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Transformers (Electrical Engineering) Solved Questions & Notes (2026) - Apex Rankers

Engineering & Technology > Electrical Engineering > Transformers

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Transformers

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Q. 1 Electrical Engineering
Difficulty: Easy (1 Mark)
An ideal transformer does not change:
A
Voltage and Current
B
Frequency and Power
✓ Correct
C
Impedance and Voltage
D
Current and Flux
💡 Step-by-Step Explanation & Concept 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).
Q. 2 Electrical Engineering
Difficulty: Easy (1 Mark)
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
✓ Correct
D
Inversely proportional to the transformation ratio
💡 Step-by-Step Explanation & Concept 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.
Q. 3 Electrical Engineering
Difficulty: Easy (1 Mark)
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)
✓ Correct
C
Leakage reactance of windings
D
Voltage regulation at full load
💡 Step-by-Step Explanation & Concept 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.
Q. 4 Electrical Engineering
Difficulty: Easy (1 Mark)
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)
✓ Correct
C
Dielectric strength of insulation oil
D
Magnetizing current
💡 Step-by-Step Explanation & Concept 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).
Q. 5 Electrical Engineering
Difficulty: Medium (1 Mark)
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
✓ Correct
C
P_cu_fl / P_i = 0.8
D
P_cu_fl / P_i = 0.64
💡 Step-by-Step Explanation & Concept 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.
Q. 6 Electrical Engineering
Difficulty: Medium (1 Mark)
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
✓ Correct
D
Zero power factor lagging
💡 Step-by-Step Explanation & Concept 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).
Q. 7 Electrical Engineering
Difficulty: Medium (1 Mark)
The maximum voltage regulation in a transformer occurs at a:
A
Lagging power factor where tan(phi) = Xeq / Req
✓ Correct
B
Leading power factor where tan(phi) = Xeq / Req
C
Unity power factor
D
Zero power factor leading
💡 Step-by-Step Explanation & Concept 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).
Q. 8 Electrical Engineering
Difficulty: Easy (1 Mark)
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
✓ Correct
C
To keep winding temperatures above ambient at all times
D
To increase leakage reactance
💡 Step-by-Step Explanation & Concept 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.
Q. 9 Electrical Engineering
Difficulty: Easy (1 Mark)
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
✓ Correct
C
KVA output to KVA input
D
Maximum efficiency multiplied by 24
💡 Step-by-Step Explanation & Concept 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.
Q. 10 Electrical Engineering
Difficulty: Medium (1 Mark)
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
✓ Correct
C
1 / k
D
k^2
💡 Step-by-Step Explanation & Concept 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).
Q. 11 Electrical Engineering
Difficulty: Medium (1 Mark)
What is the primary operational hazard of leaving the secondary winding of an energized Current Transformer (CT) open-circuited?
A
The primary winding will immediately burn out
B
The secondary EMF rises to a dangerously high voltage due to core saturation, posing fatal electric shock and insulation breakdown risks
✓ Correct
C
The CT core demagnetizes permanently
D
The primary line current drops to zero
💡 Step-by-Step Explanation & Concept Rationale
In a CT, secondary ampere-turns oppose primary ampere-turns (N2*I2 opposes N1*I1). If the secondary is opened, secondary MMF becomes zero, and the entire large line current acts as magnetizing current. This drives the core deep into saturation, inducing lethal peak voltages (thousands of volts) across the secondary terminals.
Q. 12 Electrical Engineering
Difficulty: Easy (1 Mark)
The Buchholz relay is a protective device used in oil-immersed transformers to detect:
A
External short circuits on overhead transmission lines
B
Incipient internal faults (such as insulation breakdown, inter-turn faults, and core heating) producing gas accumulation
✓ Correct
C
Overvoltage caused by lightning strikes
D
Excessive load power factor
💡 Step-by-Step Explanation & Concept Rationale
A Buchholz relay is installed in the pipe connecting the main transformer tank to the conservator tank. Gas bubbles produced by slow internal arcing or overheating rise and trigger an alarm. Severe faults produce a violent oil surge that trips the circuit breaker.
