Transformers
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📘 Comprehensive Syllabus & Examination Guide
Transformers
Official curriculum roadmap, subject/topic distribution, negative marking rules, pacing guidelines, and solved sample questions.
🎯 Mapped Subjects & Topic Question Distribution
Total Question Pool
100%
35 MCQs
Combined Active Syllabus
Transformers
35 MCQs
Topic Pool
📊 Question Pool Structure
35 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 Transformers, 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.
Practice with the interactive player below to evaluate your speed and accuracy under real exam pressure. Every question features full mathematical formulas, step-by-step worked solutions, and conceptual explanations vetted by Apex Rankers Academy subject matter specialists.
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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
Transformers
Easy • Electrical Engineering
An ideal transformer does not change:
A
Voltage and Current
B
Frequency and Power
C
Impedance and Voltage
D
Current and Flux
✓ Correct Answer: B - Frequency and Power
📖 Step-by-Step Solution & Conceptual Rationale:
A transformer is a static electromagnetic device that transfers electrical energy from one circuit to another without changing the frequency (f1 = f2) and with 100% efficiency in an ideal case (P_in = P_out, V1*I1 = V2*I2).
Sample Question 2
Transformers
Easy • Electrical Engineering
The EMF per turn in the primary winding of a transformer compared to the EMF per turn in its secondary winding is:
A
Greater than the secondary EMF per turn
B
Less than the secondary EMF per turn
C
Exactly equal to the secondary EMF per turn
D
Inversely proportional to the transformation ratio
✓ Correct Answer: C - Exactly equal to the secondary EMF per turn
📖 Step-by-Step Solution & Conceptual Rationale:
From Faraday's law, induced EMF per turn is E1/N1 = E2/N2 = 4.44 * f * phi_m. Since the same mutual magnetic flux links both windings, the induced EMF per turn is identical in both primary and secondary windings.
Sample Question 3
Transformers
Easy • Electrical Engineering
The open-circuit (OC) test on a transformer is performed to determine:
A
Full-load copper loss and equivalent resistance
B
Core (iron) loss and shunt branch parameters (Rc and Xm)
C
Leakage reactance of windings
D
Voltage regulation at full load
✓ Correct Answer: B - Core (iron) loss and shunt branch parameters (Rc and Xm)
📖 Step-by-Step Solution & Conceptual Rationale:
The open-circuit test is conducted at rated voltage and frequency, typically on the Low Voltage (LV) side with the HV side open. Because no-load current is tiny (2-5% of rated), series copper loss is negligible, and the wattmeter reading directly measures the core (hysteresis + eddy current) loss.
Sample Question 4
Transformers
Easy • Electrical Engineering
The short-circuit (SC) test on a transformer is conducted to measure:
A
Core (iron) losses at rated voltage
B
Full-load copper losses and equivalent series impedance (Req and Xeq)
C
Dielectric strength of insulation oil
D
Magnetizing current
✓ Correct Answer: B - Full-load copper losses and equivalent series impedance (Req and Xeq)
📖 Step-by-Step Solution & Conceptual Rationale:
The SC test is performed at reduced voltage (typically 5-10% of rated) on the High Voltage (HV) side with the LV side shorted, circulating full-load rated current. Because the applied voltage and core flux are very low, iron losses are negligible, and the input power represents the full-load copper loss (I^2 * Req).
Sample Question 5
Transformers
Medium • Electrical Engineering
A 100 kVA transformer has a maximum efficiency of 98% at 80% of full load. What is the ratio of full-load copper loss (P_cu_fl) to iron loss (P_i)?
