100+ Free RVUNL JE Practice Questions

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2026 Statistics

Key Facts: RVUNL JE Exam

140 Q / 200 M

Total Questions & Marks

RVUNL Exam Pattern

120 minutes

Exam Duration

RVUNL Exam Pattern

60% Weight

Core Engineering Discipline

Part A Syllabus

18% Weight

Rajasthan General Knowledge

Part B Syllabus

INR 1,600 / 1,400

Application Fee (UR / Reserved)

RVUNL Notification

25% Penalty

Negative Marking Deduction

RVUNL Notification

The RVUNL Junior Engineer (JE) exam consists of 140 questions (200 marks) split into Part A (60 Q, 120 M) and Part B (80 Q, 80 M) to be completed in 120 minutes. Negative marking is 1/4 of the question marks. The application fee is INR 1600 (General) / INR 1400 (Reserved).

Sample RVUNL JE Practice Questions

Try these sample questions to test your RVUNL JE exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1A DC shunt generator has a shunt field resistance of 100 ohms and generates an open-circuit voltage of 220 V at a speed of 1000 rpm. If the speed is increased to 1200 rpm, what will be the new open-circuit voltage generated? Assume the magnetic circuit is linear and ignore armature resistance.

A.264 V

B.220 V

C.183 V

D.316 V

Explanation: For a DC shunt generator with a linear magnetic circuit, the generated EMF (Eg) is directly proportional to both the flux (which is proportional to the field current If) and the speed (N). At open-circuit, the generated EMF is Eg = k * N * If. The field current is If = Eg / Rf. Substituting If gives Eg = k * N * (Eg / Rf), which simplifies to a critical speed relationship. However, if we assume open-circuit voltage is generated from residual magnetism and linear build-up, Eg is proportional to speed N. Thus, Eg2 / Eg1 = N2 / N1. So, Eg2 = 220 * (1200 / 1000) = 264 V.

2Which of the following faults is characterized by the presence of only zero-sequence current in the neutral circuit of a three-phase power system?

A.Single Line-to-Ground (LG) fault

B.Three-phase symmetrical fault

C.Line-to-Line (LL) fault

D.Double Line-to-Ground (LLG) fault

Explanation: In a three-phase system, a Single Line-to-Ground (LG) fault involves a single phase conducting to ground. The current flowing through the neutral circuit is equal to 3 * I0, where I0 is the zero-sequence current. Symmetrical faults do not have zero-sequence components, and Line-to-Line faults do not involve ground, meaning zero-sequence currents are absent. While LLG faults also contain zero-sequence currents, the LG fault is the classic fault where the neutral current is exclusively composed of the sum of the zero-sequence currents from the faulted phase.

3The primary reason for using silica gel in the breather of an oil-filled transformer is to:

A.Absorb moisture from the air entering the conservator tank

B.Filter out dust particles from the entering air

C.Cool the hot transformer oil during heavy loading

D.Prevent the leakage of transformer oil into the atmosphere

Explanation: Transformer oil acts as both an insulator and a coolant. When the transformer heats up and cools down, it breathes air in and out. The silica gel in the breather absorbs moisture from the incoming air to prevent it from dissolving in the transformer oil, which would otherwise drastically lower the oil's dielectric strength.

4An open-loop control system is preferred over a closed-loop control system under which of the following conditions?

A.When the system operating environment is highly stable and free of external disturbances

B.When high accuracy and precise output tracking are required

C.When the system parameters are subject to significant variation over time

D.When the system has a high sensitivity to parameter variations

Explanation: Open-loop systems do not use feedback, meaning they cannot correct for disturbances or changes in system parameters. However, they are simpler, cheaper, and inherently stable. They are preferred when the system is highly stable, disturbances are negligible, and the cost of feedback sensors is not justified.

5Which type of instrument can be used to measure both AC and DC quantities without recalibration?

A.Moving-Iron (MI) instrument

B.Permanent Magnet Moving Coil (PMMC) instrument

C.Induction type instrument

D.Electrostatic instrument for low voltages only

Explanation: Moving-Iron (MI) instruments rely on the magnetic force generated by a coil carrying the current to attract or repel iron vanes. Because the deflecting torque is proportional to the square of the current, the deflection is in the same direction regardless of current polarity. Thus, MI instruments can measure both AC and DC RMS values directly.

6In a linear circuit, the open-circuit voltage at the terminals is 24 V and the short-circuit current is 4 A. If a load resistance of 6 ohms is connected across these terminals, what is the power dissipated in the load?

