Free GATE EE mock test — 65 questions, GATE marking
One GATE EE paper in the real exam format: 65 questions across one timed section, 3 hr on a server-controlled clock.
About the real exam India's largest GATE branch with the deepest PSU pipeline — your gateway to M.Tech at IIT/IISc/NITs and PSU jobs at NTPC, PGCIL, BHEL and 30+ others.
- 65 Questions
- 3 hr Duration
- 1 Sections
- 100 Scored out of
Free, reversible until you begin · one attempt per account · no retake, no card.
The paper, start to finish
Sample questions with worked answers
8 original practice questions written by Myndaq in the GATE EE style. They are not taken from this mock paper or from official past papers. Try each one, then open the answer.
- Question 1General AptitudeOne correct answer
Tap P alone can fill an empty tank in 10 hours and tap Q alone can fill it in 15 hours. Both taps are opened together, and after 3 hours tap P is closed. How many more hours does tap Q alone take to fill the rest of the tank?
- 3
- 7.5
- 10.5
- 12
Show answer and explanation
Answer: B
Together the taps fill 1/10 + 1/15 = 3/30 + 2/30 = 1/6 of the tank per hour, so in 3 hours they fill 3 × 1/6 = 1/2 of the tank.
The remaining 1/2 is filled by Q alone at 1/15 per hour, which takes (1/2) ÷ (1/15) = 7.5 hours.
3 hours wrongly keeps both taps running for the second half. 10.5 hours counts only P's 3/10 in the first 3 hours (leaving 7/10 for Q), and 12 hours counts only Q's 1/5 (leaving 4/5). Only 7.5 hours uses the combined rate correctly.
- Question 2Engineering MathematicsType a numeric answer
Two fair six-sided dice are rolled. Given that at least one of the dice shows a 5, the probability that the sum of the two numbers is 8 is ______ (rounded off to two decimal places).
Show answer and explanation
Answer: 0.18 to 0.19
The sample space has 36 equally likely ordered outcomes. Outcomes with at least one 5: 6 with the first die showing 5, plus 6 with the second die showing 5, minus the outcome (5, 5) counted twice, giving 11.
Among these 11, the sum is 8 only for (5, 3) and (3, 5); (4, 4), (2, 6) and (6, 2) also sum to 8 but contain no 5.
P = 2/11 ≈ 0.1818, which rounds to 0.18. Using 12 in the denominator (double-counting (5, 5)) or the unconditional 5/36 are the usual slips.
- Question 3Electric CircuitsType a numeric answer
An ideal 24 V DC source in series with a 6 Ω resistor is connected across a 12 Ω resistor. A variable load resistor R is connected in parallel with the 12 Ω resistor. The maximum power, in W, that can be delivered to R is ______.
Show answer and explanation
Answer: 16
Remove R and find the Thevenin equivalent seen at its terminals.
Open-circuit voltage: Vth = 24 × 12/(6 + 12) = 16 V. Thevenin resistance (source shorted): Rth = 6 × 12/(6 + 12) = 4 Ω.
By the maximum power transfer theorem, R = Rth = 4 Ω, and Pmax = Vth²/(4 Rth) = 16²/(4 × 4) = 256/16 = 16 W.
Check: with R = 4 Ω the parallel combination is 12 × 4/16 = 3 Ω, the source current is 24/(6 + 3) = 8/3 A, the load voltage is 8 V, and 8²/4 = 16 W. Using 24²/(4 × 6) = 24 W ignores the 12 Ω shunt resistor.
- Question 4Signals and SystemsOne or more correct answers
A continuous-time system has input x(t) and output y(t) = t · x(t). Which of the following statements about this system is/are TRUE?
- The system is linear.
- The system is time-invariant.
- The system is causal.
- The system is BIBO stable.
Show answer and explanation
Answer: A, C
Linear: for input a x1(t) + b x2(t) the output is t(a x1(t) + b x2(t)) = a t x1(t) + b t x2(t), so superposition holds. A is true.
Time-invariant: the input x(t - t0) gives t · x(t - t0), whereas the original output delayed by t0 is (t - t0) · x(t - t0). These differ, so the system is time-varying. B is false.
Causal: y(t) depends only on x at the same instant t, never on future values (the system is in fact memoryless). C is true.
BIBO stable: the bounded input x(t) = 1 gives y(t) = t, which grows without bound. D is false.
