Q301. The torque of an induction motor is maximum when
A) Slip = 0
B) Slip = 1
C) Rotor reactance = Rotor resistance
D) Supply frequency is maximum
Answer: C
Explanation: Maximum torque occurs when rotor reactance equals rotor resistance at standstill frequency.
Q302. In an induction motor, if supply frequency increases while keeping voltage constant, torque will
A) Increase
B) Decrease
C) Remain constant
D) Become zero
Answer: B
Explanation: Torque ∝ V²/f², so increasing frequency decreases torque if voltage is constant.
Q303. Crawling in induction motor occurs due to
A) High slip
B) Harmonics in air-gap flux
C) Bearing friction
D) Voltage unbalance
Answer: B
Explanation: Crawling is caused by 7th harmonic torque which causes motor to run at 1/7th synchronous speed.
Q304. In a 4-pole, 50 Hz induction motor, the synchronous speed is
A) 3000 rpm
B) 1500 rpm
C) 1000 rpm
D) 750 rpm
Answer: B
Explanation: Ns = 120f/P = 120×50/4 = 1500 rpm.
Q305. The slip of an induction motor at standstill is
A) 0
B) 1
C) Between 0 and 1
D) Negative
Answer: B
Explanation: Slip s = (Ns – N)/Ns = (Ns – 0)/Ns = 1 at standstill.
Q306. The slip at no-load in an induction motor is approximately
A) 0.001
B) 0.01
C) 0.1
D) 1.0
Answer: A
Explanation: At no-load, rotor speed ≈ synchronous speed, so slip ≈ 0.001.
Q307. Starting torque of a squirrel cage induction motor is
A) Low
B) High
C) Medium
D) Very high
Answer: A
Explanation: Squirrel cage motors have low starting torque due to fixed rotor resistance.
Q308. Double cage induction motors are used to obtain
A) High efficiency
B) High starting torque
C) High speed
D) Low starting current
Answer: B
Explanation: Double cage design provides high starting torque and good running performance.
Q309. The rotor of an induction motor always runs
A) At synchronous speed
B) Below synchronous speed
C) Above synchronous speed
D) Alternately above and below
Answer: B
Explanation: Rotor speed < synchronous speed to induce current and produce torque.
Q310. Slip speed is the difference between
A) Rotor and shaft speed
B) Rotor and synchronous speed
C) Stator and rotor speed
D) None of these
Answer: B
Explanation: Slip speed = Ns – N, difference between synchronous and rotor speed.
Q311. The power factor of an induction motor at no-load is
A) Unity
B) High
C) Low
D) Zero
Answer: C
Explanation: At no-load, current is mostly magnetizing, hence power factor is low.
Q312. The rotor input power of an induction motor is
A) Air-gap power
B) Shaft power
C) Copper loss
D) Mechanical loss
Answer: A
Explanation: Power transferred across air gap = rotor input power.
Q313. The starting torque of an induction motor is proportional to
A) Voltage
B) Voltage²
C) Current
D) Current²
Answer: B
Explanation: Torque ∝ V², so doubling voltage quadruples torque.
Q314. If rotor copper loss is 400 W and slip is 4%, rotor input power is
A) 400 W
B) 10,000 W
C) 16,000 W
D) 100 W
Answer: B
Explanation: Rotor copper loss = s × rotor input ⇒ 400 = 0.04 × Pin ⇒ Pin = 10,000 W.
Q315. In an induction motor, air-gap flux is
A) Proportional to voltage
B) Proportional to frequency
C) Independent of voltage
D) Proportional to current
Answer: A
Explanation: Flux ∝ V/f; if frequency constant, flux ∝ voltage.
Q316. For constant torque, if supply frequency is doubled, the voltage should be
A) Halved
B) Doubled
C) Constant
D) Four times
Answer: B
Explanation: To maintain constant flux, V ∝ f.
Q317. The equivalent circuit of an induction motor resembles
A) Transformer
B) DC motor
C) Alternator
D) Amplifier
Answer: A
Explanation: Induction motor is essentially a rotating transformer.
Q318. Stator copper loss is proportional to
A) Current
B) Current²
C) Slip
D) Speed
Answer: B
Explanation: Copper loss = I²R, proportional to square of current.
Q319. A 6-pole, 50 Hz motor runs at 970 rpm. Slip is
A) 0.03
B) 0.05
C) 0.02
D) 0.01
Answer: A
Explanation: Ns = 1000 rpm, s = (1000–970)/1000 = 0.03.
Q320. Which component decides the torque direction?
A) Rotor current phase
B) Stator current
C) Supply voltage
D) Power factor
Answer: A
Explanation: Torque direction depends on phase of rotor current relative to stator flux.
Q321. In a slip-ring induction motor, external resistance is added to
A) Increase speed
B) Decrease speed
C) Increase starting torque
D) Reduce torque
Answer: C
Explanation: External rotor resistance improves starting torque.
Q322. Plugging is a method of
A) Starting
B) Speed control
C) Braking
D) Power factor correction
Answer: C
Explanation: Plugging reverses phase sequence, causing braking torque.
Q323. Speed control of induction motor by pole changing is possible in
A) Squirrel cage motor
B) Slip-ring motor
C) Both
D) None
Answer: A
Explanation: In squirrel cage motors, pole-changing method (Dahlander connection) is used.
Q324. Slip of an induction motor during regenerative braking is
A) Positive
B) Negative
C) Zero
D) Infinite
Answer: B
Explanation: When rotor speed > synchronous, slip becomes negative.
Q325. Rotor frequency at standstill is
A) f/2
B) 0
C) f
D) 2f
Answer: C
Explanation: At standstill, slip = 1 ⇒ rotor frequency = supply frequency.