A.24 W

B.96 W

C.16 W

D.36 W

Explanation: First, find the Thevenin equivalent parameters. The Thevenin voltage (Vth) is the open-circuit voltage, Vth = 24 V. The Thevenin resistance (Rth) is Vth / Isc = 24 V / 4 A = 6 ohms. When a load resistance RL = 6 ohms is connected, the current IL through the load is IL = Vth / (Rth + RL) = 24 / (6 + 6) = 2 A. The power dissipated in the load is P = IL^2 * RL = (2)^2 * 6 = 4 * 6 = 24 W.

7The skin effect in an AC transmission line conductor leads to which of the following changes in the conductor's effective parameters?

A.Increase in effective resistance and decrease in internal self-inductance

B.Decrease in effective resistance and increase in internal self-inductance

C.Increase in both effective resistance and internal self-inductance

D.Decrease in both effective resistance and internal self-inductance

Explanation: Skin effect causes AC current to flow primarily near the outer surface (skin) of the conductor due to higher inner flux linkages causing higher inner reactance. This reduces the effective cross-sectional area through which current flows, thereby increasing the AC resistance (Rac > Rdc). Since less current flows through the center, the internal magnetic flux linkage decreases, leading to a decrease in the internal self-inductance.

8A 220 V DC shunt motor runs at 1000 rpm while drawing an armature current of 20 A. The armature resistance is 0.5 ohms. If the load torque is doubled, what will be the new operating speed of the motor? Assume constant field flux and ignore rotational losses.

A.952 rpm

B.900 rpm

C.1000 rpm

D.976 rpm

Explanation: Since torque T = k * phi * Ia and flux phi is constant, doubling the torque means the armature current must double: Ia2 = 2 * Ia1 = 2 * 20 = 40 A. The back EMF in the first case is Eb1 = V - Ia1 * Ra = 220 - (20 * 0.5) = 210 V. The back EMF in the second case is Eb2 = V - Ia2 * Ra = 220 - (40 * 0.5) = 200 V. Since Eb is proportional to speed N under constant flux: N2 = N1 * (Eb2 / Eb1) = 1000 * (200 / 210) = 952.38 rpm.

9A control system is defined by the state-space equation dx/dt = Ax + Bu, where A = [[0, 1], [-2, -3]] and B = [[0], [1]]. What are the eigenvalues of the system matrix A, which determine the poles of the closed-loop system?

A.-1 and -2

B.1 and 2

C.0 and -3

D.-1.5 and -1.5

Explanation: The eigenvalues of matrix A are found by solving the characteristic equation det(sI - A) = 0. sI - A = [[s, -1], [2, s + 3]]. The determinant is det(sI - A) = s(s + 3) - (-1)(2) = s^2 + 3s + 2 = 0. Factoring this quadratic equation gives (s + 1)(s + 2) = 0. Therefore, the eigenvalues are s = -1 and s = -2.

10A single-phase half-controlled bridge rectifier is supplied by a 230 V, 50 Hz AC source. If the firing angle (alpha) of the thyristors is set to 60 degrees, what is the average output DC voltage? (Assume continuous conduction and ideal devices).

A.155.3 V

B.207.1 V

C.103.5 V

D.146.4 V

Explanation: For a single-phase half-controlled bridge rectifier, the average output voltage Vo is given by the formula: Vo = (Vm / pi) * (1 + cos(alpha)), where Vm is the peak value of the AC input voltage. Vm = 230 * sqrt(2) = 325.27 V. For alpha = 60 degrees, cos(60) = 0.5. Thus, Vo = (325.27 / pi) * (1 + 0.5) = (325.27 / 3.14159) * 1.5 = 103.53 * 1.5 = 155.3 V.

About the RVUNL JE Exam

The RVUNL Junior Engineer (JE) recruitment exam is conducted for positions in Rajasthan power utilities (RVUN, RVPN, JVVN, AVVN, JdVVN). Part A (60% weight) assesses professional engineering knowledge in Electrical, Mechanical, or Civil engineering. Part B (40% weight) covers General Knowledge (with a heavy emphasis on Rajasthan history, geography, and culture), Reasoning, Mathematics, everyday Science, Hindi, and English. The exam requires a strong balance of quick numerical solving in technical subjects and broad familiarity with Rajasthan state General Knowledge. Scoring is penalised by 25% negative marking for wrong answers.

Assessment

The Computer-Based Test (CBT) contains two parts: Part A (Core Engineering Discipline, 60 questions at 2 marks each) and Part B (General Aptitude, English, Hindi, Math, and Rajasthan GK, 80 questions at 1 mark each). Duration is 120 minutes. Penalty for wrong answers is 1/4th of the question's marks (0.5 marks for Part A, 0.25 marks for Part B).

