100+ Free TSGENCO AE Practice Questions

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Key Facts: TSGENCO AE Exam

100 MCQs

Total questions in the written exam

TSGENCO Syllabus

120 mins

Time limit for the written paper

TSGENCO Pattern

80 / 20

Marks split between Section A and Section B

TSGENCO Syllabus

Negative marking points for wrong answers

TSGENCO Guidelines

100%

Selection weightage on the written exam

TSGENCO Process

TSGENCO AE is a state-level competitive exam for electrical engineering recruitment in Telangana. The 120-minute test features 100 MCQs (80 core technical, 20 general) with no negative marking. Application fee is ₹700 for General and ₹400 for Telangana reserved candidates.

Sample TSGENCO AE Practice Questions

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

1A network consists of linear resistors and ideal voltage sources. If the open-circuit voltage across two terminals is 12 V and the short-circuit current is 3 A, what is the Thevenin equivalent resistance of the network looking into these terminals?

A.4 Ω

B.36 Ω

C.0.25 Ω

D.8 Ω

Explanation: According to Thevenin's theorem, any linear bilateral active network can be replaced by an equivalent circuit containing a voltage source in series with a resistance. The Thevenin equivalent resistance (Rth) is calculated as the ratio of the open-circuit voltage (Voc) to the short-circuit current (Isc) across the terminals. Here, Rth = Voc / Isc = 12 V / 3 A = 4 Ω.

2A DC source has an internal resistance of 10 Ω and is connected to a variable load resistor RL. What is the value of RL for which maximum power is transferred to the load, and what is the efficiency of power transfer under this condition?

A.RL = 10 Ω, Efficiency = 50%

B.RL = 5 Ω, Efficiency = 50%

C.RL = 10 Ω, Efficiency = 100%

D.RL = 0 Ω, Efficiency = 0%

Explanation: The Maximum Power Transfer Theorem states that maximum power is transferred from a source to a load when the load resistance equals the internal resistance of the source (RL = Rth). Thus, RL = 10 Ω. Under this condition, the power lost in the source's internal resistance is equal to the power delivered to the load. Therefore, the total power generated is split equally, resulting in an efficiency of exactly 50%.

3When applying the Superposition Theorem to a linear circuit containing both independent and dependent sources, which of the following statements is correct?

A.Only independent sources are deactivated one at a time; dependent sources are never deactivated.

B.Both independent and dependent sources are deactivated one at a time.

C.Dependent sources are deactivated one at a time; independent sources are left untouched.

D.All sources, including dependent ones, must be replaced by their internal resistances simultaneously.

Explanation: In the Superposition Theorem, independent sources are deactivated (voltage sources replaced by short circuits, current sources by open circuits) one by one to find the individual responses. Dependent sources, however, represent active elements controlled by circuit variables (voltage or current) elsewhere in the circuit. Deactivating them would alter the constraint relationships of the circuit. Thus, dependent sources must remain active and unchanged throughout the superposition process.

4A series RL circuit with R = 5 Ω and L = 2 H is connected to a constant DC voltage source of 10 V at t = 0. What is the expression for the current i(t) through the circuit for t ≥ 0, assuming zero initial current?

A.i(t) = 2(1 - e^{-2.5t}) A

B.i(t) = 2(1 - e^{-0.4t}) A

C.i(t) = 10(1 - e^{-2.5t}) A

D.i(t) = 2e^{-2.5t} A

Explanation: The transient response of a series RL circuit connected to a DC voltage V is given by i(t) = (V/R)(1 - e^{-t/τ}) A, where the time constant τ = L/R. Here, R = 5 Ω, L = 2 H, and V = 10 V. The steady-state current is V/R = 10/5 = 2 A. The time constant is τ = 2/5 = 0.4 s, so 1/τ = 2.5. Substituting these values gives i(t) = 2(1 - e^{-2.5t}) A.

5A series RLC circuit has a resistance R = 2 Ω, inductance L = 8 mH, and capacitance C = 1 μF. What is the quality factor (Q-factor) of this circuit at resonance?

