Induction Motors MCQ [Free PDF] – 14 Objective Questions with Answers

The angle φ between stator voltage and stator current is the power-factor angle; input power factor = cos φ.

Maximum torque at start (s = 1) needs R₂ = X₂. A cage rotor’s resistance is fixed and low, and no external resistance can be added, so its maximum torque occurs at a slip below 1.

Pull-out (breakdown, maximum) torque is the peak of the torque-speed curve; in the paper’s figure the peak is labelled c.

External rotor resistance can be added through the slip rings, giving high starting torque and limited speed control — so slip-ring motors suit those duties.

For a given supply, the value of maximum torque is approximately independent of rotor resistance, while the slip at which it occurs is proportional to rotor resistance. Adding rotor resistance in a slip-ring motor can improve starting torque and shift maximum torque toward standstill. Key relation: s_max≈R₂/X₂ (referred form, simplified); T∝sR₂/(R₂²+(sX₂)²).

Exam focus: Rotor resistance shifts the torque-slip curve horizontally; it does not materially raise Tmax. Common trap: Maximum torque does not occur at synchronous speed because slip is zero there.

The torque-slip curve changes shape when rotor resistance is changed. The given initial s_max=0.2 is not needed to prove the qualitative relationships. Doubling R2 doubles the approximate slip at maximum torque until model limits; Tmax remains unchanged. Key relation: Approximate condition: s_max ∝ R2/Xeq. At s=1, starting torque rises as rotor resistance is increased toward the optimum R2≈Xeq.

Exam focus: Rotor resistance shifts the torque-slip curve horizontally, not vertically at Tmax. Common trap: Saying maximum torque occurs at synchronous speed; torque is zero at s=0.

C — Less than 1. In normal stable operation the full-load torque is below the breakdown (maximum) torque, so T_FL/T_max < 1 — typically about 0.4–0.5.

C — Increasing the rotor resistance. Higher rotor resistance shifts the maximum-torque slip upward while the maximum torque magnitude stays about the same, so the torque at full-load slip rises relative to Tmax.

An induction motor is essentially a transformer with a rotating, short-circuited secondary. The stator acts like the primary and the rotor like the secondary, with power crossing the air gap electromagnetically.

D — Induction generator. It generates when driven above synchronous speed and needs no separate DC rotor-excitation supply, although magnetizing reactive power is still required.

C — To reduce the supply voltage. At start the delta-rated winding is connected in star, so each phase gets V_L/√3; starting current and torque fall to one-third of direct-on-line values.

An induction motor requires appreciable magnetizing current because its magnetic circuit includes an air gap, whose reluctance is much larger than the nearly closed iron path of a transformer. Hence its no-load current is generally higher. Both assertion and reason are true, but the reason statement as printed is not the complete/correct explanation expected by the item, leading to option B.

Rotor-current frequency is proportional to slip: at standstill it equals supply frequency, and at synchronous speed it becomes zero. Rotor frequency fr=s f=0.05×50=2.5 Hz. At synchronous speed s=0, fr=0; at standstill s=1, fr=f. Key relation: s=(N_s−N)/N_s; f_r=s f_s.

Exam focus: Standstill: s=1; synchronous speed: s=0. Common trap: Rotor frequency is not independent of slip.

D — Less than 1. In a normal squirrel-cage induction motor the starting torque is lower than the maximum (breakdown) torque, which occurs at a slip well below 1.