Time Limit

120 minutes

Passing Score

Minimum qualifying marks: 40% for General/UR, 30% for SC/ST/PwBD. Selection is strictly on merit.

Exam Fee

INR 1,600 (General) / INR 1,400 (Reserved) (Rajasthan Rajya Vidyut Utpadan Nigam Limited (RVUNL))

RVUNL JE Exam Content Outline

60%

Part A — Professional Knowledge (Core Engineering)

Core concepts, principles, and calculations in candidate's selected engineering discipline (Electrical, Mechanical, Civil, etc.)

18%

Part B — General Knowledge of Rajasthan

Geography, History, Art & Culture, Heritage, Polity, and Economy of Rajasthan

Part B — General Knowledge of India/World & Science

Current affairs, Indian history, polity, geography, and general everyday science

Part B — Reasoning & Mental Ability

Analytical reasoning, number series, coding-decoding, blood relations, and logical sequences

Part B — Mathematics

Secondary (Class X) and Higher Secondary (Class XII) level mathematics including algebra, geometry, calculus, and probability

Part B — General Hindi

Class X level Hindi grammar, vocabulary, idioms, and sentence correction

Part B — General English

Class X level English grammar, sentence structure, active/passive voice, tenses, and synonyms/antonyms

How to Pass the RVUNL JE Exam

What You Need to Know

Passing score: Minimum qualifying marks: 40% for General/UR, 30% for SC/ST/PwBD. Selection is strictly on merit.
Assessment: The Computer-Based Test (CBT) contains two parts: Part A (Core Engineering Discipline, 60 questions at 2 marks each) and Part B (General Aptitude, English, Hindi, Math, and Rajasthan GK, 80 questions at 1 mark each). Duration is 120 minutes. Penalty for wrong answers is 1/4th of the question's marks (0.5 marks for Part A, 0.25 marks for Part B).
Time limit: 120 minutes
Exam fee: INR 1,600 (General) / INR 1,400 (Reserved)

Keys to Passing

Complete 500+ practice questions
Score 80%+ consistently before scheduling
Focus on highest-weighted sections
Use our AI tutor for tough concepts

RVUNL JE Study Tips from Top Performers

1Focus heavily on Part A (60% of marks) since engineering concepts and numerical calculations are critical for obtaining a high score.

2Thoroughly study Rajasthan General Knowledge (18% of marks) since state-specific history, geography, polity, and culture represent the largest part of the non-technical section.

3Practice numerical speed and formula memorization for both engineering topics and the mathematics section (Class X/XII level).

4Minimize guessing due to the 25% negative marking penalty. Accuracy is key in CBT tests with tight time frames.

5Take timed mock tests simulating the sectional 60-minute split for Part A and Part B to adapt to the exam-day interface.

Frequently Asked Questions

What is the exam pattern of the RVUNL JE Exam?

The exam is an online Computer-Based Test of 140 multiple-choice questions for 200 marks, to be completed in 2 hours. Part A has 60 questions from core engineering (120 marks, 2 marks each). Part B has 80 questions from reasoning, mathematics, general knowledge, science, Hindi, and English (80 marks, 1 mark each).

Is there any negative marking in the RVUNL JE Exam?

Yes, there is negative marking. For each wrong answer, 1/4th of the marks assigned to that question will be deducted (0.5 marks deducted for Part A, 0.25 marks deducted for Part B).

What is the application fee for RVUNL Junior Engineer?

The application fee is INR 1,600 for General/UR candidates (with family income >= 2.5 Lakhs) and INR 1,400 for reserved category candidates (SC, ST, OBC, BC, MBC, EWS, or General with family income < 2.5 Lakhs).

Does this practice question bank cover both Part A and Part B?

Yes. The OpenExamPrep practice question bank includes exactly 100 questions designed proportionally to the exam weights: 60 questions covering Core Engineering (split into Electrical, Mechanical, and Civil) and 40 questions covering non-technical topics (Rajasthan GK, India/World GK, reasoning, mathematics, Hindi, and English).

What are the eligibility requirements for RVUNL JE?

Candidates must have a B.E./B.Tech degree or a 3-year Diploma in a relevant engineering branch from a recognized university. The age limit is 18 to 28 years, with relaxations for reserved categories. Candidates must also have working knowledge of written Hindi (Devnagri script) and Rajasthan culture.

Is there a sectional time limit in the RVUNL JE Exam?

Historically, the CBT has featured separate sectional timings (60 minutes for Part A and 60 minutes for Part B). Candidates cannot switch between the two parts once the time for a section expires, making prompt time management critical.

100+ Free RVUNL JE Practice Questions

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