Q. 13 Electrical Engineering
Difficulty: Easy (1 Mark)
The silica gel breather attached to a transformer conservator tank serves to:
A
Filter solid dust particles from the transformer oil
B
Absorb atmospheric moisture from the air entering the conservator during thermal breathing
✓ Correct
C
Cool down the hot circulating oil
D
Neutralize acid formation in the oil
💡 Step-by-Step Explanation & Concept Rationale
As transformer temperature rises and falls, oil expands and contracts, causing air to breathe in and out. The silica gel crystals absorb moisture from incoming air to maintain the high dielectric breakdown strength of the insulating oil.
Q. 14 Electrical Engineering
Difficulty: Easy (1 Mark)
Fresh, dry silica gel in a transformer breather has which color, and what color does it turn when saturated with moisture?
A
Pink when dry, Deep Blue when wet
B
Deep Blue when dry, Pink (or whitish) when saturated with moisture
✓ Correct
C
Yellow when dry, Green when wet
D
White when dry, Black when wet
💡 Step-by-Step Explanation & Concept Rationale
Dry cobalt chloride-impregnated silica gel is deep blue. When it absorbs moisture, it chemically transitions to hydrated cobalt chloride, turning light pink, indicating it needs reactivation by heating.
Q. 15 Electrical Engineering
Difficulty: Medium (1 Mark)
Which three-phase transformer connection is capable of converting a 3-phase supply into a balanced 2-phase supply for electric arc furnaces?
A
Open-Delta (V-V) Connection
B
Scott-T (or Teaser) Connection
✓ Correct
C
Star-Zigzag Connection
D
Delta-Star Connection
💡 Step-by-Step Explanation & Concept Rationale
The Scott connection utilizes two single-phase transformers: a main transformer with center-tapped primary (50% tap) and a teaser transformer tapped at 86.6% (sqrt(3)/2) of primary turns, effectively converting 3-phase to balanced 2-phase or vice versa.
Q. 16 Electrical Engineering
Difficulty: Medium (1 Mark)
In an Open-Delta (V-V) transformer bank where one transformer of a 3-transformer Delta-Delta bank is removed, the remaining two transformers can deliver what percentage of the original bank's rated kVA capacity?
A
66.7%
B
57.7% (1 / sqrt(3))
✓ Correct
C
50.0%
D
86.6%
💡 Step-by-Step Explanation & Concept Rationale
The V-V bank capacity is sqrt(3) * V * I = sqrt(3) * S_single. Since the original 3-transformer delta bank capacity was 3 * S_single, the ratio is (sqrt(3) * S_single) / (3 * S_single) = 1 / sqrt(3) ≈ 57.7% (operating at 86.6% of the two remaining transformers' combined rating).
Q. 17 Electrical Engineering
Difficulty: Hard (1 Mark)
The tertiary winding in a large Star-Star (Y-Y) power transformer is connected in Delta primarily to:
A
Provide a path for circulating triplen harmonic (3rd, 9th, etc.) currents, stabilizing the neutral and ensuring sinusoidal phase voltages
✓ Correct
B
Increase primary leakage reactance
C
Eliminate secondary line resistance
D
Reduce the total iron weight of the core
💡 Step-by-Step Explanation & Concept Rationale
In ungrounded Y-Y transformers, 3rd harmonic currents cannot flow, causing severe distortion in phase voltages and neutral oscillation. A closed Delta tertiary winding provides a circulating path for 3rd harmonic magnetizing currents, eliminating distortion.
Q. 18 Electrical Engineering
Difficulty: Medium (1 Mark)
For two 3-phase transformers to operate successfully in parallel, which condition is ABSOLUTELY ESSENTIAL and non-negotiable?
A
Identical kVA ratings
B
Identical phase sequence, zero relative phase displacement (same vector group), and identical voltage ratios
✓ Correct
C
Identical magnetizing reactance
D
Equal core cross-sectional areas
💡 Step-by-Step Explanation & Concept Rationale
If phase sequence is incorrect or vector groups have phase displacement (e.g., Yd1 vs Yd11), a massive phase difference occurs across the secondary terminals, creating dead short-circuit circulating currents that will immediately destroy the transformers.
Q. 19 Electrical Engineering
Difficulty: Medium (1 Mark)
When two transformers of different kVA ratings but equal per-unit leakage impedances operate in parallel, how do they share a common load?
A
Equally (50% each)
B
In direct proportion to their individual kVA ratings
✓ Correct
C
Inversely proportional to their kVA ratings
D
The smaller transformer carries the entire load
💡 Step-by-Step Explanation & Concept Rationale
When per-unit impedances based on their own ratings are equal (Z_pu1 = Z_pu2), the ohmic impedances are inversely proportional to their kVA ratings. Consequently, load current divides such that each transformer carries load strictly proportional to its kVA capacity without overloading.
