Power Transformer Basics — What Every New Engineer Needs to Know

You just joined the electrical engineering team. Someone hands you a transformer nameplate photo and asks, "Can we connect this to the new 33 kV feeder?" This article is for that moment. No AI-transformer-models-in-NLP digressions. No PhD-level math. Just the stuff that keeps a substation from catching fire.

What a Transformer Does (In One Sentence)

A transformer changes voltage and current levels in an AC circuit while (ideally) passing power through unchanged. It does this without moving parts — purely through magnetic coupling between two coils wrapped around a shared iron core.

That's it. Everything else — impedance matching, isolation, phase shifting, tap-changing — is an elaboration on that one idea.

Faraday's Law: The Only Equation That Matters

When Michael Faraday discovered electromagnetic induction in 1831, he gave us the equation that makes every transformer possible:

V = 4.44 × f × N × Φ_max

Where:

• V = induced voltage (RMS) in a coil
• f = supply frequency (Hz)
• N = number of turns in the coil
• Φ_max = peak magnetic flux in the core (Weber)
• 4.44 = a constant derived from the sine wave form factor (√2 × π)

This single equation explains:

• Why transformers are frequency-sensitive. Run a 50 Hz transformer at 60 Hz, and you get 20% more voltage per turn. If your supply voltage stays the same, the flux drops 17% and the core operates well below saturation — inefficient but safe. Run that same transformer at 50 Hz on a 60 Hz supply without adjusting voltage, and you oversaturate the core by 20%, drawing massive magnetizing current.
• Why high-voltage windings have more turns. To produce 33 kV with the same core flux that produces 11 kV, you need 3× the turns.
• Why transformer weight scales roughly with kVA. More power means more flux, which means more iron and more copper.

Turns Ratio and Voltage Ratio

For an ideal transformer (no losses):

V_p / V_s = N_p / N_s = I_s / I_p

Primary voltage / secondary voltage = primary turns / secondary turns = secondary current / primary current.

A real transformer deviates slightly from the ideal ratio. The turns ratio is adjusted during design to compensate for voltage drop in the winding impedance at rated load. This is why open-circuit (no-load) secondary voltage is typically 2-5% higher than the nameplate rated voltage for a distribution transformer. The manufacturer referred to this as the "turns ratio compensation."

The Equivalent Circuit: What Actually Happens Inside

A real transformer is not a perfect magnetic coupling. The standard equivalent circuit models it as:

• An ideal transformer (perfect turns ratio, no losses)
• In parallel with the primary:
• R_c (core loss resistance): represents hysteresis and eddy current losses in the iron core — always present when the transformer is energized, even at zero load
• X_m (magnetizing reactance): represents the reactive power needed to magnetize the core. This is what draws the inrush current at energization
• In series with both windings (referred to one side):
• R_eq (equivalent resistance): represents the I²R losses in both windings — hence "copper loss" or "load loss"
• X_eq (equivalent leakage reactance): represents the magnetic flux that leaks out of the core path and doesn't couple both windings. This is what limits short-circuit current and determines voltage regulation

Why X_eq matters in the real world: A transformer with Z = 6% (typical for a 10 MVA unit) will pass approximately 16.7× rated current during a bolted secondary fault. The X/R ratio of that impedance determines the asymmetrical peak — an X/R of 10 means the first peak of fault current could reach 2.55× the symmetrical RMS value. Your circuit breaker and your busbar bracing must survive that.

Losses: Where Your Money Goes

Iron Loss (No-Load Loss, Core Loss) — P₀

Present as soon as the transformer is energized. Two components:

• Hysteresis loss (P_h): Energy consumed in continuously reversing the magnetic domains in the core steel. Proportional to frequency and B_max^1.6 (approximately). This is why grain-oriented silicon steel is used — it has sharply defined magnetic domains that flip with minimal energy.
• Eddy current loss (P_e): Circulating currents induced in the core steel itself by the changing magnetic flux. Proportional to frequency², B_max², and lamination thickness². This is why the core is laminated — 0.23 mm to 0.30 mm thick sheets, each insulated from the next by a thin oxide or varnish coating.

