Transformer: EMF Equation, Equivalent Circuit, Phasor Diagram, Per-Unit System, Losses, Efficiency & Regulation (Solved Problems + 15 MCQs)

The transformer is the single most important machine in a power system: electricity generated at 11–20 kV is stepped up to 132, 230 or 400 kV for transmission and then stepped down to 33 kV, 11 kV and finally 400/230 V for your home. That is why transformer questions appear in almost every PGCB, BPDB, BREB/PBS, DPDC, DESCO, NWPGCL, EGCB and BCS (EEE) exam and viva. This guide takes you from the basic idea to the EMF equation, losses, efficiency, regulation and exam-style numericals — with animations and 15 MCQs.

⚡ Key takeaways

• A transformer is a static device that transfers AC power from one circuit to another by mutual induction, changing voltage and current but not frequency.
• EMF equation: E = 4.44 f N Φ_m volts.
• Ratio: V₁/V₂ = N₁/N₂ = I₂/I₁ (ideal transformer).
• Core (iron) loss = hysteresis + eddy current, almost constant. Copper loss = I²R, varies with load².
• Maximum efficiency occurs when copper loss = core loss.
• Rated in kVA, not kW; cannot work on DC.

📑 Contents

1. What is a transformer?
2. Construction
3. Working principle (animated)
4. EMF equation – derivation
5. Turns ratio, step-up and step-down
6. Ideal vs practical transformer
7. Equivalent circuit – derivation
8. Phasor diagrams
9. Losses in a transformer
10. Efficiency and maximum efficiency
11. Voltage regulation
12. OC and SC tests
13. Per-unit system
14. Solved problems
15. One-line exam facts
16. Practice MCQs
17. বাংলায় দ্রুত রিভিশন

1. What Is a Transformer?

A transformer is a static electrical machine that transfers electrical energy from one AC circuit (primary) to another (secondary) through a common magnetic core, by the principle of mutual induction. During the transfer, the voltage and current levels change, but the frequency remains the same and the power remains almost the same (losses are only 1–3%).

• Step-up transformer: V₂ > V₁ (N₂ > N₁) — used at generating stations.
• Step-down transformer: V₂ < V₁ (N₂ < N₁) — used at grid sub-stations and distribution (e.g. 11 kV/0.415 kV).
• Since it has no rotating parts, there is no friction or windage loss, which makes its efficiency very high (95–99.5%).

2. Construction

Main parts

• Core: made of thin laminations (about 0.23–0.35 mm) of cold-rolled grain-oriented (CRGO) silicon steel (≈3% Si), insulated from each other by varnish. Silicon raises resistivity and lowers hysteresis loss; lamination reduces eddy current loss. It provides a low-reluctance path for flux.
• Windings: copper (or aluminium) coils — primary connected to the supply and secondary to the load. The LV winding is placed next to the core and the HV winding outside it, so that less insulation is needed.
• Transformer oil: mineral oil that both insulates and cools.
• Conservator (allows oil expansion), breather with silica gel (absorbs moisture from incoming air), Buchholz relay (gas-actuated protection for internal faults in oil-filled transformers with a conservator), bushings, tap changer (adjusts the turns ratio), radiators and explosion vent / pressure relief device.

Core type vs shell type

Feature	Core type	Shell type
Arrangement	Windings surround the core	Core surrounds the windings
Magnetic circuit	Single magnetic path	Two parallel paths (double magnetic circuit)
Limbs	2 limbs (1-phase)	3 limbs; windings on the central limb
Cooling & repair	Easier — windings exposed	Harder
Leakage flux	More	Less
Typical use	High-voltage power transformers	Low-voltage, high-current transformers

3. Working Principle (Animated)

The transformer works on Faraday’s law of electromagnetic induction:

1. An alternating voltage V₁ is applied to the primary. It drives a small magnetising current that produces an alternating flux Φ in the core.
2. This flux links both windings. Because it keeps changing, it induces an EMF in each turn: e = −N dΦ/dt.
3. The EMF induced in the primary (E₁) almost balances the applied voltage (back EMF). The EMF induced in the secondary (E₂) appears at its terminals and drives current through the load.
4. Since the same flux links every turn, the EMF per turn is the same in both windings, so E₁/E₂ = N₁/N₂.