A
P_cu_fl / P_i = 1.0
B
P_cu_fl / P_i = 1 / 0.8^2 = 1 / 0.64 ≈ 1.56
C
P_cu_fl / P_i = 0.8
D
P_cu_fl / P_i = 0.64
✓ Correct Answer: B - P_cu_fl / P_i = 1 / 0.8^2 = 1 / 0.64 ≈ 1.56
📖 Step-by-Step Solution & Conceptual Rationale:
Maximum efficiency occurs when variable copper loss equals constant iron loss: x^2 * P_cu_fl = P_i. At x = 0.8: 0.8^2 * P_cu_fl = P_i -> 0.64 * P_cu_fl = P_i -> P_cu_fl / P_i = 1 / 0.64 = 1.5625.
Sample Question 6
Transformers
Medium • Electrical Engineering
The condition for zero voltage regulation in a practical power transformer occurs at:
A
Unity power factor
B
Lagging power factor with tan(phi) = Req / Xeq
C
Leading power factor with tan(phi) = Req / Xeq
D
Zero power factor lagging
✓ Correct Answer: C - Leading power factor with tan(phi) = Req / Xeq
📖 Step-by-Step Solution & Conceptual Rationale:
Voltage regulation is approximate to (I2 * Req * cos(phi) +- I2 * Xeq * sin(phi)) / V2. For zero regulation, Req * cos(phi) - Xeq * sin(phi) = 0 (which requires a leading PF), yielding tan(phi) = Req / Xeq (leading).
Sample Question 7
Transformers
Medium • Electrical Engineering
The maximum voltage regulation in a transformer occurs at a:
A
Lagging power factor where tan(phi) = Xeq / Req
B
Leading power factor where tan(phi) = Xeq / Req
C
Unity power factor
D
Zero power factor leading
✓ Correct Answer: A - Lagging power factor where tan(phi) = Xeq / Req
📖 Step-by-Step Solution & Conceptual Rationale:
Differentiating the regulation expression with respect to phi and setting to zero yields the maximum regulation condition: tan(phi) = Xeq / Req at a lagging power factor, meaning cos(phi) = Req / Zeq (lagging).
Sample Question 8
Transformers
Easy • Electrical Engineering
Why are distribution transformers designed to achieve maximum efficiency at 50% to 70% of full load rather than at 100% full load?
A
They operate under heavy overload throughout the entire day
B
They remain connected to the supply 24 hours a day but operate at light or average load (30-60%) for most hours
C
To keep winding temperatures above ambient at all times
D
To increase leakage reactance
✓ Correct Answer: B - They remain connected to the supply 24 hours a day but operate at light or average load (30-60%) for most hours
📖 Step-by-Step Solution & Conceptual Rationale:
Distribution transformers are energized continuously (24 hours/day) with fluctuating consumer loads, averaging 50-70% capacity. Designing core loss to be very low and setting maximum efficiency around average load optimizes all-day energy efficiency.
Sample Question 9
Transformers
Easy • Electrical Engineering
All-day efficiency of a transformer is defined as the ratio of:
A
Output power in kW to Input power in kW at full load
B
Total energy output in kWh over 24 hours to Total energy input in kWh over 24 hours
C
KVA output to KVA input
D
Maximum efficiency multiplied by 24
✓ Correct Answer: B - Total energy output in kWh over 24 hours to Total energy input in kWh over 24 hours
📖 Step-by-Step Solution & Conceptual Rationale:
All-day efficiency (or commercial energy efficiency) is calculated on an energy basis: eta_allday = (Energy Output in kWh over 24 hours) / (Energy Input in kWh over 24 hours). It is vital for distribution transformers.
Sample Question 10
Transformers
Medium • Electrical Engineering
In an autotransformer with transformation ratio k = V2 / V1 (where k < 1), the fraction of power transferred inductively through magnetic coupling is:
A
k
B
1 - k
C
1 / k
D
k^2
✓ Correct Answer: B - 1 - k
📖 Step-by-Step Solution & Conceptual Rationale:
In an autotransformer, total power is transferred via two mechanisms: conductive power = k * P_total, and inductive power = (1 - k) * P_total. The copper saving is k * (copper of 2-winding transformer).
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