Single-phase transformer: construction and equivalent circuit

Why it matters

The transformer is the most common machine in any plant: it steps voltage up for transmission, down for distribution, isolates control circuits and feeds every instrument power supply. Its equivalent circuit is the model you use to predict voltage drop, losses and efficiency, and the same circuit (with a rotating secondary) becomes the induction-motor model later.

Key ideas

Construction

• Two windings (primary connected to the supply, secondary to the load) on a common laminated core of cold-rolled grain-oriented silicon steel, 0.25–0.5 mm laminations insulated from each other. Silicon raises resistivity and lowers hysteresis loss; laminations cut eddy-current paths.
• Core type: windings surround the limbs of the core; the flux has one path; easier to insulate and repair; used for high-voltage units.
• Shell type: the core surrounds the windings (windings on the central limb, flux divides into two outer limbs); better mechanical support and lower leakage; used for low-voltage, high-current units.
• Windings are concentric (LV next to the core, HV outside) or sandwiched, to reduce leakage flux and insulation needs. Large units use oil for cooling and insulation.

Principle. An alternating voltage V1 drives a small magnetising current that sets up an alternating mutual flux φ = Φm·sin ωt in the core. This flux links both windings and induces EMFs e = −N·dφ/dt in each (Faraday). The rms value is E = 4.44·f·N·Φm, so E1/E2 = N1/N2. A transformer cannot work on DC: with no changing flux there is no back EMF, the current is limited only by the small winding resistance, and the winding overheats.

Ideal transformer. No winding resistance, no leakage flux, infinite core permeability and no core loss. Then V1/V2 = N1/N2 = a, I1/I2 = N2/N1 = 1/a, and V1·I1 = V2·I2 (volt-amperes are conserved). Impedance seen from the primary: Z' = a²·Z_load.

Practical transformer — the four imperfections and how each appears in the equivalent circuit

1. Winding resistances R1, R2 — series elements; they cause copper loss I²R.
2. Leakage flux that links only one winding — series leakage reactances X1, X2.
3. Core loss (hysteresis + eddy current) — a shunt resistance Rc (or R0) drawing the active component Iw of the no-load current.
4. Finite permeability — a shunt magnetising reactance Xm (or X0) drawing the magnetising component Iμ. The no-load current I0 = Iw + Iμ (phasor sum, Iμ lagging V1 by 90°) is only about 2–5 % of rated current, and its power factor is low (0.1–0.3).

Referring quantities. To draw one circuit without an ideal transformer, move secondary quantities to the primary: V2' = a·V2, I2' = I2/a, R2' = a²·R2, X2' = a²·X2. Power and losses are unchanged by referring. Referring to the secondary divides impedances by a².

Approximate equivalent circuit. Because I0 is small, the shunt branch is moved to the supply terminals. Then the series parts combine: R01 = R1 + a²·R2 and X01 = X1 + a²·X2 (referred to primary), or R02 = R2 + R1/a² and X02 = X2 + X1/a² (referred to secondary). This is the model used for regulation and efficiency in the next topic and whose parameters come from the OC and SC tests.

Formulas

• E = 4.44·f·N·Φm — rms induced EMF (V); f in Hz, N turns, Φm peak flux (Wb) = Bm·A_net. Sinusoidal flux only.
• a = N1/N2 = E1/E2 ≈ V1/V2 = I2/I1 — turns ratio (ideal or approximately at load).
• R2' = a²·R2 ; X2' = a²·X2 ; Z_L' = a²·Z_L — secondary quantities referred to the primary (Ω).
• R01 = R1 + a²·R2 ; X01 = X1 + a²·X2 ; Z01 = √(R01² + X01²) — equivalent series impedance referred to primary (Ω).
• R02 = R01/a² ; X02 = X01/a² — the same referred to secondary (Ω).
• I0 = √(Iw² + Iμ²) ; Iw = I0·cos φ0 ; Iμ = I0·sin φ0 — no-load current components (A).
• Rc = V1/Iw ; Xm = V1/Iμ — shunt branch (Ω).
• I_rated = S / V — rated current (A) for rating S (VA) on the side of voltage V.

Worked examples

Example 1 (standard): turns and currents. Given: 25 kVA, 2300/230 V, 50 Hz single-phase transformer; net core area 0.0225 m²; maximum flux density not to exceed 1.2 T.

1. Maximum flux allowed: Φm = Bm·A = 1.2 × 0.0225 = 0.027 Wb.
2. Secondary turns: N2 = E2/(4.44·f·Φm) = 230/(4.44 × 50 × 0.027) = 38.4. Round up to 40 so that Bm stays below the limit.
3. Primary turns: N1 = a·N2 = 10 × 40 = 400 turns; N2 = 40 turns.
4. Actual flux: Φm = 230/(4.44 × 50 × 40) = 0.0259 Wb, so Bm = 0.0259/0.0225 = 1.15 T.
5. Rated currents: I1 = 25000/2300 = 10.9 A; I2 = 25000/230 = 108.7 A.

Example 2 (GATE level): referring the series impedance. Given: 10 kVA, 2000/200 V transformer; R1 = 3 Ω, X1 = 6 Ω, R2 = 0.03 Ω, X2 = 0.06 Ω. Find the equivalent impedance referred to each side and the full-load copper loss.

1. a = 2000/200 = 10, a² = 100.
2. R01 = R1 + a²·R2 = 3 + 100 × 0.03 = 6 Ω; X01 = X1 + a²·X2 = 6 + 100 × 0.06 = 12 Ω; Z01 = √(6² + 12²) = 13.4 Ω.
3. Referred to LV: R02 = 6/100 = 0.06 Ω, X02 = 12/100 = 0.12 Ω.
4. Full-load HV current I1 = 10000/2000 = 5 A. Copper loss = I1²·R01 = 25 × 6 = 150 W. Check from the LV side: I2 = 50 A, 50² × 0.06 = 150 W — the same, as it must be.
5. Full-load impedance drop on the HV side = 5 × 13.4 = 67 V, i.e. 3.35 % of 2000 V.

Common mistakes

• Writing I1/I2 = N1/N2. Currents are in the inverse ratio of turns: the high-voltage side carries the smaller current.
• Reading "turns ratio 1:5" as step-down. N1:N2 = 1:5 is a step-up transformer; with I1 = 5 A, I2 = 1 A.
• Referring impedance with a instead of a² (impedance scales with the square of the turns ratio).
• Adding R1 and R2 directly without referring R2 to the same side.
• Using peak flux density with rms voltage inconsistently; 4.44 already contains the √2 and 2π factors.
• Rounding turns down, which pushes flux density above its design limit.
• Thinking the no-load current is zero; it is small but supplies core loss and magnetisation.

For GATE IN

Expect one-step questions on the EMF equation, turns, currents and referred impedances, and conceptual MCQs on core versus shell type, the meaning of each equivalent-circuit element and why a transformer cannot run on DC. Practise referring an impedance through the transformer (impedance matching, a²·Z_L), which also appears in instrumentation coupling problems.

Quick check

1. A 4:1 transformer has 240 V on the primary. What is the secondary voltage?
2. A 10 Ω load on the secondary of a 5:1 step-down transformer looks like what impedance from the primary?
3. Which equivalent-circuit element represents core loss?
4. Which construction has the core surrounding the windings?

Answers: 1. 60 V. 2. 5² × 10 = 250 Ω. 3. The shunt resistance Rc (R0). 4. Shell type.