Animation 1 – Alternating flux (red dots, size ∝ |Φ|) circulates in the core. Φ lags V₁ by 90°, and E₂ lags Φ by 90°.

⚠ Why a transformer does not work on DCWith DC, the flux is constant, so dΦ/dt = 0 and no EMF is induced in the secondary. Worse, without a back EMF the primary current is limited only by the small winding resistance (I = V/R), so the winding overheats and can burn out.

4. EMF Equation of a Transformer – Derivation

Let the flux in the core be sinusoidal: Φ = Φ_m sin ωt, where ω = 2πf.

1. Induced EMF in a winding of N turns: e = −N dΦ/dt = −N ω Φ_m cos ωt = N ω Φ_m sin(ωt − 90°)
2. Maximum value: E_max = N ω Φ_m = 2π f N Φ_m
3. RMS value: E_rms = E_max/√2 = (2π/√2) f N Φ_m

E = 4.44 f N Φ_m = 4.44 f N B_m A_i volts

So E₁ = 4.44 f N₁ Φ_m and E₂ = 4.44 f N₂ Φ_m. Here B_m is the maximum flux density (T) and A_i is the net iron area of the core (m²). The result also shows that the induced EMF lags the flux by 90°.

💡 Exam shortcutEMF per turn = E/N = 4.44 f Φ_m, and it is the same for both windings. If a question gives “volts per turn”, multiply by N to get the winding voltage.

5. Turns Ratio, Step-Up and Step-Down

E₁/E₂ = V₁/V₂ = N₁/N₂ = I₂/I₁ = a

Many Indian/Bangladeshi textbooks define the transformation ratio K = N₂/N₁ = V₂/V₁. K > 1 → step-up; K < 1 → step-down. Because input VA ≈ output VA (V₁I₁ = V₂I₂), the side with higher voltage carries lower current — that is why the HV winding uses thinner wire with more turns.

Impedance transfer: an impedance Z₂ on the secondary appears on the primary as Z₂′ = Z₂ (N₁/N₂)² = Z₂/K². This is used for impedance matching (e.g. an audio output transformer matching an amplifier to a loudspeaker).

6. Ideal vs Practical Transformer

Ideal transformer	Practical transformer
Winding resistance = 0	Windings have resistance R₁, R₂ → copper loss
No leakage flux (all flux links both windings)	Some leakage flux → leakage reactance X₁, X₂
Core has infinite permeability, zero core loss	Core needs magnetising current and has hysteresis + eddy losses
Efficiency 100%, regulation 0	Efficiency 95–99.5%, regulation a few percent

At no load, a practical transformer draws a small no-load current I₀ (about 2–5% of rated current) with two components: the magnetising component I_μ (lags V₁ by 90°, produces flux) and the core-loss (working) component I_w (in phase with V₁, supplies iron loss). The no-load power factor is therefore very low (about 0.1–0.2).

7. Equivalent Circuit of a Transformer – Derivation

To calculate currents, losses, efficiency and regulation we replace the real transformer by an electric circuit that behaves the same way at its terminals. The idea is simple: start from an ideal transformer and add one circuit element for each imperfection of a practical one.

Step 1 — Put every imperfection outside an ideal transformer

• Winding resistances R₁, R₂ in series with each winding → copper loss I²R.
• Leakage reactances X₁, X₂ in series → leakage flux that links only its own winding.
• Exciting (shunt) branch R₀ ∥ X₀ across the primary → carries the no-load current I₀. R₀ draws the core-loss component I_w, X₀ draws the magnetising component I_μ:

R₀ = V₁ / I_w     X₀ = V₁ / I_μ

Fig. 7(a) — A practical transformer = an ideal transformer + winding resistances (R₁, R₂) + leakage reactances (X₁, X₂) + exciting branch (R₀ ∥ X₀).

Step 2 — Write the circuit equations

V₁ = E₁ + I₁(R₁ + jX₁)   |   E₂ = V₂ + I₂(R₂ + jX₂) E₂ / E₁ = N₂ / N₁ = K   |   I₁ = I₀ + I₁′,   I₁′ = K I₂

I₁′ is the load component of primary current: it flows only to cancel the mmf of I₂ (N₁I₁′ = N₂I₂).

