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GATE electrical machines: 5 practice questions with answers

By the PrepMatic team · Updated 9 Oct 2026 · 4 min read

Electrical Machines is one of the ten sections of the GATE EE paper and a dependable source of numerical questions: efficiency, regulation, speed, torque and losses, all from a handful of equations. The method is to write the equation that connects the quantities given and asked, then substitute with care for units. The five original questions below cover transformers, DC machines, induction motors, alternators and autotransformers.

What GATE tests in electrical machines

The GATE 2027 EE paper lists Electrical Machines as a section, and the topic list was revised, so confirm yours on the official syllabus page. The standard content is single-phase and three-phase transformers (equivalent circuits, tests, regulation, efficiency, parallel operation, autotransformers), DC machines, three-phase induction motors, synchronous machines and, usually, special machines and drives.

Typical questions give a rating and a few test or operating values, then ask for efficiency, regulation, a load at maximum efficiency, a speed after a change, rotor losses or torque. The real paper has MCQ, MSQ and NAT questions. All five below are MCQ.

Practice questions

Question 1

A 100 kVA single-phase transformer has an iron loss of 1 kW and a full-load copper loss of 4 kW. The load, in kVA, at which its efficiency is maximum is:

• A. 100
• B. 75
• C. 50
• D. 25

Question 2

A 220 V DC shunt motor has an armature resistance of 0.5 Ω. At one load it draws an armature current of 20 A and runs at 1000 rpm. The load increases until the armature current is 40 A, with the supply voltage and field flux unchanged. The speed becomes approximately:

• A. 952 rpm
• B. 1000 rpm
• C. 905 rpm
• D. 875 rpm

Question 3

A three-phase induction motor develops 9.6 kW of mechanical power at a slip of 0.04. Neglecting mechanical losses, the rotor copper loss is:

• A. 384 W
• B. 416 W
• C. 960 W
• D. 400 W

Question 4

A three-phase alternator has a rated terminal voltage of 2000 V per phase. When the load is removed at the same field current, the terminal voltage rises to 2400 V per phase. The voltage regulation is:

• A. 16.7%
• B. 20%
• C. 24%
• D. 83.3%

Question 5

An ideal 100 kVA autotransformer steps 400 V down to 300 V. The part of the 100 kVA that is transferred by transformer action, with the rest transferred conductively, is:

• A. 25 kVA
• B. 33.3 kVA
• C. 75 kVA
• D. 100 kVA

Answers and explanations

1. C. Efficiency is maximum when the copper loss equals the iron loss. Copper loss varies as the square of the load fraction x, so x² × 4 kW = 1 kW, which gives x = 0.5. The load is 0.5 × 100 = 50 kVA. At that point the efficiency at unity power factor is 50 / (50 + 2) = 96.2%.
2. A. The back emf is Eb = V − Ia Ra, which is 220 − 20 × 0.5 = 210 V at first and 220 − 40 × 0.5 = 200 V after the load increase. With constant flux, speed is proportional to Eb, so N2 = 1000 × 200 / 210 = 952 rpm.
3. D. Mechanical power is (1 − s) times the air-gap power, so the air-gap power is 9.6 / 0.96 = 10 kW. The rotor copper loss is s times the air-gap power = 0.04 × 10 = 0.4 kW = 400 W. Option A (384 W) multiplies s by the mechanical power, which is the wrong base.
4. B. Regulation = (E0 − V) / V × 100 = (2400 − 2000) / 2000 × 100 = 20%. The base is the rated (full-load) terminal voltage, not the no-load voltage. Option A divides by E0.
5. A. In an autotransformer with input V1 and output V2, the fraction of the rated kVA transferred by transformer action is 1 − V2 / V1 = 1 − 300 / 400 = 0.25. So 25 kVA is transformed and 75 kVA is conducted directly. This is why an autotransformer is smaller than a two-winding transformer of the same rating when the voltage ratio is close to 1.

How to prepare electrical machines

• Learn the equivalent circuits well enough to redraw them from memory, including the approximate transformer circuit and the per-phase induction motor circuit. Every calculation rests on one of them.
• Memorise the key relations as short chains: transformer efficiency and the maximum-efficiency condition, DC motor speed N proportional to Eb / flux, induction motor power flow (air-gap power, mechanical power = (1 − s) × air-gap power, rotor copper loss = s × air-gap power), and alternator regulation.
• Practise the standard tests: open-circuit and short-circuit tests for the transformer parameters, no-load and blocked-rotor tests for the induction motor, and OC/SC tests for the alternator.
• Know the torque-slip curve of the induction motor, how speed control and starting methods change it, and the effect of rotor resistance.
• For synchronous machines, draw the phasor diagram for lagging, leading and unity power factor, and understand V-curves and the power-angle equation.
• Keep a unit checklist: per-phase versus line quantities, kVA versus kW, and star versus delta connections cause more lost marks here than any concept.

For the machine-side link to the grid, move to power systems. The GATE electrical preparation guide places Electrical Machines in the plan, and control systems covers the feedback side of motor drives.

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