Test: Electrical Machines- 2 - Electrical Engineering (EE) MCQ

# Test: Electrical Machines- 2 - Electrical Engineering (EE) MCQ

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## 20 Questions MCQ Test GATE Electrical Engineering (EE) Mock Test Series 2025 - Test: Electrical Machines- 2

Test: Electrical Machines- 2 for Electrical Engineering (EE) 2024 is part of GATE Electrical Engineering (EE) Mock Test Series 2025 preparation. The Test: Electrical Machines- 2 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Electrical Machines- 2 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Electrical Machines- 2 below.
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Test: Electrical Machines- 2 - Question 1

### The starting current of a 3ϕ induction motor is 5 times the rated current, while the rated slip is 4%. The ratio of starting torque to full-load torque is

Detailed Solution for Test: Electrical Machines- 2 - Question 1

Test: Electrical Machines- 2 - Question 2

### A single-phase transformer when supplied from 220V, 50 Hz has eddy current loss of 50W. If the transformer is connected to a voltage of 330V, 50 Hz, the eddy current loss will be

Detailed Solution for Test: Electrical Machines- 2 - Question 2

Test: Electrical Machines- 2 - Question 3

### If Pc and Psc represent core and full-load ohmic losses respectively, the maximum kVA delivered to load corresponding to maximum efficiency is equal to rated kVA multiplied by

Test: Electrical Machines- 2 - Question 4

A d.c. series motor is accidently connected to single-phase a.c. supply. The torque produced will be

Detailed Solution for Test: Electrical Machines- 2 - Question 4

In dc series motor, the ac currents through the field & armature windings will always be in the same direction. So torque will be unidirectional but pulsating due to ac.

Test: Electrical Machines- 2 - Question 5

The ‘synchronous-impedance method’ of finding the voltage regulation by a cylindrical rotor alternator is generally considered as

Detailed Solution for Test: Electrical Machines- 2 - Question 5

As unsaturated value of Zs is more than the saturated value, voltage regulation computed by emf method is much higher than the actual value. It is because of this reason that the emf method is called a pessimistic method.

Test: Electrical Machines- 2 - Question 6

Generally the no-load losses of an electrical machine is represented in its equivalent circuit by a

Detailed Solution for Test: Electrical Machines- 2 - Question 6

No-load losses (essentially core losses not F & W losses) are low and depend upon the applied voltage in an induction machine.

Test: Electrical Machines- 2 - Question 7

The power factor of a synchronous motor

Detailed Solution for Test: Electrical Machines- 2 - Question 7

Refer, inverted V-curves related to V-curves.

Test: Electrical Machines- 2 - Question 8

A 6-pole, 3-phase alternator running at 1000 rpm supplies to a 8-pole, 3-phase induction motor which has a rotor current of frequency 2 Hz. The speed at which the motor operates is

Detailed Solution for Test: Electrical Machines- 2 - Question 8

Test: Electrical Machines- 2 - Question 9

For maximum current during ‘Slip Test’ on a synchronous machine, the armature mmf aligns along

Test: Electrical Machines- 2 - Question 10

A 3-phase 50 MVA 10kV generator has a reactance of 0.2 ohm per phase. Hence the per-unit value of the reactance on a base of 100 MVA 25 kV will be

Test: Electrical Machines- 2 - Question 11

The results of a ‘Slip-Test’ for determining direct-axis (Xd) and quadrature-axis (Xq) reactances of a star-connected, salient-pole alternator are given below:

Phase values: Vmax = 108V; Vmin = 96V, Imax = 12A, Imin = 10A. Hence the two reactances will be

Detailed Solution for Test: Electrical Machines- 2 - Question 11

Test: Electrical Machines- 2 - Question 12

A two-winding transformer is used as an auto-transformer. The kVA rating of the auto-transformer compared to the twowinding transformer will be

Detailed Solution for Test: Electrical Machines- 2 - Question 12

Answer would depend upon the voltage ratio

and type of auto-transformer
i.e.., set-up or step-down.
For step-up transformer

Where a < 1

Let, N1 = N2 (Ideal two winding transformer)   For step-up transformer

Test: Electrical Machines- 2 - Question 13

A 20 kVA, 2000/200V, 1-phase transformer has name-plate leakage impedance of 8%. Voltage required to be applied on the high-voltage side to circulate full-load current with the low-voltage winding short-circuited will be

Detailed Solution for Test: Electrical Machines- 2 - Question 13

V = 0.08 × 1 pu
∴ V actual = 0.08 × 2000 = 160V

Test: Electrical Machines- 2 - Question 14

The full-load copper-loss and iron loss of a transformer are 6400W and 5000W respectively. The copper-loss and ironloss at half load will be, respectively

Detailed Solution for Test: Electrical Machines- 2 - Question 14

Copper loss = (1/2)2 × 6400 = 1600W Iron loss does not depend upon the load but upon the applied voltage.

Test: Electrical Machines- 2 - Question 15

In a 100 kVA, 1100/220V, 50 Hz single-phase transformer with 2000 turns on the high-voltage side, the open-circuit test result gives 200V, 91 A, 5kW on low-voltage side. The coreloss component of current is approximately

Detailed Solution for Test: Electrical Machines- 2 - Question 15

Test: Electrical Machines- 2 - Question 16

For a given torque, reducing the diverter-resistance of a d.c. series motor

Detailed Solution for Test: Electrical Machines- 2 - Question 16

reducing the diverter-resistance reduces the ϕ, so for a given load torque, Ia would increase.

would also increase.

Test: Electrical Machines- 2 - Question 17

Possible three-to-three phase transformer connection for parallel operation is

Detailed Solution for Test: Electrical Machines- 2 - Question 17

To ensure co-phasor primary and secondary windings.

Test: Electrical Machines- 2 - Question 18

A 4 kVA, 400/200V single-phase transformer has resistance of 0.02 p.u. and reactance of 0.06 p.u. Its actual resistance and reactance referred to high voltage side are respectively

Detailed Solution for Test: Electrical Machines- 2 - Question 18

Test: Electrical Machines- 2 - Question 19

Two 10 kV/440V, 1-phase transformers of ratings 600 kVA and 350 kVA are connected in parallel to share a load of 800 kVA. The reactances of the transformers, referred to the secondary side are 0.0198Ω and 0.0304Ω respectively (resistances negligible). The load shared by the two transformers will be respectively

Detailed Solution for Test: Electrical Machines- 2 - Question 19

Test: Electrical Machines- 2 - Question 20

The current drawn by a 120V d.c. motor with back e.m.f. of 110V and armature resistance of 0.4 ohm is

Detailed Solution for Test: Electrical Machines- 2 - Question 20

Eb = Vt – Ia Ra
∴ 110 = 120 – Ia × 0.4   ⇒ Ia = 25A

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