Step 3 — Refer the secondary to the primary

We want one circuit without the ideal transformer, so every secondary quantity is replaced by an equivalent primary-side value that gives the same power / loss. For the resistance:

I₁′² R₂′ = I₂² R₂  ⇒  R₂′ = R₂ (I₂ / I₁′)² = R₂ / K²

The same reasoning (equal reactive power) gives X₂′, and the voltage scales with the turns ratio:

X₂′ = X₂ / K²    Z_L′ = Z_L / K²    V₂′ = V₂ / K    I₂′ = K I₂

Fig. 7(b) — Exact equivalent circuit referred to the primary (K = N₂/N₁).

Step 4 — Approximate: move the exciting branch to the input

I₀ is only 2–5 % of rated current, so its drop in R₁ and X₁ is negligible. Shifting R₀ ∥ X₀ to the supply terminals puts R₁ in series with R₂′ and X₁ in series with X₂′:

R₀₁ = R₁ + R₂′ = R₁ + R₂ / K²     X₀₁ = X₁ + X₂ / K² Z₀₁ = √(R₀₁² + X₀₁²)

Step 5 — Same circuit referred to the secondary

Multiplying every primary-side impedance by K² refers it to the secondary:

R₀₂ = R₂ + K² R₁    X₀₂ = X₂ + K² X₁    Z₀₂ = K² Z₀₁

Fig. 7(c) — Approximate equivalent circuit: I₀ is only 2–5 % of rated current, so the exciting branch is moved to the input terminals and the series elements combine into R₀₁ and X₀₁.

Link to testsThe open-circuit test measures the shunt branch R₀, X₀; the short-circuit test measures the series branch R₀₁, X₀₁ (or R₀₂, X₀₂) — see section 12.

8. Phasor Diagrams of a Transformer

8.1 On no load

Take the mutual flux Φ as reference. Both induced EMFs lag Φ by 90° (E = −N dΦ/dt). The supply voltage must balance −E₁, so V₁ leads Φ by 90°. The no-load current I₀ lags V₁ by φ₀ (close to 90°):

I_w = I₀ cos φ₀  (in phase with V₁)   |   I_μ = I₀ sin φ₀  (in phase with Φ)

Fig. 8(a) — No-load phasor diagram. I₀ lags V₁ by nearly 90°: its large component I_μ magnetises the core, the small component I_w supplies core loss.

8.2 On load — complete diagram (lagging pf)

The diagram is built straight from the equivalent-circuit equations of section 7. The secondary drops are added to V₂ to obtain E₂; the primary current is the phasor sum of I₀ and the load component I₁′; the primary drops are added to −E₁ to obtain V₁.

Fig. 8(b) — Complete on-load phasor diagram for a lagging (inductive) load.

8.3 Simplified diagram referred to the secondary

Neglecting I₀ and using R₀₂, X₀₂, the diagram reduces to one drop triangle on top of V₂. The resistive drop I₂R₀₂ is parallel to I₂ and the reactive drop I₂X₀₂ leads I₂ by 90°. Projecting the triangle on V₂:

₀V₂ ≈ V₂ + I₂R₀₂ cos φ + I₂X₀₂ sin φ

Fig. 8(c) — Simplified phasor diagram referred to the secondary (exciting current neglected). The horizontal projection of the drop triangle gives the voltage regulation formula.

For a leading load the I₂X₀₂ sin φ term changes sign, so the terminal voltage can rise on load — this is exactly the regulation formula used in section 11.

