DC Machines
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📘 Comprehensive Syllabus & Examination Guide
DC 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%
30 MCQs
Combined Active Syllabus
DC Machines
30 MCQs
Topic Pool
📊 Question Pool Structure
30 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 DC 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
Showing 10 solved representative questions
📝 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
DC Machines
Easy • Electrical Engineering
The function of the commutator in a DC generator is to:
A
Collect alternating current directly for the load
B
Convert the alternating EMF (AC) induced in rotating armature conductors into direct current (DC) at the external brushes
C
Control the field flux
D
Reduce armature reaction
✓ Correct Answer: B - Convert the alternating EMF (AC) induced in rotating armature conductors into direct current (DC) at the external brushes
📖 Step-by-Step Solution & Conceptual Rationale:
Armature conductors always generate alternating current (AC) due to reversing magnetic polarities. The mechanical commutator acts as a rotating mechanical rectifier, reversing coil connections at each neutral crossing to produce unidirectional DC at the stationary carbon brushes.
Sample Question 2
DC Machines
Easy • Electrical Engineering
In a simplex lap-wound DC machine, the number of parallel paths (A) in the armature is:
A
Always equal to 2, independent of pole count
B
Equal to the number of poles (A = P)
C
Equal to P / 2
D
Equal to 2 * P
✓ Correct Answer: B - Equal to the number of poles (A = P)
📖 Step-by-Step Solution & Conceptual Rationale:
In lap winding, the number of parallel paths equals the number of poles (A = P), making it suitable for high-current, low-voltage machines. In wave winding, A = 2 regardless of poles, making it suitable for high-voltage, low-current machines.
Sample Question 3
DC Machines
Easy • Electrical Engineering
In a simplex wave-wound DC machine with 8 poles, how many parallel paths are there in the armature winding?
A
8
B
2
C
4
D
16
✓ Correct Answer: B - 2
📖 Step-by-Step Solution & Conceptual Rationale:
For any simplex wave winding, the number of parallel paths A is always equal to 2, irrespective of the number of magnetic poles P.
Sample Question 4
DC Machines
Easy • Electrical Engineering
Armature reaction in a DC generator causes:
A
Cross-magnetization (distortion) and demagnetization of main field flux
B
An increase in total terminal voltage
C
Elimination of mechanical brush friction
D
Uniform flux distribution under pole shoes
✓ Correct Answer: A - Cross-magnetization (distortion) and demagnetization of main field flux
📖 Step-by-Step Solution & Conceptual Rationale:
Armature reaction is the effect of armature MMF on the main magnetic field. It distorts the flux distribution across pole tips (cross-magnetization) and, due to magnetic saturation, reduces the total average flux per pole (demagnetization).
Sample Question 5
DC Machines
Medium • Electrical Engineering
In a DC generator under load, the Magnetic Neutral Axis (MNA) shifts:
A
Opposite to the direction of rotation
B
In the direction of rotation
C
Does not shift at all
D
Oscillates at 50 Hz
✓ Correct Answer: B - In the direction of rotation
📖 Step-by-Step Solution & Conceptual Rationale:
In a DC generator, cross-magnetizing armature flux crowds lines of force toward the trailing pole tips, shifting the resultant magnetic neutral axis (MNA) forward in the direction of rotation. In a DC motor, MNA shifts backward against rotation.
Sample Question 6
DC Machines
Medium • Electrical Engineering
Compensating windings in DC machines are installed in slots in the pole shoes and connected in series with the armature to:
A
Neutralize armature reaction under the pole faces across the entire load range
B
Increase field coil resistance
C
Provide regenerative braking
D
Reduce mechanical noise
✓ Correct Answer: A - Neutralize armature reaction under the pole faces across the entire load range
📖 Step-by-Step Solution & Conceptual Rationale:
Compensating windings carry armature current in the opposite direction to adjacent armature conductors under the pole faces. This cancels the cross-magnetizing MMF at its origin, preventing distortion, flashover across commutator segments, and severe field weakening during sudden overloads.
Sample Question 7
DC Machines
Medium • Electrical Engineering
Interpoles (commutating poles) in a DC machine are placed midway between main poles and connected in series with the armature to:
A
Compensate for voltage drop in the field regulator
B
Produce a commutating flux that cancels reactance voltage in coils undergoing commutation, providing sparkless commutation
C
Increase motor speed above base speed
D
Eliminate residual magnetism
✓ Correct Answer: B - Produce a commutating flux that cancels reactance voltage in coils undergoing commutation, providing sparkless commutation
📖 Step-by-Step Solution & Conceptual Rationale:
Interpoles produce an opposing commutating EMF in the short-circuited coil undergoing commutation, exactly neutralizing the self-induced reactance voltage (e_r = L * di/dt) and cross-magnetizing MMF in the interpolar zone, ensuring sparkless commutation at the brushes.
Sample Question 8
DC Machines
Easy • Electrical Engineering
Why must a self-excited DC shunt generator have residual magnetism in its field poles to build up voltage?
A
To keep the bearings lubricated
B
To induce an initial small EMF when rotated, which drives current through the shunt field to reinforce and build up magnetic flux
C
To prevent high inrush current
D
To avoid demagnetization of brushes
✓ Correct Answer: B - To induce an initial small EMF when rotated, which drives current through the shunt field to reinforce and build up magnetic flux
📖 Step-by-Step Solution & Conceptual Rationale:
Without residual flux, the induced EMF at startup would be zero, no field current could flow, and the generator could never build up voltage. The initial residual EMF drives current that aids the residual flux until the voltage builds up to the saturation intersection point.
Sample Question 9
DC Machines
Medium • Electrical Engineering
If a DC shunt generator fails to build up voltage, which of the following is the most probable cause?
A
Field circuit resistance is lower than the critical field resistance
B
Field circuit resistance is greater than the critical resistance (Rf > R_critical)
C
Armature resistance is too low
D
Load resistance is infinite
✓ Correct Answer: B - Field circuit resistance is greater than the critical resistance (Rf > R_critical)
📖 Step-by-Step Solution & Conceptual Rationale:
For voltage buildup, the slope of the field resistance line must be less than the initial tangent to the open-circuit magnetization curve (critical resistance). If Rf > R_critical, the field line does not intersect the saturation curve, and voltage fails to build up.
Sample Question 10
DC Machines
Easy • Electrical Engineering
The Back EMF (Eb) in a DC motor is mathematically related to terminal voltage V, armature current Ia, and armature resistance Ra by:
A
Eb = V + Ia * Ra
B
Eb = V - Ia * Ra
C
Eb = Ia * Ra - V
D
Eb = V / (Ia * Ra)
✓ Correct Answer: B - Eb = V - Ia * Ra
📖 Step-by-Step Solution & Conceptual Rationale:
By KVL across a DC motor armature: V = Eb + Ia * Ra (plus brush drop). Hence, Back EMF Eb = V - Ia * Ra. Eb acts as a self-regulating governor; as motor speed changes, Eb adjusts Ia to match the mechanical load torque.
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