100+ Free UPRVUNL AE Practice Questions

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2026 Statistics

Key Facts: UPRVUNL AE Exam

200 MCQs

Total questions in written CBT

Official UPRVUNL Syllabus

3 hours

Time limit for the CBT exam

Official UPRVUNL Guidelines

75% / 25%

Weight split between Technical and Non-Technical sections

Official UPRVUNL Syllabus

65%

Minimum B.E./B.Tech aggregate percentage required for eligibility

Official UPRVUNL Eligibility

-0.25

Negative marking penalty for every incorrect answer

Official UPRVUNL Marking Scheme

UPRVUNL AE is a state-level recruitment CBT for engineers in Uttar Pradesh. The 3-hour exam features 200 MCQs (150 technical, 50 non-technical) with a -0.25 negative marking penalty. General fee is ₹1,180; SC/ST fee is ₹826.

Sample UPRVUNL AE Practice Questions

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

1Three resistors of values 6 Ω, 12 Ω, and 4 Ω are connected in parallel. What is the equivalent resistance of this parallel combination?

A.2 Ω

B.1.5 Ω

C.22 Ω

D.4 Ω

Explanation: The equivalent resistance R of a parallel combination is calculated using the formula 1/R = 1/R1 + 1/R2 + 1/R3. Substituting the values gives 1/R = 1/6 + 1/12 + 1/4 = 2/12 + 1/12 + 3/12 = 6/12 = 1/2. Therefore, the equivalent resistance R is 2 Ω.

2Which component of core loss in a transformer is minimized by using thin laminations on the core?

A.Eddy current loss

B.Hysteresis loss

C.Copper loss

D.Friction and windage loss

Explanation: Eddy current loss is caused by circulating currents induced in the core by the alternating magnetic flux. Because eddy current loss is directly proportional to the square of the lamination thickness, using thin insulated steel laminations increases the resistance to these currents, thereby minimizing the loss.

3What is the typical range of surge impedance for an overhead three-phase transmission line?

A.400 Ω to 500 Ω

B.40 Ω to 50 Ω

C.4000 Ω to 5000 Ω

D.0.4 Ω to 0.5 Ω

Explanation: The surge impedance of an overhead transmission line is given by Z_c = sqrt(L/C). For standard three-phase overhead lines, the typical values of inductance L and capacitance C result in a surge impedance ranging between 400 Ω and 500 Ω.

4What are the type and order of the open-loop transfer function G(s)H(s) = K / [s²(s + 1)(s + 5)]?

A.Type 2, Order 4

B.Type 4, Order 2

C.Type 2, Order 3

D.Type 3, Order 4

Explanation: The 'type' of a system is defined by the number of open-loop poles at the origin (s = 0), which is 2 in this case due to the s² term. The 'order' is the highest power of s in the characteristic equation or denominator polynomial when expanded, which is 4 (s² * s * s = s⁴). Therefore, the system is Type 2 and Order 4.

5How many PN junctions does a standard silicon-controlled rectifier (SCR) possess?

A.Three PN junctions

B.Two PN junctions

C.Four PN junctions

D.One PN junction

Explanation: A silicon-controlled rectifier (SCR) is a four-layer semiconductor device of PNPN structure. This structure creates three distinct PN junctions (J1, J2, and J3) in series between the anode and cathode terminals.

6What is the resonant frequency of a series RLC circuit where R = 10 Ω, L = 0.1 H, and C = 10 μF?

A.159.2 Hz

B.1000 Hz

C.50.3 Hz

D.500 Hz

Explanation: The resonant frequency f_r of a series RLC circuit is given by f_r = 1 / (2 * pi * sqrt(L * C)). Substituting the given values: L * C = 0.1 * 10 * 10^-6 = 10^-6. Taking the square root gives 10^-3. Thus, f_r = 1 / (2 * pi * 10^-3) = 1000 / (2 * pi) ≈ 159.15 Hz.

7In a linear circuit, the Thevenin equivalent voltage is 20 V and the Thevenin resistance is 5 Ω. What is the maximum power that can be delivered to a variable load resistor connected across this network?

A.20 W

B.80 W

C.40 W

D.10 W

Explanation: According to the Maximum Power Transfer Theorem, maximum power is delivered to the load when the load resistance R_L equals the Thevenin resistance R_th (R_L = 5 Ω). The maximum power P_max is calculated using the formula: P_max = V_th² / (4 * R_th) = 20² / (4 * 5) = 400 / 20 = 20 W.

8A 4-pole, 3-phase induction motor is connected to a 50 Hz supply. If the rotor frequency of the motor is 2 Hz, what is the slip and the actual speed of the motor?

A.Slip = 0.04, Speed = 1440 rpm

B.Slip = 0.02, Speed = 1470 rpm

C.Slip = 0.05, Speed = 1425 rpm

D.Slip = 0.04, Speed = 1500 rpm

Explanation: The synchronous speed is N_s = 120f / P = (120 * 50) / 4 = 1500 rpm. The rotor frequency is f_r = s * f, where s is the slip. Thus, s = f_r / f = 2 / 50 = 0.04 (or 4%). The actual rotor speed N is N = N_s * (1 - s) = 1500 * (1 - 0.04) = 1500 * 0.96 = 1440 rpm.