9. Losses in a Transformer

Loss	Cause	Formula / depends on	How to reduce
Hysteresis loss (core)	Repeated magnetic reversal of the core	Ph = η Bm1.6 f V (Steinmetz)	Silicon steel / CRGO with narrow hysteresis loop
Eddy current loss (core)	Circulating currents induced in the core	Pe = Ke Bm² f² t² V	Thin insulated laminations (t small), high-resistivity steel
Copper loss	I²R heating of windings	Pcu = I₁²R₁ + I₂²R₂ ∝ (load)²	Thicker conductors, good design
Stray loss	Leakage flux inducing currents in tank/clamps	Small	Magnetic shielding
Dielectric loss	Insulation (oil, paper)	Very small; rises as insulation ages	Good-quality insulation

Core (iron) loss = hysteresis + eddy current loss. It depends on voltage and frequency, which are nearly constant, so it is called the constant loss. Copper loss depends on load current, so it is the variable loss. Amorphous-metal cores (see Types of Solids) cut no-load core loss dramatically, and soft magnetic materials are chosen for their narrow hysteresis loop.

10. Efficiency and Condition for Maximum Efficiency

η = Output / (Output + Core loss + Cu loss) = x S cosφ / (x S cosφ + P_i + x² P_cu,FL)

where S = rated kVA, x = fraction of full load, P_i = core loss, P_cu,FL = full-load copper loss. Differentiating η with respect to load current and setting it to zero gives:

η is maximum when Copper loss = Core loss ⇒ x = √(P_i / P_cu,FL)

The animated graph below plots the numerical example from section 14 (100 kVA, P_i = 1 kW, P_cu,FL = 1.5 kW, pf 0.8). Watch the efficiency peak exactly where the copper-loss curve crosses the core-loss line:

Animation 2 – Efficiency vs load. Maximum efficiency occurs at the load where copper loss equals core loss.

All-day (energy) efficiency = output energy in kWh / input energy in kWh over 24 hours. Distribution transformers are energised all day but lightly loaded much of the time, so they are designed with low core loss and reach maximum efficiency at about 50–75% load. Power transformers in generating stations run near full load and are designed for maximum efficiency near full load.

11. Voltage Regulation

Voltage regulation is the change in secondary terminal voltage from no load to full load, expressed as a percentage of the full-load voltage (primary voltage held constant):

% Regulation = (V_2,NL − V_2,FL) / V_2,FL × 100 % R ≈ (I₂R₀₂ cosφ ± I₂X₀₂ sinφ) / V₂ × 100 (+ lagging, − leading)

Lower regulation is better. With a leading power factor the regulation can become zero or even negative (terminal voltage rises on load). Zero regulation occurs when tan φ = R₀₂/X₀₂ (leading pf).

12. Open-Circuit and Short-Circuit Tests

	Open-circuit (no-load) test	Short-circuit test
Connection	Rated voltage on LV side, HV side open	Reduced voltage (5–10% of rated) on HV side, LV side shorted
Wattmeter reads	Core (iron) loss	Full-load copper loss
Gives	R0, X0 (shunt branch)	Req, Xeq, Zeq (series branch)
Why that side?	Rated LV voltage is easy to supply; current is small	Rated HV current is small and easy to measure

13. Per-Unit (pu) System

In power-system work, voltages, currents and impedances are expressed as a fraction of a chosen base value instead of in volts, amperes and ohms:

Per-unit value = Actual value / Base value    (dimensionless; × 100 gives per cent)

Step 1 — Choose two bases, derive the rest

Pick a common apparent-power base S_base (normally the rated kVA) for the whole transformer, and the rated voltage of each side as its V_base. Current and impedance bases follow from them:

I_base = S_base / V_base     Z_base = V_base / I_base = V_base² / S_base

For three-phase: I_base = S_base(3φ) / (√3 V_L,base) and Z_base = V_L,base² / S_base(3φ) — the √3 disappears from pu calculations.

Fig. 13(a) — Choose one S_base for the whole transformer and the rated voltage of each side as V_base; the per-unit impedance then comes out the same on both sides.

Step 2 — Why the per-unit impedance is the same on both sides

Z_pu,1 = Z₀₁ / Z_base1 = Z₀₁ S_base / V₁² Z_pu,2 = Z₀₂ / Z_base2 = (K² Z₀₁) S_base / (K V₁)² = Z₀₁ S_base / V₁² = Z_pu,1 ✔

The K² in the referred impedance is cancelled by the K² in the base impedance, so a transformer has one per-unit impedance and the ideal transformer vanishes from the pu circuit.