Q326. Induction motor cannot run at synchronous speed because
A) No rotor current is induced
B) Torque becomes zero
C) Slip is zero
D) All of these
Answer: D
Explanation: At synchronous speed, no emf induced → no current → no torque.
Q327. The starting current of an induction motor is
A) Low
B) Same as full-load current
C) 5–7 times full-load current
D) Zero
Answer: C
Explanation: High inrush current due to low rotor impedance at standstill.
Q328. In star-delta starter, starting current is reduced to
A) 1/√3 of direct starting current
B) 1/3 of full-load current
C) 1/√3 of line current
D) 1/3 of voltage
Answer: A
Explanation: Star-delta reduces line voltage by 1/√3 per phase, so current reduces accordingly.
Q329. The torque-slip curve of induction motor is
A) Linear
B) Parabolic
C) Non-linear
D) Exponential
Answer: C
Explanation: It’s non-linear; linear at low slip, hyperbolic beyond max torque.
Q330. The speed regulation of induction motor is
A) Poor
B) Excellent
C) Infinite
D) Depends on load
Answer: B
Explanation: Speed regulation is good since slip variation is small.
Q331. For constant load, slip increases with
A) Voltage decrease
B) Voltage increase
C) Frequency increase
D) Rotor resistance decrease
Answer: A
Explanation: Lower voltage reduces torque, causing higher slip for same load.
Q332. The torque at synchronous speed is
A) Maximum
B) Zero
C) Minimum but not zero
D) Constant
Answer: B
Explanation: No rotor current at synchronous speed, hence zero torque.
Q333. The magnetizing component of current in an induction motor is
A) In phase with voltage
B) In phase with flux
C) Lags voltage by 90°
D) Leads voltage by 90°
Answer: C
Explanation: Magnetizing current lags stator voltage by 90°.
Q334. The main disadvantage of rotor resistance control is
A) Low efficiency
B) High cost
C) Complex design
D) Poor torque
Answer: A
Explanation: Energy is wasted as heat in external resistors, reducing efficiency.
Q335. The frequency of rotor current at 2% slip and 50 Hz supply is
A) 1 Hz
B) 2 Hz
C) 50 Hz
D) 25 Hz
Answer: A
Explanation: fr = s × f = 0.02 × 50 = 1 Hz.
Q336. Induction motor works on
A) Faraday’s law
B) Lenz’s law
C) Both A and B
D) Ampere’s law
Answer: C
Explanation: Operation depends on electromagnetic induction (Faraday + Lenz laws).
Q337. A deep-bar rotor improves
A) Power factor
B) Starting torque
C) Efficiency
D) Cooling
Answer: B
Explanation: Deep bars increase effective resistance at start, improving torque.
Q338. The core losses in an induction motor are practically
A) Constant
B) Variable with load
C) Proportional to slip
D) Zero
Answer: A
Explanation: Core losses depend on voltage and frequency, nearly constant.
Q339. The synchronous wattmeter method is used to measure
A) Rotor copper loss
B) Input power
C) Air-gap power
D) Torque
Answer: D
Explanation: Torque can be measured by synchronous wattmeter method.
Q340. The rotor speed in regenerative braking is
A) Less than Ns
B) Greater than Ns
C) Equal to Ns
D) Independent of Ns
Answer: B
Explanation: During regeneration, rotor runs faster than synchronous speed.
Q341. In an induction motor, torque is proportional to
A) Slip
B) (Slip)/(Rotor reactance²)
C) (Slip × voltage²)/(Rotor impedance²)
D) (Voltage²)/Slip
Answer: C
Explanation: Torque ∝ sE₂²R₂/(R₂²+(sX₂)²).
Q342. For small slip values, torque is
A) Constant
B) ∝ slip
C) ∝ 1/slip
D) ∝ slip²
Answer: B
Explanation: At low slip, torque ≈ K×s since R₂ dominates sX₂.
Q343. For high slip values, torque is
A) ∝ 1/slip
B) ∝ slip
C) Constant
D) Zero
Answer: A
Explanation: At large slip, sX₂ >> R₂ ⇒ torque ∝ 1/s.
Q344. Circle diagram is used to determine
A) Power
B) Efficiency
C) Torque
D) All of these
Answer: D
Explanation: Circle diagram gives complete performance including torque, PF, and efficiency.
Q345. Slip rings in induction motor are short-circuited when
A) Starting
B) Running
C) Stopping
D) During maintenance
Answer: B
Explanation: External resistance is cut out after start; rings shorted during normal run.
Q346. The voltage induced in rotor bars is due to
A) Stator current
B) Magnetic field rotation
C) Load torque
D) Shaft speed
Answer: B
Explanation: Rotating magnetic field induces emf in rotor bars.
Q347. The main cause of unbalanced currents in induction motor is
A) Voltage unbalance
B) Rotor resistance
C) Bearing failure
D) Load variation
Answer: A
Explanation: Unequal phase voltages lead to unbalanced stator currents.
Q348. The air-gap power in induction motor is
A) Mechanical power
B) Rotor input
C) Shaft power
D) Electrical loss
Answer: B
Explanation: Power crossing air gap = rotor input.
Q349. If slip = 0.05 and rotor copper loss = 500 W, then mechanical power developed =
A) 25 W
B) 475 W
C) 9500 W
D) 5000 W
Answer: C
Explanation: Pmech = (1–s)×Pg = (1–0.05)×(500/0.05) = 9500 W.
Q350. Induction motor speed control using frequency variation is
A) Efficient
B) Inefficient
C) Obsolete
D) Not possible
Answer: A
Explanation: V/f control is most efficient modern method for variable-speed drives.
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