What is Topic 19 in GATE?

Topic 19 is a topic in the Subject Specific syllabus of GATE. It carries a weight of 3% in the exam. Our notes cover the key concepts, formulas, and problem-solving approaches you need to master this topic.

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Electrical Machines — Transformers

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

GATE Weightage: ~5–8 marks/year (Electrical branch); equivalent circuit and efficiency/voltage regulation are most frequently tested.

Ideal Transformer:

V₁/V₂ = N₁/N₂ = a (turns ratio)
I₁/I₂ = N₂/N₁ = 1/a (current ratio, neglecting magnetizing current)
V₁I₁ = V₂I₂ (power conservation)

Equivalent Circuit (referred to primary):

V₁ → [R₁] → [jX₁] → (+) → [R_c || jX_m] → [R₂'/jX₂'] → [a²R_L] → (−)
                             |_____|         |_______|
                           magnetizing       referred
                             branch           secondary

Open-Circuit Test (OCC): Measures core losses → R_c, X_m (LV side shorted, HV open) Short-Circuit Test (SCT): Measures copper losses → R_eq, X_eq (LV side shorted, rated current flows)

Voltage Regulation (VR): VR = (E₂ – V₂)/V₂ × 100% (at full load, lagging PF)

Efficiency: η = (Output power) / (Output + losses) = P_out / (P_out + P_core + P_cu)

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Construction and Operating Principle

A transformer transfers electrical energy between circuits at different voltage levels using electromagnetic induction.

Core Type vs Shell Type

Feature	Core Type	Shell Type
Construction	Windings surround core limbs	Core surrounds windings
Flux path	Each limb has its own winding	Central limb with windings
Mechanical strength	Lower	Higher
Cooling	Easier	More difficult
Use	High voltage, high capacity	Low voltage, distribution

Ideal Transformer Equations

EMF equation: E = 4.44 f N Φ_m = 4.44 f N B_m A_c

Where:

f = frequency (Hz)
N = number of turns
Φ_m = maximum flux (Wb) = B_m · A_c
B_m = maximum flux density (T)
A_c = core cross-sectional area (m²)

Voltage ratio: V₁/V₂ = N₁/N₂ = a (a > 1 for step-down, a < 1 for step-up)

Current ratio: I₁/I₂ = N₂/N₁ = 1/a

Power: S₁ = V₁I₁ = V₂I₂ = S (apparent power conserved in ideal case)

Equivalent Circuit — Complete Model

The approximate equivalent circuit referred to primary (all secondary quantities divided by a²):

V₁ → R₁ + jX₁ → → (+) → R_c || jX_m → R₂' + jX₂' → V₂' → (−)

Parameters:

R₁, X₁ = primary winding resistance and leakage reactance
R₂’, X₂’ = secondary resistance/reactance referred to primary
R_c = core loss resistance (represents hysteresis + eddy current losses)
X_m = magnetizing reactance (represents magnetizing current)

Key insight: The magnetizing branch draws a small current I_m even at no-load (mostly reactive). The core loss component I_c is in phase with V₁.

Open-Circuit Test (OCT)

Procedure: Rated voltage applied to primary, secondary open-circuited. Measure I_oc, P_oc, V_oc.

From OCT:

Core loss P_core = P_oc (at rated voltage)
I_c = P_oc / V₁ (core loss component)
I_m = √(I_oc² – I_c²) (magnetizing component)
R_c = V₁ / I_c
X_m = V₁ / I_m

Typical values: I_oc = 5–10% of rated current; P_oc = small (few hundred watts for small transformers)

GATE Tip: OCT is performed at rated voltage to measure core losses. The LV side is typically shorted in this test (not the HV side).

Short-Circuit Test (SCT)

Procedure: Low voltage applied to primary such that rated current flows, secondary short-circuited.

From SCT:

Copper loss at rated current P_cu(rated) = P_sc
Total series impedance Z_eq = V_sc / I_rated
R_eq = P_sc / I_rated²
X_eq = √(Z_eq² – R_eq²)

Important: P_sc represents full-load copper loss at rated current. Actual copper loss at any load = P_sc × (I/I_rated)²

Voltage Regulation

Definition: Change in secondary voltage from no-load to full-load at a given power factor, expressed as percentage of full-load rated voltage.

VR = (E₂ – V₂)/V₂ × 100%

Where E₂ = induced EMF at no-load (same as rated secondary voltage), V₂ = actual terminal voltage at load.