Step 3 — Per-unit impedance = short-circuit voltage

In the SC test, rated current flows when V_sc = I_rated Z₀₁. Dividing by rated voltage:

V_sc / V_rated = I_rated Z₀₁ / V_rated = Z₀₁ / Z_base = Z_pu

So a nameplate "impedance 5 %" means 5 % of rated voltage circulates full-load current with the secondary shorted. Likewise R_pu = full-load copper loss in pu, and regulation ≈ R_pu cos φ ± X_pu sin φ.

Step 4 — Changing the base

Z_pu,new = Z_pu,old × (S_base,new / S_base,old) × (V_base,old / V_base,new)²

Example: a 500 kVA transformer with Z = 0.05 pu on its own rating becomes 0.05 × (100 000 / 500) = 10 pu on a 100 MVA system base (same voltage).

Advantages of the pu system	Typical transformer values
• Same pu impedance from HV or LV side — no referring needed • Values fall in a narrow range, so errors are easy to spot • Machines of very different ratings are compared directly • √3 and turns ratios drop out of 3-φ network calculations • Ideal for computer load-flow and fault studies	• Leakage impedance: 0.04–0.10 pu (distribution ≈ 0.04–0.06) • Rpu ≪ Xpu for large units • No-load current: 0.02–0.05 pu • Full-load copper loss ≈ Rpu (e.g. 0.01 pu = 1 %)

14. Solved Problems

Problem 1 — EMF equation: number of turns

QuestionA 50 Hz single-phase transformer has a net core area of 400 cm² and maximum flux density of 1.2 T. Find the number of turns for a 3300 V / 230 V rating.

Given	To find
f = 50 Hz, A = 400 cm², Bm = 1.2 T V₁ = 3300 V, V₂ = 230 V	N₁, N₂

Fig. P1 — Core-type transformer: the same flux Φ_m links both windings, so both see the same EMF per turn.

Step 1 — Convert the core area to m²

A = 400 cm² = 400 × 10⁻⁴ m² = 0.04 m²

Step 2 — Maximum flux

Φ_m = B_m × A = 1.2 × 0.04 = 0.048 Wb

Step 3 — EMF per turn (same for both windings)

E / N = 4.44 f Φ_m = 4.44 × 50 × 0.048 = 10.656 V/turn

Step 4 — Primary turns

N₁ = E₁ / (E/N) = 3300 / 10.656 = 309.7 ≈ 310 turns

Step 5 — Secondary turns

N₂ = E₂ / (E/N) = 230 / 10.656 = 21.6 ≈ 22 turns

Step 6 — Check

Turns are rounded up to whole numbers. With 310 and 22 turns the no-load secondary voltage is 3300 × 22/310 = 234 V — slightly above 230 V, which also allows for the voltage drop on load.

Answer — Problem 1N₁ = 310 turns  |  N₂ = 22 turns  |  EMF per turn = 10.656 V

Problem 2 — Full-load and maximum efficiency

QuestionA 100 kVA transformer has a core loss of 1 kW and a full-load copper loss of 1.5 kW. Find (a) its full-load efficiency at 0.8 pf lagging and (b) the load and value of maximum efficiency at the same pf.

Given	To find
S = 100 kVA, Pi = 1 kW, Pcu,FL = 1.5 kW, pf = 0.8	ηFL; load for ηmax; ηmax

Fig. P2 — Core loss happens in the shunt branch (constant); copper loss happens in the series resistance (varies as load²).

(a) Full-load efficiency

Step 1 — Output power

P_out = S × cos φ = 100 × 0.8 = 80 kW

Step 2 — Total losses at full load

P_loss = P_i + P_cu,FL = 1 + 1.5 = 2.5 kW

Step 3 — Input power

P_in = P_out + P_loss = 80 + 2.5 = 82.5 kW

Step 4 — Efficiency

η = P_out / P_in × 100 = 80 / 82.5 × 100 = 96.97 %

(b) Maximum efficiency

Step 5 — Condition: copper loss = core loss

At a fraction x of full load, copper loss = x² P_cu,FL. Setting it equal to P_i:

x² × 1.5 = 1  ⇒  x = √(1 / 1.5) = 0.8165

Step 6 — Load at maximum efficiency

S_ηmax = x × 100 = 81.65 kVA

Step 7 — Output and losses at that load

P_out = 81.65 × 0.8 = 65.32 kW    P_loss = 1 + (0.8165² × 1.5) = 1 + 1 = 2 kW

Step 8 — Maximum efficiency

η_max = 65.32 / (65.32 + 2) × 100 = 65.32 / 67.32 × 100 = 97.03 %

Answer — Problem 2(a) η_FL = 96.97 %  |  (b) η_max = 97.03 % at 81.65 kVA (81.65 % of full load)

