Magnetic Circuits and Single-Phase Transformers
Magnetic circuits and reluctance, EMF equation of a transformer, equivalent circuit, open and short-circuit tests, voltage regulation, losses, efficiency and condition for maximum efficiency.
The transformer is the simplest AC machine: two coils linked by a common magnetic core. Its analysis uses magnetic circuits and an equivalent electrical circuit.
Magnetic circuits
- MMF F = N I (ampere-turns) drives flux through reluctance R = l/(μA): Φ = F/R (like Ohm's law).
- An air gap has a far higher reluctance than iron, so most of the MMF is used in the gap.
- Core losses: hysteresis P_h = k_h f B_m^1.6 V (Steinmetz) and eddy current P_e = k_e f² B_m² t². Laminations reduce eddy loss.
EMF equation and ideal transformer
E = 4.44 f N Φ_m (RMS)
Ideal: V₁/V₂ = N₁/N₂ = a and I₁/I₂ = 1/a. An impedance Z₂ on the secondary appears as a² Z₂ on the primary.
Equivalent circuit and tests
- Series: winding resistances and leakage reactances (R_eq, X_eq referred to one side).
- Shunt: core-loss resistance R_c and magnetising reactance X_m.
- Open-circuit test (on the LV side, rated voltage): gives the core (iron) loss and R_c, X_m.
- Short-circuit test (on the HV side, rated current, reduced voltage): gives the full-load copper loss and R_eq, X_eq, Z_eq.
Voltage regulation
Change in secondary voltage from no load to full load, as a fraction of the full-load voltage. Approximately, in per unit:
Regulation ≈ ε_r cos φ ± ε_x sin φ (+ for lagging, − for leading pf)
It can be zero or negative for a leading pf. Maximum regulation occurs when the load angle equals the impedance angle.
Efficiency
η = output/(output + core loss + copper loss).
- Core loss is constant; copper loss varies as the square of the load.
- Maximum efficiency when copper loss = core loss, at a load fraction x = √(P_i/P_cu,fl).
- All-day efficiency = energy output/energy input over 24 hours; distribution transformers are designed for maximum efficiency at light load.
Worked example
A 20 kVA transformer has core loss 200 W and full-load copper loss 400 W, at pf 0.8.
- Full-load efficiency = 16,000/(16,000 + 600) = 96.39%.
- Maximum efficiency at x = √(200/400) = 0.707 of full load, with η = 11,314/(11,314 + 400) = 96.59%.
Practice
A transformer has 2% resistance and 4% reactance (per unit). What is its approximate regulation at 0.8 pf lagging, in percent?
Show answer and explanation
Answer: 4Reg ≈ 2 × 0.8 + 4 × 0.6 = 1.6 + 2.4 = 4%.
For the worked example, what is the full-load efficiency in percent? (Give 2 decimal places.)
Show answer and explanation
Answer: 96.39η = 16,000/16,600 = 96.39%.
A 4 Ω load on the 200 V side of a 2000/200 V transformer appears as what resistance (Ω) on the 2000 V side?
Show answer and explanation
Answer: 400a = 10, so the referred resistance is a² × 4 = 400 Ω.
The open-circuit test on a transformer mainly measures:
Show answer and explanation
Answer: AWith no load current, copper loss is negligible, so the wattmeter reads the core loss.