- Question 5Electrical MachinesOne correct answer
A 3-phase, 4-pole, 50 Hz induction motor runs at 1440 rpm on full load. The power crossing the air gap to the rotor is 20 kW. The rotor copper loss is
- 768 W
- 800 W
- 833 W
- 19200 W
Show answer and explanation
Answer: B
Synchronous speed Ns = 120 f/P = 120 × 50/4 = 1500 rpm. Slip s = (1500 - 1440)/1500 = 0.04.
The air-gap power divides as rotor copper loss = s × Pag and gross mechanical power = (1 - s) × Pag. So rotor copper loss = 0.04 × 20000 = 800 W, and the gross mechanical power is 19200 W.
768 W wrongly multiplies the slip by the mechanical power, 833 W wrongly uses s/(1 - s) × Pag, and 19200 W is the gross mechanical power, not the loss.
- Question 6Power SystemsType a numeric answer
A 20 MVA, 11 kV, three-phase synchronous generator has a subtransient reactance of 0.25 pu on its own rating. It is running at no load with rated terminal voltage when a bolted three-phase short circuit occurs at its terminals. Neglecting resistance, the magnitude of the initial symmetrical fault current, in kA, is ______ (rounded off to two decimal places).
Show answer and explanation
Answer: 4.18 to 4.22
At no load the internal subtransient EMF equals the terminal voltage, 1.0 pu. The fault current is therefore I = 1.0/0.25 = 4 pu.
Base current = S/(√3 × V) = 20 × 10^6/(√3 × 11 × 10^3) = 20000/19.053 ≈ 1049.7 A.
Fault current = 4 × 1049.7 ≈ 4198.9 A ≈ 4.20 kA.
Forgetting the √3 in the base current gives about 7.27 kA, which is the typical error.
- Question 7Control SystemsOne correct answer
A unity negative feedback system has the open-loop transfer function G(s) = K/(s(s + 4)), with K > 0. The value of K for which the closed-loop system has a damping ratio of 0.5 is
- 4
- 8
- 16
- 64
Show answer and explanation
Answer: C
The closed-loop characteristic equation is 1 + G(s) = 0, that is s² + 4s + K = 0. Comparing with s² + 2ζωn s + ωn² = 0 gives ωn² = K and 2ζωn = 4.
With ζ = 0.5, ωn = 4/(2 × 0.5) = 4 rad/s, so K = ωn² = 16.
K = 4 gives ωn = 2 and ζ = 1 (critical damping). K = 64 gives ωn = 8 and ζ = 0.25. K = 8 gives ζ = 2/√8 ≈ 0.71. Only K = 16 gives ζ = 0.5.
- Question 8Power ElectronicsType a numeric answer
A single-phase fully controlled thyristor bridge rectifier is fed from a 230 V (rms), 50 Hz supply. The load is highly inductive, so the load current is continuous and ripple-free. For a firing angle of 60°, the average output voltage, in V, is ______ (rounded off to one decimal place).
Show answer and explanation
Answer: 103.3 to 103.8
For a single-phase fully controlled bridge with continuous load current, Vdc = (2Vm/π) cos α, where Vm is the peak supply voltage.
Vm = 230 × √2 ≈ 325.27 V, so 2Vm/π = 650.54/3.1416 ≈ 207.07 V.
With cos 60° = 0.5, Vdc ≈ 207.07 × 0.5 ≈ 103.5 V.
The formula (Vm/π)(1 + cos α), which gives about 155.3 V, applies to a resistive load or a half-controlled bridge where the output voltage cannot go negative, so it does not apply here.
What you get back
You get your raw marks and per-section scores. No scaled score, band, rank or percentile is estimated for this paper.
- Scored out of 100 raw marks, with per-section totals.
- Wrong answers can cost marks where the paper defines it - the per-section marks in the paper map are the marks the scorer applies.
How the real GATE EE levels are read
Top GATE EE rank-holders typically score 75-90 marks (out of 100) for AIR under 100, which unlocks IISc Bangalore, IIT Bombay/Delhi/Madras M.Tech admission and shortlisting at top-tier PSUs (PGCIL, NTPC E1 grade). Marks of 55-75 with AIR 100-1000 generally place candidates into IIT M.Tech across other branches or mid-tier PSU interviews. Qualifying cut-off (for valid GATE scorecard) is around 25 marks; approximately 15-17% of EE candidates qualify each year.
Context only - this free mock returns: You get your raw marks and per-section scores. No scaled score, band, rank or percentile is estimated for this paper.