Problem 3 — Current ratio

QuestionAn 11 kV / 415 V distribution transformer supplies a secondary current of 139 A. Find the primary current (ideal transformer).

Given	To find
V₁ = 11 000 V, V₂ = 415 V, I₂ = 139 A	I₁

Fig. P3 — Ampere-turns balance: the HV side carries a much smaller current than the LV side.

Step 1 — Power balance of an ideal transformer

V₁ I₁ = V₂ I₂  ⇒  I₁ / I₂ = V₂ / V₁ = K

Step 2 — Transformation ratio

K = V₂ / V₁ = 415 / 11 000 = 0.03773

Step 3 — Primary current

I₁ = K × I₂ = 0.03773 × 139 = 5.24 A

Step 4 — Check with volt-amperes

V₂ I₂ = 415 × 139 = 57.7 kVA    V₁ I₁ = 11 000 × 5.24 = 57.7 kVA ✔

139 A is in fact the full-load line current of a 100 kVA three-phase 11 kV/415 V unit (100 000 / (√3 × 415) = 139 A); the ratio of line currents is the same, so I₁ ≈ 5.25 A in that case too.

Answer — Problem 3I₁ ≈ 5.24 A (the HV side carries 1/26.5 of the LV current)

Problem 4 — Per-unit impedance

QuestionA 50 kVA, 2400/240 V single-phase transformer has an equivalent impedance referred to the HV side of Z₀₁ = 1.152 + j4.608 Ω. Find (a) the base values on both sides, (b) the per-unit impedance seen from each side, (c) the HV voltage needed in the short-circuit test, (d) the full-load copper loss and (e) the approximate regulation at 0.8 pf lagging.

Given	To find
S = 50 kVA, V₁ = 2400 V, V₂ = 240 V Z₀₁ = 1.152 + j4.608 Ω	Bases, Zpu (both sides), Vsc, Pcu,FL, regulation

Fig. P4 — (a) Ohmic circuit referred to the HV side; (b) the identical circuit in per-unit.

Step 1 — Base values

S_base = 50 kVA    V_base1 = 2400 V    V_base2 = 240 V I_base1 = 50 000 / 2400 = 20.83 A    I_base2 = 50 000 / 240 = 208.3 A Z_base1 = 2400² / 50 000 = 115.2 Ω    Z_base2 = 240² / 50 000 = 1.152 Ω

Step 2 — Per-unit impedance from the HV side

Z_pu = (1.152 + j4.608) / 115.2 = 0.01 + j0.04 pu

Step 3 — Refer to LV side and repeat

K = 240 / 2400 = 0.1  ⇒  Z₀₂ = K² Z₀₁ = 0.01 × (1.152 + j4.608) = 0.01152 + j0.04608 Ω Z_pu = (0.01152 + j0.04608) / 1.152 = 0.01 + j0.04 pu  — identical ✔

Step 4 — Short-circuit test voltage

|Z_pu| = √(0.01² + 0.04²) = 0.04123 pu (4.12 %) V_sc = 0.04123 × 2400 = 98.95 V  (applied on HV side to circulate 20.83 A)

Step 5 — Full-load copper loss

P_cu,FL = R_pu × S_base = 0.01 × 50 = 0.5 kW    (check: 20.83² × 1.152 = 500 W ✔)

Step 6 — Approximate regulation at 0.8 pf lagging

Reg ≈ R_pu cos φ + X_pu sin φ = 0.01 × 0.8 + 0.04 × 0.6 = 0.032 pu = 3.2 %

Answer — Problem 4Z_pu = 0.01 + j0.04 pu from both sides  |  V_sc = 98.95 V  |  P_cu,FL = 500 W  |  Regulation ≈ 3.2 %

Problem 5 — Equivalent circuit and regulation

QuestionA 10 kVA, 2000/200 V single-phase transformer has R₁ = 5.5 Ω, X₁ = 12 Ω, R₂ = 0.045 Ω and X₂ = 0.1 Ω. Find the equivalent resistance and reactance referred to each side, and the full-load regulation and secondary terminal voltage at 0.8 pf lagging.