Q. 20 Electrical Engineering
Difficulty: Hard (1 Mark)
Transformer magnetizing inrush current occurs upon energizing because:
A
The load demands instantaneous reactive power
B
Core flux attempts to double (flux doubling effect) when energized at voltage zero-crossing, driving the core deep into magnetic saturation
✓ Correct
C
The oil temperature is cold
D
Secondary winding is open-circuited
💡 Step-by-Step Explanation & Concept Rationale
Since flux is the integral of voltage, switching on at a voltage zero-crossing requires flux to rise from 0 to 2*phi_m (flux doubling). In saturation, core permeability drops to near air permeability, resulting in inrush currents up to 6-10 times rated full-load current (predominantly 2nd harmonic).
Q. 21 Electrical Engineering
Difficulty: Medium (1 Mark)
Harmonic analysis shows that transformer magnetizing inrush current is predominantly rich in which harmonic?
A
3rd harmonic
B
2nd harmonic
✓ Correct
C
5th harmonic
D
7th harmonic
💡 Step-by-Step Explanation & Concept Rationale
Transformer inrush current has a strong unidirectional DC offset and asymmetric wave shape rich in the 2nd harmonic. Modern differential protective relays use 2nd Harmonic Restraint to distinguish inrush from internal short-circuit faults.
Q. 22 Electrical Engineering
Difficulty: Easy (1 Mark)
In a core-type transformer, the low-voltage (LV) winding is placed closest to the iron core, with the high-voltage (HV) winding on the outside. Why?
A
To reduce core eddy current losses
B
To minimize the thickness and cost of insulating materials required between winding and grounded core
✓ Correct
C
To improve cooling efficiency of the LV winding
D
To increase mutual inductance
💡 Step-by-Step Explanation & Concept Rationale
Insulating the winding against the grounded core requires dielectric insulation proportional to operating voltage. Placing the LV winding closest to the core minimizes insulation thickness and clearance distance, substantially lowering material costs and physical size.
Q. 23 Electrical Engineering
Difficulty: Medium (1 Mark)
What is the standard test method used to determine the dielectric breakdown strength of transformer insulating oil?
A
Megger insulation resistance test at 500 V
B
Oil Breakdown Voltage (BDV) test across a 2.5 mm spark gap using spherical or mushroom electrodes
✓ Correct
C
Kelvin double bridge measurement
D
Open circuit test
💡 Step-by-Step Explanation & Concept Rationale
Under IEC 60156 standards, insulating oil BDV is tested across a calibrated 2.5 mm electrode gap. Good transformer oil must withstand at least 30 kV (and above 50 kV for high-voltage power transformers) without dielectric flashover.
Q. 24 Electrical Engineering
Difficulty: Hard (1 Mark)
Sumpner's Test (Back-to-Back Test) on two identical transformers is conducted to determine:
A
Only the transformation ratio
B
Full-load temperature rise and total losses under full-load conditions without drawing large net power from the supply
✓ Correct
C
Mechanical strength of transformer tank
D
Insulation resistance of bushings
💡 Step-by-Step Explanation & Concept Rationale
Sumpner's test connects two identical transformers with primaries in parallel and secondaries in series opposition. It allows operating both at rated flux and rated full-load current simultaneously, consuming only the internal losses (core + copper), enabling precise heat run testing.
Q. 25 Electrical Engineering
Difficulty: Hard (1 Mark)
A 50 Hz transformer is operated on a 60 Hz supply of the same rated voltage. What will be the effect on its core losses?
A
Hysteresis loss decreases, while eddy current loss remains essentially constant
✓ Correct
B
Both hysteresis and eddy current losses increase
C
Both hysteresis and eddy current losses decrease to zero
D
Total core loss increases by 20%
💡 Step-by-Step Explanation & Concept Rationale
With constant voltage V, Bm is proportional to V/f = 1/f. Hysteresis loss P_h ∝ f * Bm^1.6 ∝ f * (1/f)^1.6 = f^-0.6 (decreases). Eddy current loss P_e ∝ f^2 * Bm^2 ∝ f^2 * (1/f)^2 = constant. Thus, total core loss decreases.
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