Home Subjects Electrical Engg (EEE)

Electrical Engg (EEE)

Electrical machines, power systems, circuits

321 Q 7 Topics Take Mock Test
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Difficulty: All Easy Medium Hard 221–230 of 321
Topics in Electrical Engg (EEE)
Q.221 Medium Electrical Machines
An induction motor operating at slip s = 0.04 has rotor frequency of 2 Hz. Find the supply frequency.
A 40 Hz
B 50 Hz
C 60 Hz
D 48 Hz
Correct Answer:  B. 50 Hz
EXPLANATION

Rotor frequency = s × f. Therefore, 2 = 0.04 × f, giving f = 50 Hz

Test
Q.222 Medium Electrical Machines
A DC shunt motor shows increased speed. What could be the reason?
A Increased load
B Increased shunt field current
C Decreased shunt field current
D Increased armature resistance
Correct Answer:  C. Decreased shunt field current
EXPLANATION

Speed = (V - IaRa)/(kΦ). Decreasing field current reduces flux Φ, increasing motor speed.

Test
Q.223 Medium Electrical Machines
A 100 kVA transformer has 1000 primary turns and 100 secondary turns. The turns ratio and voltage ratio are:
A k=0.1, voltage ratio=10:1
B k=10, voltage ratio=10:1
C k=0.1, voltage ratio=1:10
D k=100, voltage ratio=100:1
Correct Answer:  A. k=0.1, voltage ratio=10:1
EXPLANATION

Turns ratio k = N2/N1 = 100/1000 = 0.1. Voltage ratio V1/V2 = N1/N2 = 10/1 or 10:1 (step-down).

Test
Q.224 Medium Electrical Machines
In a synchronous motor, the damper winding is used to:
A Reduce noise
B Provide starting torque (asynchronous torque)
C Reduce harmonics
D Improve power factor
Correct Answer:  B. Provide starting torque (asynchronous torque)
EXPLANATION

Damper windings provide asynchronous starting torque to accelerate the rotor near synchronous speed before synchronous operation begins.

Test
Q.225 Medium Electrical Machines
What is the relationship between slip and rotor frequency in an induction motor?
A fr = s × fs
B fr = s × (1-s) × fs
C fr = (1-s) × fs
D fr = fs/s
Correct Answer:  A. fr = s × fs
EXPLANATION

Rotor frequency fr = slip × stator frequency = s × fs, where fs = 50 Hz for 50 Hz supply.

Test
Q.226 Medium Electrical Machines
The power factor of an induction motor at full load is typically:
A 0.7-0.8 lagging
B 0.9-0.95 leading
C 0.85-0.95 lagging
D Unity (1.0)
Correct Answer:  C. 0.85-0.95 lagging
EXPLANATION

Induction motors operate at lagging power factor (0.85-0.95) due to reactive magnetizing current required for flux production.

Test
Q.227 Medium Electrical Machines
A DC series motor is suitable for:
A Constant speed operation
B Applications requiring high starting torque
C Precision positioning
D Light load conditions
Correct Answer:  B. Applications requiring high starting torque
EXPLANATION

DC series motors have high starting torque (proportional to load current) and are used in electric vehicles, cranes, and hoists.

Test
Q.228 Medium Electrical Machines
Which method is NOT used for starting a three-phase induction motor?
A Direct on line starting
B Star-delta starting
C Autotransformer starting
D Frequency variation starting
Correct Answer:  D. Frequency variation starting
EXPLANATION

Standard starting methods are DOL, star-delta, autotransformer, and rotor resistance. Frequency variation is used for speed control, not starting.

Test
Q.229 Medium Electrical Machines
The starting current of a 3-phase induction motor is typically how many times the full-load current?
A 2-3 times
B 4-7 times
C 10-12 times
D 15-20 times
Correct Answer:  B. 4-7 times
EXPLANATION

Induction motors draw 4-7 times full-load current during starting because rotor speed is zero and rotor resistance appears as reactance.

Test
Q.230 Medium Electrical Machines
The efficiency of a transformer is maximum when:
A Core loss = Copper loss
B Core loss = 2 × Copper loss
C Copper loss = 0
D Core loss = 0
Correct Answer:  A. Core loss = Copper loss
EXPLANATION

Transformer efficiency is maximum when core loss (Pi) equals copper loss (Pcu), i.e., Pi = Pcu.

Test
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