PracticeVideosBlogFlashcardsEspañol
All Practice Exams

100+ Free PSPCL JE Practice Questions

Pass your PSPCL Junior Engineer (JE) Exam exam on the first try — instant access, no signup required.

✓ No registration✓ No credit card✓ No hidden fees✓ Start practicing immediately
Top merit basis Pass Rate
100+ Questions
100% Free

Loading practice questions...

Same family resources

Explore More India PSU Engineering Recruitment Exams

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

BEL Probationary Engineer exam

Practice Questions

Coal India Management Trainee (CIL MT)

Practice Questions

ISRO Scientist/Engineer — Civil

Practice Questions

ISRO Scientist/Engineer — Computer Science

Practice Questions

ISRO Scientist/Engineer — Electrical

Practice Questions

ISRO Scientist/Engineer — Mechanical

Practice Questions
2026 Statistics

Key Facts: PSPCL JE Exam

100

Total Questions

PSPCL Official

70%

Technical Weight

Electrical / Civil

45%

Qualifying Mark

General Category

0.25

Negative Marking

Per Incorrect Answer

₹1,416

Application Fee

General/OBC

120 mins

Exam Duration

2 Hours

The PSPCL JE (Junior Engineer) exam is a highly competitive state-level computer-based test in Punjab. It consists of 100 questions (70 technical, 30 non-technical) to be completed in 2 hours. Candidates need a minimum of 45% (General) or 35% (Reserved) to qualify, with final selection based on merit. Registration requires a diploma or engineering degree in the relevant discipline and passing Class 10 Punjabi.

Sample PSPCL JE Practice Questions

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

1What is the equivalent resistance of three resistors of 6 ohms, 12 ohms, and 4 ohms connected in parallel?
A.22 ohms
B.2 ohms
C.4.5 ohms
D.6 ohms
Explanation: To find the equivalent resistance (Rp) of resistors in parallel, use the formula 1/Rp = 1/R1 + 1/R2 + 1/R3. Here, 1/Rp = 1/6 + 1/12 + 1/4 = (2 + 1 + 3)/12 = 6/12 = 1/2. Therefore, Rp = 2 ohms. The equivalent resistance of a parallel circuit is always smaller than the smallest individual resistor in the circuit.
2According to Kirchhoff's Voltage Law (KVL), what is the algebraic sum of all the voltage drops and EMFs in a closed loop equal to?
A.Infinity
B.Zero
C.The sum of all currents
D.The source voltage only
Explanation: Kirchhoff's Voltage Law (KVL) states that the algebraic sum of all potential differences (EMFs and voltage drops) around any closed loop in a circuit is equal to zero. This is a direct consequence of the law of conservation of energy, which dictates that energy cannot be created or destroyed within a closed circuit loop.
3In a network, if we double the voltage across a resistor, what happens to the power dissipated by the resistor?
A.It remains unchanged
B.It doubles
C.It increases four times
D.It decreases to half
Explanation: Power dissipated by a resistor is given by the formula P = V^2 / R. Since the power is directly proportional to the square of the voltage, doubling the voltage (2V) will result in P' = (2V)^2 / R = 4 * (V^2 / R) = 4P. Thus, the power increases by four times.
4Which of the following theorems states that any linear, bilateral network containing generators and impedances can be replaced by an equivalent circuit containing a single voltage source in series with a single impedance?
A.Norton's Theorem
B.Thevenin's Theorem
C.Superposition Theorem
D.Maximum Power Transfer Theorem
Explanation: Thevenin's Theorem states that any linear, active, bilateral network with terminal pair A and B can be replaced by a single voltage source (Vth) in series with a single resistance (Rth). The voltage source is equal to the open-circuit voltage across terminals A and B, and the resistance is the equivalent resistance of the network looking back from terminals A and B with all independent sources deactivated.
5A circuit has a Thevenin equivalent voltage of 24 V and a Thevenin equivalent resistance of 6 ohms. For what load resistance will the maximum power be transferred to the load?
A.12 ohms
B.3 ohms
C.6 ohms
D.0 ohms
Explanation: The Maximum Power Transfer Theorem states that maximum power is transferred from a source to a load when the load resistance (RL) is equal to the internal resistance of the source (Rth). Therefore, RL = Rth = 6 ohms.
6What is the maximum power transferred to the load in the previous question (Vth = 24 V, Rth = 6 ohms, RL = 6 ohms)?
A.96 W
B.24 W
C.48 W
D.12 W
Explanation: The formula for maximum power transferred to a load is Pmax = Vth^2 / (4 * Rth). Playout: Pmax = 24^2 / (4 * 6) = 576 / 24 = 24 W. Alternatively, calculate the load current IL = Vth / (Rth + RL) = 24 / (6 + 6) = 2 A. Then P = IL^2 * RL = 2^2 * 6 = 24 W.
7Which of the following elements is bilateral in nature?
A.Diode
B.Transistor
C.Resistor
D.Silicon Controlled Rectifier (SCR)
Explanation: A bilateral element is one whose characteristics are independent of the direction of current flowing through it. Resistors, inductors, and capacitors are bilateral elements because they permit current flow in both directions equally. Diodes, transistors, and SCRs are unilateral, meaning their behavior changes depending on the polarity of the applied voltage.
8A delta network consists of three equal resistors of R ohms. What is the resistance of each branch in its equivalent star network?
A.3R ohms
B.R / 3 ohms
C.R ohms
D.R / 9 ohms
Explanation: For a delta-star transformation where all delta resistors are equal (R_delta = R), the equivalent star resistor is given by R_star = R_delta / 3. Therefore, each resistor in the star network has a value of R / 3. Conversely, star-to-delta transformation of equal resistors yields R_delta = 3 * R_star.
9If two capacitors of 10 microfarads and 20 microfarads are connected in series across a 60 V DC supply, what is the charge on each capacitor?
A.600 microcoulombs
B.400 microcoulombs
C.200 microcoulombs
D.1200 microcoulombs
Explanation: In a series connection of capacitors, the charge (Q) on each capacitor is the same and is equal to the total charge. First, find the equivalent capacitance (Ceq): 1/Ceq = 1/10 + 1/20 = 3/20, which gives Ceq = 20/3 microfarads. The charge is Q = Ceq * V = (20/3 microfarads) * 60 V = 400 microcoulombs.
10According to Faraday's Law of Electromagnetic Induction, what determines the magnitude of the induced EMF in a conductor?
A.The resistance of the conductor
B.The length of the conductor
C.The rate of change of magnetic flux linkage
D.The temperature of the surrounding medium
Explanation: Faraday's Law states that the magnitude of the induced EMF (e) in a circuit is directly proportional to the time rate of change of magnetic flux (phi) linking the circuit, expressed as e = -N(d phi / dt), where N is the number of turns.