Given	To find
S = 10 kVA, V₁ = 2000 V, V₂ = 200 V R₁ = 5.5 Ω, X₁ = 12 Ω, R₂ = 0.045 Ω, X₂ = 0.1 Ω, pf 0.8 lag	R₀₁, X₀₁, R₀₂, X₀₂, regulation, V₂ on full load

Fig. P5 — Approximate equivalent circuit referred to the secondary (exciting branch neglected for regulation).

Step 1 — Transformation ratio

K = V₂ / V₁ = 200 / 2000 = 0.1  ⇒  K² = 0.01

Step 2 — Referred to the primary

R₀₁ = R₁ + R₂ / K² = 5.5 + 0.045 / 0.01 = 5.5 + 4.5 = 10 Ω X₀₁ = X₁ + X₂ / K² = 12 + 0.1 / 0.01 = 12 + 10 = 22 Ω

Step 3 — Referred to the secondary

R₀₂ = R₂ + K² R₁ = 0.045 + 0.01 × 5.5 = 0.1 Ω    (= K² R₀₁ ✔) X₀₂ = X₂ + K² X₁ = 0.1 + 0.01 × 12 = 0.22 Ω    (= K² X₀₁ ✔)

Step 4 — Full-load secondary current

I₂ = S / V₂ = 10 000 / 200 = 50 A

Step 5 — Voltage drop (from the phasor diagram, Fig. 8(c))

cos φ = 0.8, sin φ = 0.6 Drop ≈ I₂(R₀₂ cos φ + X₀₂ sin φ) = 50 × (0.1 × 0.8 + 0.22 × 0.6) = 50 × 0.212 = 10.6 V

Step 6 — Regulation and terminal voltage

% Regulation ≈ 10.6 / 200 × 100 = 5.3 % V₂ (full load) ≈ 200 − 10.6 = 189.4 V

Exact check: (V₂ + 10.6)² + (I₂X₀₂ cos φ − I₂R₀₂ sin φ)² = 200², with the second term = 50(0.176 − 0.06) = 5.8 V ⇒ V₂ = 189.3 V — the approximation is excellent.

Answer — Problem 5R₀₁ = 10 Ω, X₀₁ = 22 Ω  |  R₀₂ = 0.1 Ω, X₀₂ = 0.22 Ω  |  Regulation ≈ 5.3 %  |  V₂ ≈ 189.4 V

15. One-Line Exam Facts

• Transformer does not change frequency or (ideally) power.
• Rated in kVA because copper loss depends on current and core loss on voltage — neither depends on load power factor.
• Core loss is found by the OC test; copper loss by the SC test.
• Laminations reduce eddy current loss; silicon steel reduces hysteresis loss.
• Hysteresis loss ∝ f; eddy current loss ∝ f².
• Buchholz relay protects oil-immersed transformers against internal (incipient) faults; it is placed in the pipe between tank and conservator.
• Silica gel breather removes moisture from air entering the conservator.
• Conditions for parallel operation: same voltage ratio, same polarity, equal % impedance (and similar X/R ratio); for 3-phase also the same phase sequence and phase displacement (vector group).
• An auto-transformer has a single winding; it saves copper but gives no electrical isolation.
• CT secondary must never be left open-circuited (dangerous high voltage and core saturation).

16. Practice MCQs (click to see answer)

1. A transformer works on the principle of: (a) Self induction (b) Mutual induction (c) Electrostatic induction (d) Lenz’s law only

✔ (b) Mutual induction.

2. The EMF equation of a transformer is: (a) E = 4.44 f N Φm (b) E = 2.22 f N Φm (c) E = 4.44 f Φm / N (d) E = f N Φm

✔ (a) E = 4.44 f N Φm.

