AC Induction & Synchronous Machines
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📘 Comprehensive Syllabus & Examination Guide
AC Induction & Synchronous Machines
Official curriculum roadmap, subject/topic distribution, negative marking rules, pacing guidelines, and solved sample questions.
🎯 Mapped Subjects & Topic Question Distribution
Total Question Pool
100%
25 MCQs
Combined Active Syllabus
AC Induction & Synchronous Machines
25 MCQs
Topic Pool
📊 Question Pool Structure
25 MCQs across fundamental, intermediate, and advanced concept tiers.
⚡ Recommended Pacing
45 to 60 seconds per MCQ. Flag complex problems and preserve 10 minutes for final revision.
⚖️ Scoring & Negative Marking
+1 mark per correct answer. In competitive tests with negative marking, -0.25 applies for incorrect guesses.
💡 Strategic Preparation & Exam Hall Guidelines
To maximize your score on AC Induction & Synchronous Machines, candidates are advised to follow a structured three-pass approach. In the First Pass, solve all direct recall and formula-based questions within 30 seconds each to secure foundational marks. In the Second Pass, tackle multi-step analytical and quantitative reasoning problems. In the Third Pass, review marked questions and verify calculations.
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Solved Blueprint Examples
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📝 Pre-Rendered Solved Sample Questions & Detailed Solutions
Review the solved problems below to understand question phrasing, answer choices, and step-by-step solution logic prior to starting the full interactive practice drill:
Sample Question 1
AC Induction & Synchronous Machines
Easy • Electrical Engineering
The synchronous speed (Ns) of a 3-phase, 4-pole induction motor connected to a 50 Hz AC supply is:
A
3000 RPM
B
1500 RPM
C
1440 RPM
D
750 RPM
✓ Correct Answer: B - 1500 RPM
📖 Step-by-Step Solution & Conceptual Rationale:
Synchronous speed Ns = 120 * f / P = (120 * 50) / 4 = 6000 / 4 = 1500 RPM. The actual rotor speed N will be slightly less (e.g., 1440 RPM) due to slip.
Sample Question 2
AC Induction & Synchronous Machines
Easy • Electrical Engineering
The slip (s) of a 3-phase induction motor operating at standstill (rotor locked) is:
A
s = 0
B
s = 1.0 (100%)
C
s = -1.0
D
s = 0.04
✓ Correct Answer: B - s = 1.0 (100%)
📖 Step-by-Step Solution & Conceptual Rationale:
Slip is defined as s = (Ns - N) / Ns. At standstill, rotor speed N = 0, so s = (Ns - 0) / Ns = 1.0 (or 100%). At synchronous speed (N = Ns), slip is zero.
Sample Question 3
AC Induction & Synchronous Machines
Easy • Electrical Engineering
The frequency of rotor induced currents (f_r) in an induction motor running at slip s with stator supply frequency f is:
A
f_r = f
B
f_r = s * f
C
f_r = f / s
D
f_r = (1 - s) * f
✓ Correct Answer: B - f_r = s * f
📖 Step-by-Step Solution & Conceptual Rationale:
Rotor EMF is induced by relative motion between the rotating stator field (Ns) and rotor (N). Since relative speed is s * Ns, rotor frequency is f_r = s * (P * Ns / 120) = s * f. At full load (s ≈ 0.03-0.05 on 50 Hz), f_r is only 1.5 to 2.5 Hz.
Sample Question 4
AC Induction & Synchronous Machines
Medium • Electrical Engineering
In a 3-phase induction motor, maximum torque (breakdown torque) occurs at a slip s_max equal to:
A
s_max = R2 / X2
B
s_max = X2 / R2
C
s_max = R2^2 / X2
D
s_max = sqrt(R2 / X2)
✓ Correct Answer: A - s_max = R2 / X2
📖 Step-by-Step Solution & Conceptual Rationale:
From maximum power transfer principles applied to the rotor equivalent circuit, maximum developed torque occurs when rotor resistance equals stand-still rotor leakage reactance: R2 = s * X2, giving s_max = R2 / X2.