Typical iron loss for a modern 1000 kVA oil-immersed transformer (EU Tier 2): 1100 W. That's 9,636 kWh per year, at the meter, 24 hours a day, whether you draw a single amp or not.

Copper Loss (Load Loss, Winding Loss) — P_k

• I²R loss (DC loss): The straightforward resistive heating in the copper winding. Proportional to current² and conductor DC resistance.
• Eddy current loss in windings: The same principle as core eddy currents, but in the copper conductors. The leakage flux penetrating the conductor cross-section induces circulating currents. This is why large transformer windings use continuously transposed conductor (CTC) — multiple parallel strands that are individually insulated and transposed so that each strand occupies every radial position and carries equal current.

Typical load loss for a modern 1000 kVA oil-immersed transformer (EU Tier 2): 10,500 W at rated load. Unlike iron loss, this varies with the square of your load — at 50% load, it drops to 25% of the rated value.

Per-Unit System: The Engineer's Calculator

Why does everyone use per-unit (p.u.) for transformer calculations? Because it eliminates the voltage transformation step:

Z_pu (transformer) = Z_actual (Ω) / Z_base (Ω)

Where Z_base = V_rated² / S_rated.

A transformer with 6% impedance means Z = 0.06 p.u. Whether you're on the 33 kV or 11 kV side doesn't matter — 0.06 p.u. is 0.06 p.u. everywhere. The short-circuit current is simply 1 / 0.06 = 16.67 p.u. Multiply by your base current and you have the answer in amps.

How to Read a Transformer Nameplate

Here's a real-world nameplate, annotated:

RATED POWER:         2000 kVA          ← Maximum continuous apparent power
PRIMARY VOLTAGE:     11,000 V          ← Nominal HV voltage
SECONDARY VOLTAGE:   433 V             ← Nominal LV voltage (line-to-line)
FREQUENCY:           50 Hz             ← Designed for this frequency only
PHASES:              3                 ← Three-phase
VECTOR GROUP:        Dyn11             ← HV Delta, LV Star with neutral, LV lags HV by 30°
COOLING:             ONAN              ← Oil Natural, Air Natural
TEMPERATURE RISE:    65 K oil / 65 K winding   ← Per IEC 60076-2
INSULATION CLASS:    Class A (oil)     ← Mineral oil, 105°C hot-spot limit
IMPEDANCE:           6.0% @ 75°C       ← Measured at rated tap, reference temperature
NO-LOAD LOSS:        2300 W            ← Iron loss, always present
LOAD LOSS:           16,500 W @ 75°C   ← Copper loss at rated current
TOTAL MASS:          5200 kg           ← Including oil
OIL MASS:            1100 kg           ← Mineral oil weight
BIL:                 LI 75 kV / AC 28 kV (HV)   ← IEC insulation level
                     LI — / AC 3 kV (LV)
YEAR:                2024
STANDARD:            IEC 60076

The vector group "Dyn11" tells you:

• D = HV connected in Delta (each winding phase sees line-to-line voltage)
• y = LV connected in Star (wye) (each winding phase sees phase-to-neutral voltage)
• n = Neutral terminal brought out on LV side (you can connect a neutral conductor)
• 11 = Phase displacement: LV voltage lags HV voltage by 11 × 30° = 330° (or leads by 30°, depending on your convention)

The impedance of 6.0% @ 75°C means: at rated current (104.9 A on the HV side for 2000 kVA at 11 kV), the voltage drop across the transformer's internal impedance equals 6.0% of rated voltage. This is measured at 75°C because copper resistance increases with temperature.

FAQ

Q: Why does the magnetizing inrush current happen?

A: When you energize a transformer, the residual flux in the core from the last de-energization and the instantaneous voltage at the moment of closing determine the peak flux. If the voltage crosses zero at the closing instant, the flux must reach 2× the normal peak (plus residual) to maintain the voltage-flux relationship. At that flux level, the core saturates deeply, and the magnetizing impedance collapses to near the air-core value, drawing 5-12× rated current for a few cycles. This is normal. The protection relays must be set to distinguish this inrush (rich in 2nd harmonic) from a genuine fault current (fundamental frequency dominant).