A.44.72

B.89.44

C.22.36

D.10.00

Explanation: The quality factor (Q) of a series RLC circuit at resonance is given by Q = (1/R) * √(L/C). Substituting the given values: R = 2 Ω, L = 8 * 10^-3 H, and C = 1 * 10^-6 F. Thus, L/C = (8 * 10^-3) / (1 * 10^-6) = 8000. Under the square root, √8000 ≈ 89.44. Multiplying by 1/R (which is 1/2 = 0.5) gives Q = 0.5 * 89.44 = 44.72.

6In a parallel RLC resonant circuit, what happens to the net dynamic impedance and the line current at the resonant frequency?

A.Impedance is maximum (resistive), and line current is minimum.

B.Impedance is minimum (resistive), and line current is maximum.

C.Impedance is zero, and line current is infinite.

D.Impedance is maximum (reactive), and line current is zero.

Explanation: At resonance in a parallel RLC circuit, the inductive and capacitive susceptances are equal and opposite, cancelling each other out (BL = BC). This leaves only the conductance of the resistor, making the total admittance minimum and purely conductive. Consequently, the dynamic impedance (Z = 1/Y) reaches its maximum value, which is purely resistive. Since the impedance is maximum, the line current drawn from the source is at its minimum.

7For a symmetrical and reciprocal two-port network, what are the relationships between the standard network parameters?

A.Z11 = Z22 (Symmetry) and AD - BC = 1 (Reciprocity)

B.Z11 = Z22 (Symmetry) and H12 = H21 (Reciprocity)

C.Y11 = -Y22 (Symmetry) and AD - BC = 1 (Reciprocity)

D.Z12 = Z21 (Symmetry) and A = D (Reciprocity)

Explanation: A two-port network is symmetrical if the input and output ports can be interchanged without changing the port voltages and currents. In terms of Z-parameters, symmetry requires Z11 = Z22 (or A = D in ABCD parameters). A network is reciprocal if it satisfies the reciprocity theorem, which means Z12 = Z21 (or AD - BC = 1 in transmission parameters).

8Which of Maxwell's equations represents Faraday's Law of Electromagnetic Induction in differential form?

A.∇ × E = -∂B/∂t

B.∇ ⋅ D = ρ

C.∇ × H = J + ∂D/∂t

D.∇ ⋅ B = 0

Explanation: Faraday's Law states that a time-varying magnetic field induces an electromotive force (EMF), which creates an electric field. In differential form, this is expressed as the curl of the electric field intensity E being equal to the negative rate of change of the magnetic flux density B: ∇ × E = -∂B/∂t.

9In a single-phase AC circuit, a voltage v(t) = 200 sin(wt + 30°) V is applied across an impedance, resulting in a current i(t) = 10 sin(wt - 30°) A. What is the active power consumed by the circuit?

A.500 W

B.1000 W

C.866 W

D.2000 W

Explanation: The voltage and current are given in peak values: Vm = 200 V and Im = 10 A. The RMS values are V = Vm/√2 = 200/√2 V and I = Im/√2 = 10/√2 A. The phase angle of voltage is θv = 30° and that of current is θi = -30°. The phase difference is φ = θv - θi = 30° - (-30°) = 60°. The active power (P) is given by P = V * I * cos(φ) = (Vm/√2) * (Im/√2) * cos(φ) = (Vm * Im / 2) * cos(φ) = (200 * 10 / 2) * cos(60°) = 1000 * 0.5 = 500 W.

10A linear circuit contains a resistor network connected to a load. If the Norton equivalent current In is 5 A and the Norton equivalent resistance Rn is 8 Ω, what are the corresponding Thevenin voltage Vth and Thevenin resistance Rth?

A.Vth = 40 V, Rth = 8 Ω

B.Vth = 40 V, Rth = 1.6 Ω

C.Vth = 0.625 V, Rth = 8 Ω

D.Vth = 5 V, Rth = 8 Ω

Explanation: Source transformation allows us to convert a Norton equivalent circuit (current source in parallel with a resistor) to a Thevenin equivalent circuit (voltage source in series with a resistor). The Thevenin resistance is equal to the Norton resistance: Rth = Rn = 8 Ω. The Thevenin voltage is equal to the short-circuit current multiplied by the Norton resistance: Vth = In * Rn = 5 A * 8 Ω = 40 V.