Sample Question 5
AC Induction & Synchronous Machines
Medium • Electrical Engineering
The magnitude of maximum developed torque in a 3-phase induction motor is:
A
Directly proportional to rotor resistance R2
B
Independent of rotor circuit resistance R2
C
Inversely proportional to supply voltage squared
D
Directly proportional to slip
✓ Correct Answer: B - Independent of rotor circuit resistance R2
📖 Step-by-Step Solution & Conceptual Rationale:
Substituting s_max = R2/X2 into the torque equation reveals that T_max = (3 / (2 * omega_s)) * (V1^2 / (2 * X2)). Thus, T_max depends on V1^2 and X2, but is completely independent of rotor resistance R2 (R2 only shifts the slip at which T_max occurs).
Sample Question 6
AC Induction & Synchronous Machines
Medium • Electrical Engineering
Rotor skewing in squirrel-cage induction motors is employed to:
A
Prevent cogging (magnetic locking), crawling, and reduce acoustic magnetic hum
B
Increase maximum breakdown torque
C
Eliminate stator copper losses
D
Provide regenerative braking
✓ Correct Answer: A - Prevent cogging (magnetic locking), crawling, and reduce acoustic magnetic hum
📖 Step-by-Step Solution & Conceptual Rationale:
Skewing rotor conductor bars slightly along the shaft axis prevents stator and rotor tooth harmonics from locking together (cogging) during startup, suppresses harmonic synchronous torques causing crawling at 1/7th speed, and lowers electromagnetic humming noise.
Sample Question 7
AC Induction & Synchronous Machines
Hard • Electrical Engineering
Crawling in a 3-phase squirrel cage induction motor is primarily caused by:
A
High supply voltage transients
B
Harmonic space fields produced by stator winding, predominantly the 7th harmonic
C
Rotor bar open circuits
D
Unbalanced line currents
✓ Correct Answer: B - Harmonic space fields produced by stator winding, predominantly the 7th harmonic
📖 Step-by-Step Solution & Conceptual Rationale:
Stator space harmonics produce parasitic rotating fields. The 7th space harmonic rotates forward at Ns/7. Superimposing this on the fundamental torque-speed curve creates a stable dip where the motor can 'crawl' stably at slightly below 1/7th synchronous speed.
Sample Question 8
AC Induction & Synchronous Machines
Medium • Electrical Engineering
Cogging (magnetic locking) in an induction motor occurs when:
A
Supply frequency is reduced by 50%
B
The number of stator slots is equal to or an integral multiple of the number of rotor slots
C
Rotor resistance is equal to leakage reactance
D
The motor runs at 150% full load
✓ Correct Answer: B - The number of stator slots is equal to or an integral multiple of the number of rotor slots
📖 Step-by-Step Solution & Conceptual Rationale:
When stator slot count S1 equals rotor slot count S2 (or an integer ratio), the stator and rotor teeth align perfectly at standstill, creating minimum reluctance paths that magnetically lock the rotor, preventing it from starting.
Sample Question 9
AC Induction & Synchronous Machines
Medium • Electrical Engineering
A Star-Delta starter reduces the starting current of a 3-phase induction motor compared to Direct-On-Line (DOL) starting by a factor of:
A
1 / 2 (50%)
B
1 / sqrt(3) (57.7%)
C
1 / 3 (33.3%)
D
1 / 4 (25%)
✓ Correct Answer: C - 1 / 3 (33.3%)
📖 Step-by-Step Solution & Conceptual Rationale:
In star connection, phase voltage is V_line / sqrt(3). Since current per phase is proportional to voltage, phase current is reduced by 1/sqrt(3). The line current in star is equal to phase current, so line starting current is reduced to (1/sqrt(3)) * (1/sqrt(3)) = 1/3 (33.3%) of DOL delta starting current.
Sample Question 10
AC Induction & Synchronous Machines
Easy • Electrical Engineering
A wound-rotor (slip ring) induction motor allows inserting external resistance into the rotor circuit during startup to:
A
Increase starting torque while simultaneously reducing starting inrush current
B
Decrease starting torque
C
Increase synchronous speed
D
Operate without a rotating magnetic field
✓ Correct Answer: A - Increase starting torque while simultaneously reducing starting inrush current
📖 Step-by-Step Solution & Conceptual Rationale:
Adding external rotor resistance shifts the maximum torque peak toward standstill (s_max = R2/X2 = 1.0), achieving maximum possible starting torque (T_start = T_max) while increasing rotor impedance to choke starting current.
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