Q: What does "impedance voltage" mean in practice?

A: Apply a short circuit to the secondary terminals. Then raise the primary voltage from zero until rated current flows in both windings. The voltage required, expressed as a percentage of rated primary voltage, is the impedance voltage (Uk%). This is the same as the p.u. impedance. A 6% impedance transformer requires 6% of rated voltage to drive rated current into a short circuit. Higher impedance means lower fault current but worse voltage regulation under load.

Q: Can I connect two transformers in parallel?

A: Only if they match on: (1) same primary and secondary rated voltages, (2) same vector group (or compatible — Dyn11 can parallel with Dyn11, but not with YNd11), (3) impedance within ±10% of each other, expressed as percentage of their own rating, and (4) turns ratio within ±0.5% at each tap position. Different impedances cause unequal load sharing — the lower-impedance unit takes more than its share. Different vector groups cause circulating current that heats both units even at no load.

Q: Why do I see both "kVA" and "kW" on some transformer documentation?

A: Transformers are rated in kVA (apparent power = voltage × current), not kW (real power), because the thermal limit is determined by current (which causes I²R heating) and voltage (which determines dielectric stress and core flux). The load's power factor determines how many kW you can actually deliver. A 1000 kVA transformer feeding a load with 0.8 power factor can deliver 800 kW. The same transformer feeding a 1.0 power factor load can deliver 1000 kW. Never exceed the kVA rating — that's the current limit — regardless of what kW number you're drawing.

Q: What determines the physical size of a transformer for a given kVA rating?

A: (1) Voltage class — higher BIL requires larger clearances and more insulation, directly increasing core limb spacing and tank size. (2) Impedance specification — higher impedance (lower fault current) requires more leakage reactance, which means thicker winding cross-sections or larger winding spacing, increasing copper and iron weight. (3) Loss capitalization — a low-loss specification requires more copper cross-section (to reduce I²R) and higher-grade core steel (to reduce iron loss), both of which increase physical size and weight. A 10 MVA, 33/11 kV transformer can vary from 18 to 28 tonnes depending on these three parameters.

Q: Why does transformer humming get louder under load?

A: The primary hum (100 Hz or 120 Hz) comes from magnetostriction — the core steel physically expands and contracts minutely with each half-cycle of magnetic flux. This is present at any load and is inherent to the core material. Under load, an additional component appears: electromagnetic forces between current-carrying conductors in the windings. These forces vary with the square of the current and create vibrations at the same frequency. This is why a heavily loaded transformer is audibly louder than a lightly loaded one. If the noise changes character — becomes irregular, develops harmonics, or suddenly increases — that may indicate winding looseness or core delamination, not normal magnetostriction.

Q: Are there single-phase power transformers? When are they used?

A: Yes. Single-phase transformers are common in rural distribution (pole-mounted, typically 10-100 kVA), railway traction supply, and as single-phase units that form a three-phase bank when three identical units are connected together. A three-phase bank of three single-phase units has the advantage that a single-unit failure requires only one spare, not a complete three-phase replacement. The disadvantage is larger footprint, higher total losses, and more busbar connections.

References

• IEC 60076-1:2011 — Power transformers — Part 1: General
• IEC 60076-8:1997 — Power transformers — Part 8: Application guide
• IEEE Std C57.12.00-2015 — General Requirements for Liquid-Immersed Distribution, Power, and Regulating Transformers
• Fitzgerald, A.E., et al. "Electric Machinery," 7th Edition, McGraw-Hill
• Kulkarni, S.V. and Khaparde, S.A. "Transformer Engineering: Design, Technology, and Diagnostics," 2nd Edition, CRC Press

*Written from the factory floor. The nameplate tells you what the transformer can do. The equivalent circuit tells you why. One day, you'll need both.*