consumed by a 240V single-phase pure resistive load with a resistance of 12 ohms?

A.4.8 kW

B.2.4 kW

C.1.2 kW

D.9.6 kW

Explanation: Using the formula P = V^2 / R, the power is P = 240^2 / 12 = 57600 / 12 = 4800 Watts, which is 4.8 kW. For a pure resistive load, the power factor is 1.0.

9In a single-phase AC circuit, if the voltage and current are in phase, what is the power factor of the circuit?

A.0.0

B.0.5

C.1.0

D.0.8

Explanation: Power factor is defined as the cosine of the phase angle between voltage and current (cos θ). When they are in phase, the angle is 0 degrees, and cos(0) = 1.0. This indicates a purely resistive circuit.

10What is the frequency of the AC public electricity supply standard in Jamaica?

A.50 Hz

B.60 Hz

C.100 Hz

D.120 Hz

Explanation: The public electricity grid in Jamaica, operated by the Jamaica Public Service Company (JPS) and regulated by the GER, operates at a standard frequency of 50 Hz.

About the GER Category 2 Exam

Prepare for the Jamaica Government Electrical Regulator (GER) Registered Electrician Category 2 Examination. Access 100 free practice questions, detailed explanations, and study guides based on the Electricity Act 2015 and JS316 standards.

Exam sponsor: Government Electrical Regulator (GER). The requirements and fees below concern the certification or admission exam, separate from our free practice resources.

Questions

100 questions

Time Limit

3 hours

Passing Score

70%

Exam / Certification Fees

$15,000 JMD

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.

30%

Electrical Codes and Standards

JS316 standards, wiring regulations, and standards compliance

20%

Electrical Safety & Laws

Jamaica Electricity Act 2015, safety regulations, and inspector coordination

20%

Basic Electrical Theory & Math

Ohm's Law, power factor, sizing calculations, AC/DC circuit theory

15%

Wiring Methods, Grounding & Bonding

Conductor sizing, conduit fill, grounding electrode system, bonding metal structures

15%

Motors, Transformers, and Protective Devices

Transformer connections, motor control circuits, GFCIs, AFCIs, and circuit breakers

Preparing for the GER Category 2 Exam

What You Need to Know

Passing score: 70%
Exam length: 100 questions
Time limit: 3 hours
Exam / certification fees: $15,000 JMD 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

GER Category 2: Suggested Study Strategy

1Master electrical calculations including voltage drop, power factor correction, kVA-to-amps conversion, and parallel resistance.

2Thoroughly review the JS316 electrical code requirements, focusing on low-voltage (up to 1 kV) residential and commercial installations.

3Understand grounding and bonding rules: how to size the grounding electrode conductor (GEC) and main bonding jumper.

4Learn safety procedures inside out, particularly Lockout/Tagout (LOTO) protocols and NFPA 70E / JS316 safety clearance guidelines.

5Familiarize yourself with the Jamaica Electricity Act 2015, including licensing categories, penalty rules, and the inspection process.

Frequently Asked Questions

What is the scope of work for a GER Category 2 Electrician in Jamaica?

A Category 2 electrician is licensed to design, carry out, or supervise electrical installations operating at voltage levels not exceeding 1 kV (1000 Volts) for both single-phase and three-phase systems, with a maximum power capacity of 500 kVA.

What is the passing score for the GER Category 2 exam?

Candidates must successfully complete both the written theory and code examination (with a passing score typically around 70%) and a subsequent hands-on practical assessment to obtain their licensing recommendation.

What are the prerequisites to take the Category 2 Registered Electrician exam?

Applicants must be at least 18 years of age, hold a recognized Diploma in Electrical Engineering (or equivalent), and have at least five years of practical experience in electrical installation works, verified by a voluntary declaration from a registered electrician or inspector.

How much does it cost to take the GER Category 2 exam?

The written examination fee is $15,000 JMD. Additional registration, portal processing, and physical license fees are separate from the examination fee.

What codes and regulations are covered on the examination?

The exam covers the Jamaica Electricity Act 2015, the national JS316 electrical standard (which replaced JS21), and core electrical safety procedures (e.g. lockout/tagout, PPE requirements).

How is the GER Registered Electrician exam structured?

The assessment consists of two parts: a written examination covering electrical theory, calculations, and code regulations (using nominal 110/220/415V systems), followed by a practical test demonstrating installation and testing skills.

Can I retake the GER exam if I fail?

Yes, candidates who do not pass are permitted to resit the exam, typically within 6 to 12 months, subject to additional registration and the Board's guidelines.