Transformer phasor diagram, regulation and efficiency

Why it matters

A supply transformer must hold its output voltage within a few percent as load changes, and it runs 24 hours a day, so its losses cost money for its whole life. Regulation tells you how much the voltage sags (or rises) under load; efficiency, maximum efficiency and all-day efficiency tell you how to choose and load a transformer economically.

Key ideas

No-load phasor diagram. Take the mutual flux Φ as reference. The induced EMFs E1 and E2 lag Φ by 90°. The applied voltage V1 ≈ −E1, so V1 leads Φ by 90°. The no-load current I0 lags V1 by an angle φ0 close to 90°: its magnetising component Iμ is in phase with Φ (lags V1 by 90°) and its small core-loss component Iw is in phase with V1. No-load power factor cos φ0 is typically 0.1–0.3.

On-load phasor diagram (referred to the secondary). Start from the terminal voltage V2 and the load current I2 at angle φ2.

• Lagging load: I2 lags V2. Add the drops I2·R02 (parallel to I2) and I2·X02 (90° ahead of I2) to V2 to get the no-load voltage E2 ≈ V20. E2 is larger than V2, so the voltage falls on load.
• Unity power factor: the resistive drop adds directly; the reactive drop is in quadrature, so the fall is small.
• Leading load: the I·X drop points backwards, and V2 can be larger than E2 — negative regulation (voltage rises on load). On the primary side, the load component I2' = I2/a flows in addition to I0 so that the core flux stays almost constant (ampere-turn balance).

Voltage regulation. The change in secondary terminal voltage from no load to full load at constant primary voltage, as a fraction of rated voltage. Indian textbooks and GATE normally divide by the full-load (rated) value V2; some books use V20. Always state which. Using the approximate equivalent circuit, the drop is approximately I·R·cos φ + I·X·sin φ (+ for lagging, − for leading), so in per unit ε ≈ εr·cos φ ± εx·sin φ.

• Regulation is maximum when the load angle equals the impedance angle: cos φ = R/Z lagging, and then ε = Z in per unit.
• Regulation is zero when tan φ = εr/εx on a leading load, i.e. cos φ = X/Z leading.

Losses and efficiency.

• Core (iron) loss Pi: hysteresis + eddy current; depends on voltage and frequency, practically constant from no load to full load. Measured by the open-circuit test.
• Copper loss Pc: I²R in the windings; varies as the square of load. At fraction x of full load, Pc = x²·Pcu,FL. Measured by the short-circuit test.
• Efficiency η = output/(output + losses). Because losses are small, it is far more accurate to compute η from losses than from measuring input and output separately.
• Maximum efficiency occurs where variable loss equals constant loss, x²·Pcu,FL = Pi, i.e. at x = √(Pi/Pcu,FL) of full load. It is independent of power factor; for a given load the efficiency is highest at unity power factor.
• Transformers are rated in kVA because copper loss depends on current and core loss on voltage — neither on the load power factor.
• All-day (energy) efficiency = energy output over 24 h / energy input over 24 h. Distribution transformers are lightly loaded much of the day but always energised, so they are designed with low core loss and reach maximum efficiency at around half load.

Formulas

• %R = I·R02 / V2 × 100 ; %X = I·X02 / V2 × 100 — per-unit (percent) resistance and reactance drops at rated current; same value from either side.
• ε ≈ %R·cos φ ± %X·sin φ — percent regulation (+ lagging, − leading); approximate, good for small drops.
• ε = (V20 − V2) / V2 × 100 — definition, referred to full-load (rated) voltage V2.
• ε_max = %Z = √(%R² + %X²) at cos φ = R/Z lagging.
• η = x·S·cos φ / (x·S·cos φ + Pi + x²·Pcu,FL) — efficiency at fraction x of rated S (VA) and load power factor cos φ; losses in W.
• x_ηmax = √(Pi / Pcu,FL) — load fraction at maximum efficiency; kVA at ηmax = S·x.
• η_all-day = Σ(output kWh) / Σ(output kWh + 24·Pi + Σ copper-loss kWh).

Worked examples

Example 1 (standard): regulation at lagging and leading power factor. Given: 10 kVA, 2000/200 V transformer; equivalent resistance and reactance referred to HV: R01 = 6 Ω, X01 = 12 Ω.

1. Full-load HV current: I1 = 10000/2000 = 5 A.
2. %R = I1·R01/V1 × 100 = 5 × 6/2000 × 100 = 1.5 %; %X = 5 × 12/2000 × 100 = 3.0 %.
3. At 0.8 pf lagging (sin φ = 0.6): ε = 1.5 × 0.8 + 3.0 × 0.6 = 3.0 %. (An exact phasor solution gives 3.01 %.)
4. At 0.8 pf leading: ε = 1.2 − 1.8 = −0.6 % — the voltage rises on load. (Exact: −0.55 %.)
5. Maximum regulation: %Z = √(1.5² + 3²) = 3.35 %, at cos φ = R/Z = 6/13.42 = 0.447 lagging. Zero regulation at cos φ = X/Z = 0.894 leading.

Example 2 (GATE level): efficiency and maximum efficiency. Given: the same 10 kVA transformer; core loss Pi = 100 W; full-load copper loss Pcu,FL = 150 W.

1. Full load, 0.8 pf: output = 10 × 0.8 = 8 kW; losses = 100 + 150 = 250 W. η = 8000/(8000 + 250) = 96.97 %.
2. Load at maximum efficiency: x = √(100/150) = 0.8165, i.e. 8.16 kVA.
3. Maximum efficiency at unity pf: copper loss = Pi = 100 W, total loss 200 W; η = 8165/(8165 + 200) = 97.61 %.
4. Half load, unity pf: copper loss = 0.5² × 150 = 37.5 W; η = 5000/(5000 + 100 + 37.5) = 97.32 %.

Common mistakes

• Dividing by the wrong voltage in regulation; state whether the base is full-load or no-load voltage (240 → 230 V gives 4.35 % on 230 V but 4.17 % on 240 V).
• Using + sin φ for a leading load; the reactive term changes sign.
• Scaling copper loss linearly with load instead of with x².
• Applying the 0.8 pf to kVA in the denominator but not the numerator, or treating core loss as load-dependent.
• Believing maximum efficiency depends on power factor — the load fraction does not; only the value of ηmax does.
• Drawing I0 in phase with V1 on no load; it lags by nearly 90°.

For GATE IN

Typical questions: regulation from %R, %X and power factor; the power factor for zero or maximum regulation; efficiency at a given load and pf; the load at which maximum efficiency occurs, given core and copper losses; and all-day efficiency from a 24-hour load cycle. Practise working in per unit; it removes most side-referral errors.

Quick check

1. Core loss 400 W, full-load copper loss 900 W. At what fraction of full load is efficiency maximum?
2. Can voltage regulation be negative? When?
3. Why are transformers rated in kVA?
4. %R = 2 %, %X = 4 %. Regulation at unity power factor?

Answers: 1. √(400/900) = 2/3 of full load. 2. Yes, on a sufficiently leading load. 3. Both losses depend on current and voltage, not on load power factor. 4. About 2 %.