The real GATE EE
GATE Electrical Engineering (EE) is the gateway for India's top M.Tech programmes at IISc, IITs and NITs in Power Systems, Power Electronics, Control and Instrumentation, Energy Systems and Machine Drives. It is also the recruitment screen for the country's largest PSU pipeline — NTPC, PowerGrid (PGCIL), BHEL, NHPC, IOCL, ONGC, GAIL, HPCL, NLC, DVC and over 30 other central PSUs hire EE candidates primarily on their GATE score. With ~150,000 candidates annually, EE is one of the most competitive papers in the GATE family. The exam tests breadth across 10 sections — engineering mathematics, electric circuits, electromagnetic fields, signals and systems, electrical machines, power systems, control systems, measurements, analog and digital electronics, and power electronics — with 65 questions in 180 minutes for a total of 100 marks.
What the real GATE EE covers
Ten official sections from the GATE 2026 EE syllabus, fully modelled at topic level: (1) Engineering Mathematics — Linear Algebra, Calculus, ODE/PDE, Complex Variables, Probability & Statistics with Correlation/Regression (13 marks, 15.3%). (2) Electric Circuits — KCL/KVL, network theorems, transient/AC steady-state, two-port, three-phase (8 marks, 9.4%). (3) Electromagnetic Fields — electrostatics, magnetostatics, induction, capacitance, inductance (7 marks, 8.2%). (4) Signals & Systems — LTI systems, Fourier series/transform, Laplace, Z-transform, sampling (9 marks, 10.6%). (5) Electrical Machines — transformers, DC machines, induction motors, synchronous machines (11 marks, 12.9%). (6) Power Systems — generation, transmission, per-unit, load flow, fault analysis, protection, stability (11 marks, 12.9%). (7) Control Systems — modelling, time/frequency response, Routh/Nyquist/Bode/root locus, compensators, PID, state space (9 marks, 10.6%). (8) Measurements — bridges, instruments, oscilloscope, error analysis (5 marks, 5.9%). (9) Analog & Digital Electronics — diodes, amplifiers, oscillators, op-amps, logic, ADC/DAC (7 marks, 8.2%). (10) Power Electronics — SCR/MOSFET/IGBT, DC-DC, AC-DC rectifiers, DC-AC inverters, sinusoidal PWM (5 marks, 5.9%).
How people prepare for it
- 1Build a 6-9 month plan: 3-4 months for first-pass concept coverage, 2-3 months for problem-solving depth, 1-2 months for full-length mocks and revision
- 2Prioritise heavy-weight sections: Engineering Math (13), Electrical Machines (11), Power Systems (11), Signals (9), Control (9) together cover 53 of 85 core marks
- 3Practice numerical answer type (NAT) problems daily — they have no negative marking and reward accuracy
- 4Master per-unit calculations, phasor algebra, transfer function manipulation, and Bode plot construction — these recur across multiple sections
- 5Solve at least 15 years of previous-year GATE EE questions; pattern recognition is the single biggest score multiplier
- 6Take a full-length 3-hour mock paper every Sunday from month 5 onwards; review every wrong answer the same day
- 7Maintain a formula notebook — EE involves 200+ standard formulas; reviewing them daily for the last 30 days is critical
- 8Don't neglect General Aptitude — 15 free marks if practised, easily lost if ignored
Figures above are as reported for the cycle named in each row and are refreshed each cycle. Sourced, dated facts for this exam (official links on every row) are on the GATE EE course page.
How this mirrors the real format
- The paper structure is the real format: a single timed section, 65 questions in total.
- Timing is server-controlled for the full 3 hr -- closing the tab does not pause the clock, exactly like our paid mock series.
- Marking follows the published paper: the per-section marks shown in the paper map are the marks the scorer applies.
- Navigation is free within a section: you can review and change answers before you submit.
- An on-screen calculator is provided in the sections where the exam allows one.
"Real format" means the paper shape, timing, marking and navigation above - not the official exam software, its visual identity, or any affiliation with the exam body.
What differs or is limited
- This free mock follows the same paper shape, timing, marking scheme and navigation as our paid GATE Electrical Engineering mock series. It is scored automatically. It does not include tutor explanations, analysis dashboards or practice content.
- Includes the virtual calculator and free navigation, as in the real exam.
Device and setup checklist
- A stable internet connection for the full duration.
- An on-screen calculator is provided where the exam allows one.
- A laptop or desktop is recommended for the closest exam experience.
One attempt, explicitly
Adding this mock to your account is free and reversible - you can change the exam any time before you begin. Only the confirmed Begin action uses your one attempt. No retake, no reset, no card.