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Home/Electricity & Magnetism/Transformer Equivalent Circuit

Transformer Equivalent Circuit

Turns ratio with magnetizing branch, leakage reactance, copper/core losses, load regulation, and efficiency versus load.

Windings and rating

500
250
230 V
5 kVA

Equivalent circuit

1.6 %
4 %
65 W
2.8 kΩ

Load

75 %
0.85

Measured values

Turns ratio N2/N10.5000
No-load secondary E2115.0V
Loaded secondary U2112.0V
Voltage regulation2.60%
Efficiency η96.58%
Copper + core loss110.0W
No-load current I00.294A
TypeStep-down

The equivalent circuit is referred to the secondary side. Regulation uses the common approximate lagging-power-factor formula; saturation, harmonics, temperature rise, inrush, and detailed short-circuit tests are outside this teaching model.

About this model

A practical transformer is often represented by an ideal turns ratio plus an equivalent circuit: a magnetizing branch for core excitation and iron loss, and a series impedance for winding resistance and leakage reactance. This simulator keeps the turns-ratio intuition while adding no-load current, percent resistance/reactance, voltage regulation at a chosen power factor, copper loss, core loss, and an efficiency-versus-load curve. The equivalent impedance is referred to the secondary side using a rated-kVA base, so percent impedance maps directly to ohms for the selected voltage and rating.

Who it's for: Electrical machines, power systems, AC circuits, transformer testing, and introductory power engineering.

Key terms

  • Transformer equivalent circuit
  • Magnetizing branch
  • Leakage reactance
  • Voltage regulation
  • Copper loss
  • Core loss
  • Efficiency

How it works

Transformer equivalent-circuit simulator with turns ratio, magnetizing branch, leakage reactance, load regulation, copper/core loss, and efficiency versus load.

Key equations

E2/E1 = N2/N1, Zbase = E2²/Srated, Req/Xeq from percent impedance
reg ≈ Ipu(Rpu cosφ + Xpu sinφ), η = Pout/(Pout + I²Req + Pcore)

Frequently asked questions

What does the magnetizing branch represent?
The shunt branch models the current needed to establish core flux and the real power lost to hysteresis and eddy currents. It remains present even with the secondary open.
Why does the secondary voltage drop under load?
Load current flows through winding resistance and leakage reactance. The approximate regulation term Ipu(Rpu cosφ + Xpu sinφ) estimates the resulting voltage drop for a lagging power factor.
Why does the current in the primary coil change when I add a load to the secondary?
The secondary load current reflects back to the primary through the turns ratio. The source supplies output power plus copper and core losses, while the shunt branch supplies magnetizing current.
Can a transformer work with direct current (DC)?
No. A transformer relies on a changing magnetic flux to induce a voltage in the secondary coil, as described by Faraday's law. A constant DC voltage creates a steady magnetic field, not a changing one, so no voltage is induced in the secondary after the initial switch-on transient. This simulator uses AC for this reason.
What does 'ideal' mean in this context, and how do real transformers differ?
'Ideal' means the model assumes perfect efficiency with no energy losses. Real transformers have losses from wire resistance (copper losses), magnetic hysteresis, and eddy currents in the core (iron losses). They also have leakage flux (not all flux links both coils) and require a magnetizing current to establish the core's magnetic field. This simulator ignores these effects to focus on the fundamental voltage, current, and power relationships.
If I step up the voltage, why does the current step down?
This is a direct consequence of energy conservation. For a given power output (P = U₂I₂), if the voltage U₂ is increased, the current I₂ must decrease proportionally to keep the product constant. Since the input power must equal the output power (U₁I₁ = U₂I₂), the primary current I₁ adjusts accordingly. High-voltage transmission lines use this principle to reduce current, minimizing resistive power losses over long distances.