3. Which quantity does a transformer NOT change? (a) Voltage (b) Current (c) Frequency (d) Impedance level

✔ (c) Frequency.

4. Transformer core is laminated to reduce: (a) Hysteresis loss (b) Eddy current loss (c) Copper loss (d) Stray loss

✔ (b) Eddy current loss.

5. Maximum efficiency of a transformer occurs when: (a) Copper loss = core loss (b) Copper loss = 2 × core loss (c) Core loss is zero (d) Load is maximum

✔ (a) Copper loss = core loss.

6. The open-circuit test of a transformer measures: (a) Copper loss (b) Core loss (c) Total loss (d) Stray loss only

✔ (b) Core (iron) loss.

7. Transformer rating is expressed in: (a) kW (b) kVAR (c) kVA (d) HP

✔ (c) kVA.

8. If a transformer is connected to a DC supply of rated voltage, it will: (a) Work normally (b) Step up voltage (c) Burn out due to heavy current (d) Give double output

✔ (c) Burn out — no back EMF, current limited only by winding resistance.

9. Eddy current loss is proportional to: (a) f (b) f² (c) √f (d) 1/f

✔ (b) f² (and to t², the lamination thickness squared).

10. Buchholz relay is used for protection of: (a) Alternators (b) Transmission lines (c) Oil-immersed transformers (d) Induction motors

✔ (c) Oil-immersed transformers against internal faults.

11. A transformer has 500 primary and 50 secondary turns. With 230 V primary, the secondary voltage is: (a) 2300 V (b) 23 V (c) 46 V (d) 11.5 V

✔ (b) 23 V (V₂ = 230 × 50/500).

12. The no-load current of a transformer is about: (a) 2–5% of rated current (b) 25% (c) 50% (d) 100%

✔ (a) 2–5% of rated current.

13. The core of a power transformer is commonly made of: (a) Cast iron (b) CRGO silicon steel (c) Copper (d) Aluminium

✔ (b) CRGO silicon steel.

14. Voltage regulation can be negative when the load power factor is: (a) Lagging (b) Unity (c) Leading (d) Zero lagging

✔ (c) Leading.

15. Silica gel in a transformer breather is used to: (a) Cool the oil (b) Absorb moisture (c) Detect faults (d) Increase insulation thickness

✔ (b) Absorb moisture from incoming air.

17. বাংলায় দ্রুত রিভিশন

📝 পরীক্ষার আগে এক নজরে

• ট্রান্সফরমার একটি স্থির যন্ত্র যা পারস্পরিক আবেশ (mutual induction) নীতিতে AC শক্তি এক সার্কিট থেকে অন্য সার্কিটে স্থানান্তর করে; ভোল্টেজ ও কারেন্ট বদলায়, কিন্তু ফ্রিকোয়েন্সি বদলায় না।
• EMF সমীকরণ: E = 4.44 f N Φm; অনুপাত: V₁/V₂ = N₁/N₂ = I₂/I₁।
• কোর লস (হিস্টেরেসিস + এডি কারেন্ট) প্রায় স্থির; কপার লস (I²R) লোডের বর্গের সমানুপাতিক। ল্যামিনেশন এডি কারেন্ট লস কমায়, সিলিকন স্টিল হিস্টেরেসিস লস কমায়।
• সর্বোচ্চ দক্ষতা হয় যখন কপার লস = কোর লস।
• OC টেস্টে কোর লস এবং SC টেস্টে কপার লস পাওয়া যায়।
• রেটিং kVA-তে হয়; DC সাপ্লাইতে ট্রান্সফরমার কাজ করে না, বরং পুড়ে যেতে পারে।
• বুখোলজ রিলে তেলভর্তি ট্রান্সফরমারের অভ্যন্তরীণ ত্রুটি থেকে সুরক্ষা দেয়; সিলিকা জেল আর্দ্রতা শোষণ করে।

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References: B. L. Theraja, A Textbook of Electrical Technology Vol. II; J. B. Gupta, Theory & Performance of Electrical Machines; Chapman, Electric Machinery Fundamentals. Images and animations: original work by Totalecer.