9According to Peek's formula, how does the corona loss in an overhead transmission line vary with the supply frequency (f) and the operating voltage (V) relative to the disruptive critical voltage (Vc)?

A.Directly proportional to (f + 25) and (V - Vc)²

B.Directly proportional to f and (V - Vc)

C.Directly proportional to f² and (V - Vc)²

D.Inversely proportional to (f + 25) and (V - Vc)²

Explanation: Peek's formula for corona loss under fair weather conditions states that the power loss P is given by: P = (242.2 / d) * (f + 25) * sqrt(r / D) * (V - Vc)² * 10^-5 kW/km/phase. Thus, the corona loss is directly proportional to (f + 25) and the square of the difference between operating voltage and disruptive critical voltage, (V - Vc)².

10A unity feedback system has an open-loop transfer function G(s) = 10 / [s(s + 2)]. What is the steady-state error of this system when subjected to a unit ramp input?

A.0.2

B.0.5

C.0

D.Infinity

Explanation: For a unit ramp input, the steady-state error e_ss is given by e_ss = 1 / K_v, where K_v is the velocity error constant. K_v = lim (s -> 0) [s * G(s)] = lim (s -> 0) [s * 10 / (s(s + 2))] = 10 / 2 = 5. Therefore, e_ss = 1 / 5 = 0.2.

About the UPRVUNL AE Exam

The UPRVUNL Assistant Engineer (Trainee) Recruitment Exam is a competitive state-level selection process. It features a Computer-Based Test (CBT) carrying 200 marks, followed by a personal interview (20 marks) and document verification. The written CBT comprises two parts: Part I consists of 150 branch-specific technical engineering questions, while Part II consists of 50 non-technical questions covering General Hindi, General Knowledge, and Reasoning. Our practice materials focus extensively on the objective CBT stage.

Questions

100 scored questions

Time Limit

3 hours (180 minutes)

Passing Score

Minimum qualifying marks: 30% for General/OBC/EWS, 25% for SC/ST.

Exam Fee

₹1,180 for General/OBC/EWS candidates; ₹826 for SC/ST candidates. (Uttar Pradesh Rajya Vidyut Utpadan Nigam Limited (UPRVUNL))

UPRVUNL AE Exam Content Outline

75%

Technical (Part I)

150 branch-specific technical engineering questions at degree level.

25%

Non-Technical (Part II)

50 questions covering General Hindi, General Knowledge, and Reasoning.

How to Pass the UPRVUNL AE Exam

What You Need to Know

Passing score: Minimum qualifying marks: 30% for General/OBC/EWS, 25% for SC/ST.
Exam length: 100 questions
Time limit: 3 hours (180 minutes)
Exam fee: ₹1,180 for General/OBC/EWS candidates; ₹826 for SC/ST candidates.

Keys to Passing

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

UPRVUNL AE Study Tips from Top Performers

1Dedicate a significant portion of study time to Part I (Technical) as it carries 75% of the total marks (150 out of 200).

2Practice core engineering numerical calculations, including electrical circuit equations, thermodynamic cycles, structural analysis, and algorithm complexity.

3Strengthen your vocabulary, grammar, idioms, and sentence correction in General Hindi, which is a key differentiator in Part II.

4Keep updated with national and international current affairs, along with state-specific General Knowledge of Uttar Pradesh.

5Solve previous years' UPRVUNL AE papers to get familiar with the difficulty level and manage your time effectively during the 3-hour exam.

Frequently Asked Questions

What is the exam structure for UPRVUNL AE?

The Computer-Based Test (CBT) consists of 200 multiple-choice questions divided into two parts: Part I (150 technical questions in the relevant engineering branch) and Part II (50 non-technical questions covering General Hindi, General Knowledge, and Reasoning). The duration is 3 hours.

Is there negative marking in the UPRVUNL AE exam?

Yes, there is negative marking. For each correct answer, you receive 1 mark. For each incorrect response, 0.25 marks are deducted.

What is the age limit for UPRVUNL Assistant Engineer?

Candidates must be between 21 and 40 years of age. SC, ST, and OBC candidates belonging to Uttar Pradesh are eligible for an upper age limit relaxation of 5 years.

What is the minimum educational qualification required?

Candidates must hold a regular, full-time Bachelor’s Degree in Engineering (B.E./B.Tech) in Electrical, Mechanical, Civil, Electronics, Instrumentation, or Computer Science from a recognized university with at least 65% aggregate marks.

What is the application fee for UPRVUNL AE?

The application fee is ₹1,180 for General, OBC, and EWS candidates. For SC and ST candidates who are domiciled in Uttar Pradesh, the fee is ₹826. Physically Handicapped (PH) candidates of UP pay a nominal fee of ₹12.

Is knowledge of Hindi required for this post?

Yes, candidates must have working knowledge of Hindi in Devanagari script. If they have not passed high school or equivalent with Hindi, they must pass a departmental Hindi exam within 3 years of joining.

100+ Free UPRVUNL AE Practice Questions

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