About the TSGENCO AE Exam

The TSGENCO Assistant Engineer (Electrical) Recruitment Exam is a highly competitive state-level written test. It consists of a single objective paper with 100 questions to be completed in 120 minutes. There is no negative marking. Section A comprises 80 questions testing technical electrical engineering concepts at the degree level. Section B comprises 20 questions assessing general awareness, numerical ability, basic English, computer knowledge, and Telangana culture & movement. Selection is made 100% based on the written exam performance.

Questions

100 scored questions

Time Limit

2 hours (120 minutes)

Passing Score

Qualifying marks: 40% for OC, 35% for BC, 30% for SC/ST/PH.

Exam Fee

₹700 for General/Other States; ₹400 for Telangana reserved categories. (Telangana State Power Generation Corporation Limited (TSGENCO))

TSGENCO AE Exam Content Outline

80%

Section A: Technical (Electrical)

80 questions testing circuits, machines, power systems, control systems, electronics, measurements, and power plants.

20%

Section B: General Awareness & Aptitude

20 questions testing numerical/analytical ability, GK, computer basics, English, and Telangana culture/movement.

How to Pass the TSGENCO AE Exam

What You Need to Know

Passing score: Qualifying marks: 40% for OC, 35% for BC, 30% for SC/ST/PH.
Exam length: 100 questions
Time limit: 2 hours (120 minutes)
Exam fee: ₹700 for General/Other States; ₹400 for Telangana reserved categories.

Keys to Passing

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

TSGENCO AE Study Tips from Top Performers

1Dedicate 80% of your prep time to core Electrical Engineering topics, as Section A accounts for 80 of the 100 total marks.

2Review electric circuits, machines (transformers, induction motors, DC machines), and power systems (transmission lines, fault analysis, switchgear) thoroughly.

3Solve numerical problems regularly to build speed; since there are 100 questions to be solved in 120 minutes, quick calculation is essential.

4Spend regular time learning the history and cultural aspects of Telangana, including the Telangana movement, as these questions are unique to the state exam.

5Ensure you practice basic computer concepts, such as MS Office shortcuts, networking basics, and hardware/software terms, to secure easy marks in Section B.

Frequently Asked Questions

What is the exam structure for TSGENCO AE?

The exam is a single objective written paper consisting of 100 multiple-choice questions (MCQs). The duration is 2 hours (120 minutes). Section A covers technical subjects (80 questions), and Section B covers general aptitude and general awareness (20 questions).

Is there negative marking in the TSGENCO AE written test?

No, there is no negative marking for incorrect answers. Each correct response is awarded 1 mark.

What is the application fee for TSGENCO AE?

The application processing fee is ₹400 for all candidates. General/Other States candidates must also pay an examination fee of ₹300, making their total ₹700. Candidates belonging to SC, ST, BC, EWS, and PH categories of Telangana are exempted from the exam fee, paying only the ₹400 processing fee.

What are the eligibility requirements for TSGENCO AE Electrical?

Candidates must hold a regular B.E. / B.Tech / A.M.I.E. degree in Electrical Engineering or Electrical & Electronics Engineering from a recognized university. The age limit is 18 to 44 years, with relaxations for reserved categories.

How are candidates selected for the TSGENCO Assistant Engineer post?

Selection is 100% based on the marks secured in the objective written examination. There is no personal interview, physical test, or typing test. Tie-breakers are resolved using age (older candidate preferred).

What topics are covered in the General Section (Section B)?

Section B comprises 20 questions covering: 1) Analytical and Numerical Ability, 2) General Awareness, 3) English Language proficiency, 4) Basic Computer Knowledge, and 5) Telangana Culture and Movement history.

100+ Free TSGENCO AE Practice Questions

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