About the PSPCL JE Exam

Competitive recruitment exam for Junior Engineers in Punjab's state power sector. Consists of 100 questions split into 70% technical (Electrical/Civil) and 30% non-technical (Reasoning, English, Punjab GK/Punjabi).

Questions

100 scored questions

Time Limit

2 hours

Passing Score

45%

Exam Fee

₹1,416 (General) / ₹885 (SC/PwBD) (Punjab State Power Corporation Limited (PSPCL))

PSPCL JE Exam Content Outline

70%

Professional/Technical Knowledge

Discipline-specific concepts (Electrical Engineering circuits, machines, power systems, electronics)

10%

Reasoning Ability

Analytical and logical reasoning questions, puzzles, and sequences

10%

General English

Grammar, prepositions, error detection, active/passive voice, and vocabulary

10%

General Awareness / GK of Punjab

Punjab state history, geography, culture, and Punjabi language concepts

How to Pass the PSPCL JE Exam

What You Need to Know

  • Passing score: 45%
  • Exam length: 100 questions
  • Time limit: 2 hours
  • Exam fee: ₹1,416 (General) / ₹885 (SC/PwBD)

Keys to Passing

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

PSPCL JE Study Tips from Top Performers

1Focus heavily on Power Systems and Electrical Machines, as they form the core of the 70-mark technical section.
2Do not neglect the non-technical sections (Reasoning, English, and Punjab GK/Punjabi) which account for 30 marks.
3Ensure you are familiar with the geography, rivers, dams, and history of Punjab for the General Awareness section.
4Practice quantitative problems in network theorems and electrical machines to solve them quickly during the exam.
5Take timed mock tests to manage your time effectively under the 120-minute limit and avoid excessive negative marks from guessing.

Frequently Asked Questions

What is the exam pattern for PSPCL Junior Engineer (JE)?

The exam is a Computer-Based Test (CBT) consisting of 100 multiple-choice questions. It includes 70 questions on discipline-specific technical knowledge (such as Electrical Engineering) and 30 questions covering General Knowledge, Reasoning, English, and Punjabi. The exam duration is 2 hours (120 minutes).

Is there negative marking in the PSPCL JE exam?

Yes, there is negative marking. For every incorrect answer, 0.25 mark (one-fourth of the question's weight) is deducted. Correct answers are awarded 1 mark.

What are the eligibility criteria for PSPCL JE Electrical?

Candidates must have a full-time regular 3/4 years Diploma in Electrical or Electrical and Electronics Engineering with at least 60% marks, or a B.E./B.Tech. in the same discipline with at least 50% marks. Additionally, candidates must have passed Punjabi at the Matriculation (Class 10) level.

What is the application fee for the PSPCL JE exam?

For general and OBC candidates, the application fee is ₹1,416 (inclusive of GST). For SC and PwBD (disabled) candidates, the application fee is ₹885 (inclusive of GST).

What is the passing score for the PSPCL JE exam?

The minimum qualifying score in the online test is 45% for General category candidates and 35% for Reserved category candidates. However, the final selection is based strictly on merit ranking against the available vacancies.