Regulation at Different Power Factors

For a lagging load (inductive):

Lagging PF: VR > 0 (voltage drops, E₂ > V₂) — most common case
Unity PF: Small positive VR
Leading PF: VR can be negative (voltage rises due to capacitive effect)

Formula: VR ≈ (I_a R_eq cosφ ± I_r X_eq sinφ) / V₂ × 100%

The ± depends on load power factor:

Lagging: + I_r X_eq sinφ (adds to regulation)
Leading: – I_r X_eq sinφ (can subtract)

Efficiency

η = Output Power / Input Power = P_out / (P_out + P_core + P_cu)

At any load level k (fraction of full load):

Core loss P_core = constant (independent of load)
Copper loss P_cu(k) = k² × P_cu(rated)

Maximum efficiency occurs when: P_core = P_cu → k² P_cu(rated) = P_core

Solving for k: k_max = √(P_core / P_cu(rated))

Design principle: Transformers are designed so that maximum efficiency occurs near 75–80% of full load, since they rarely operate at full load continuously.

Autotransformers

An autotransformer has a single winding common to both primary and secondary circuits (partially shared).

Advantages:

Lower cost (less copper, smaller core)
Higher efficiency (lower losses)
Better voltage regulation

Disadvantages:

No isolation between primary and secondary (safety hazard)
Higher fault currents due to low impedance

Voltage ratio: Same as two-winding: V₁/V₂ = N₁/N₂ = a

Capacity: For same VA rating, autotransformer is smaller but the advantage decreases as voltage ratio approaches 1.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Three-Phase Transformer Connections

Common Configurations

Connection	Primary	Secondary	Use case
Star-Star (Y-Y)	Y	Y	Neutral available both sides
Star-Delta (Y-Δ)	Y	Δ	Step-down, distribution
Delta-Star (Δ-Y)	Δ	Y	Step-up at generating stations
Delta-Delta (Δ-Δ)	Δ	Δ	No neutral; heavy industrial

Phase Shift in Star-Delta and Delta-Star

Y-Y: 0° phase shift
Y-Δ: Secondary lags primary by 30° (or +30°, depending on connection)
Δ-Y: Secondary leads primary by 30°

GATE frequently tests the 30° phase shift relationship in star-delta transformations.

Parallel Operation of Transformers

For parallel operation, conditions required:

Same voltage ratio (or as close as possible, to prevent circulating currents)
Same polarity (proper terminal connections)
Same phase sequence (essential — wrong sequence causes short circuit)
Impedance ratios proportional to VA ratings (for load sharing)

When two transformers share load:

S₁/S₂ = (Z₂/Z₁) for proportional loading by impedance
Each transformer carries load proportional to its rating divided by its impedance

Circulating current I_c = (V₁ – V₂) / (Z₁ + Z₂) when voltage ratios differ slightly.

Per-Unit System

Per-unit simplifies three-phase transformer calculations:

Base values: S_base (VA), V_base (LV or HV side), I_base = S_base/(√3 V_base), Z_base = V_base²/S_base

Impedance in pu = Actual Z / Z_base

For three-phase transformers, per-unit impedance is the same on both sides when referred to appropriate base values.

Harmonics in Transformers

Magnetizing Current Harmonics

When operated near saturation, magnetizing current contains:

3rd harmonic (triple-n, absent in Y with neutral point)
5th, 7th harmonics

Harmonic Effects

Third harmonic: Circulates in Δ-connected windings, adds to core losses
Fifth, seventh: Cause additional core and copper losses
K-factor: Rating for harmonic current heating (used for non-linear loads)

Thermal and Loading Considerations

Temperature rise depends on:

Core loss (constant, all day)
Copper loss (varies with load²)

Loading guides:

Overload capability: 150% for short duration (emergency)
Continuous overload reduces life expectancy

Thermal model: Time constant τ (typically 5–10 minutes for oil-filled transformers).

Example Problem — Voltage Regulation

A 100 kVA, 2000/200 V transformer has R_eq = 0.02 pu, X_eq = 0.05 pu. Find voltage regulation at 0.8 PF lagging, full load.

Solution:

Base: S = 100 kVA, V₂ = 200 V
I₂_full = S/(√3 × 200) = 100000/(346.4) = 288.7 A
I₂_pu = 1.0 (full load)
Regulation formula: VR = (I_pu R_eq cosφ + I_pu X_eq sinφ) × 100%
cosφ = 0.8, sinφ = 0.6
VR = (1 × 0.02 × 0.8 + 1 × 0.05 × 0.6) × 100%
VR = (0.016 + 0.030) × 100% = 4.6%

Example Problem — Maximum Efficiency

A 500 kVA transformer has core loss = 2 kW, full-load copper loss = 5 kW. At what load does η_max occur?