Electrical Machines

Kapp Diagram Calculator

Compute secondary voltage, voltage drop, and regulation for single-phase transformers under inductive or capacitive loads — using the infinite bus model and short-circuit test data. Includes live phasor diagram.

Voltage Regulation Phasor Diagram Short-Circuit Test Infinite Bus Power Factor
Results Real-time
Secondary Voltage U2
—
V
Voltage Drop |ΔV|
—
V
Regulation e
—
%
Req
—
O
Xeq
—
O
Load angle f
—
°

Scales the ΔV triangle independently — useful when Vcc is small.

Kapp Diagram

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About This Simulator

This simulator models the voltage regulation of a single-phase transformer using the infinite bus assumption: the primary voltage U1 is held constant at the grid voltage regardless of load, which is the standard approach for distribution transformers connected to a stiff network.

The equivalent circuit is derived from the short-circuit test, which yields the total series impedance Zeq referred to one side. From the short-circuit voltage Vcc and angle fcc, the impedance is split into:

Zeq = Vcc / In  ·  Req = Zeq · cos fcc  ·  Xeq = Zeq · sin fcc

The secondary voltage is computed as the complex difference U2 = U1 - Zeq · I, where the load current phasor I is constructed from its magnitude and power factor angle. Voltage regulation is defined as: e = (|U1| - |U2|) / |U2| × 100 %. A positive e indicates a voltage drop (inductive loads); a negative e indicates a voltage rise (capacitive loads).

How to Use

  1. Enter the grid voltage U1. This is the RMS voltage applied to the primary winding, typically 220 V or 380 V for low-voltage systems.
  2. Set nominal and load currents. In is the rated current from the nameplate; I is the actual operating current. Set both equal for full-load regulation.
  3. Configure the load. Enter cos f and select whether the load is inductive (lagging) or capacitive (leading).
  4. Enter the short-circuit test data. Vcc can be entered as a percentage of U1 (most common in datasheets) or as an absolute voltage. fcc is the impedance angle from the test: cos fcc = Pcc / (Vcc · In).
  5. Read the results. The phasor diagram updates in real time. U1 is the fixed horizontal reference; U2 is the secondary voltage; ?V (dashed) connects their tips; I is the load current (shown at an independent scale).

Understanding the Results

Secondary Voltage U2

The RMS magnitude of the secondary terminal voltage under the specified load. For an ideal transformer U2 = U1; real transformers show a drop under inductive loads or a slight rise under capacitive loads.

Voltage Drop |ΔV|

The magnitude of the complex drop across Zeq: |ΔV| = |Zeq · I|. This is not simply |U1| - |U2|, because the phasors are not collinear — the diagram makes the vector triangle clear. For inductive loads the drop has a significant quadrature component that increases regulation.

Voltage Regulation e

Regulation measures how much U2 changes from no-load to full-load, relative to U2: e = (|U1| - |U2|) / |U2| × 100 %. Distribution transformers are designed for e = 5 %. A negative value means capacitive compensation is actually boosting the secondary voltage above U1.

Equivalent Impedance Req and Xeq

The total series resistance and leakage reactance referred to one winding, extracted from the short-circuit test. Xeq dominates in most power transformers (fcc ≈ 70–85°), which is why inductive loads produce significantly more regulation than resistive ones at the same current.

Phasor Diagram

U1 (primary, blue) is fixed as the horizontal reference. U2 (green) is the secondary voltage. ΔV (amber) is the drop across Zeq, drawn from the tip of U2 to close the triangle at the tip of U1. The current phasor I (grey, dashed) is shown at an independent scale. Angles are geometrically correct but the phasor magnitudes are scaled for visibility.

Frequently Asked Questions

Why does the secondary voltage rise under capacitive load?
With a leading (capacitive) load, the current phasor is ahead of the voltage. The reactive component of the voltage drop across Xeq partially opposes the resistive drop and can produce a net secondary voltage greater than U1. This is the same effect as the Ferranti rise on lightly loaded long transmission lines. Capacitor banks are sometimes deliberately used to exploit this and boost voltage at the end of a distribution feeder.
What is fcc and how is it measured?
fcc is the phase angle between Vcc and In during the short-circuit test. It is derived from the active power consumed: cos fcc = Pcc / (Vcc · In). For most power transformers fcc is between 70° and 85°, indicating a highly inductive Zeq dominated by leakage flux rather than winding resistance.
Why is Vcc usually expressed as a percentage?
Expressing Vcc as a fraction of U1 makes the parameter independent of the voltage ratio and allows direct comparison across different voltage levels. A 4 % Vcc means the impedance drops 4 % of U1 when rated current flows. IEC 60076 and IEEE C57 both standardize on this per-unit convention; nameplate values are almost always given this way.
What is the difference between voltage drop and voltage regulation?
Voltage drop |ΔV| = |Zeq · I| is the magnitude of the phasor across the series impedance. Voltage regulation e compares the primary and secondary magnitudes: e = (|U1| - |U2|) / |U2| × 100 %. Because U1, U2, and ΔV form a triangle (not a straight line), |ΔV| ≠ |U1| - |U2|. The phasor diagram makes this geometry explicit.
What are the limitations of this simulator?
This model uses the simplified series equivalent circuit and assumes: (1) the core magnetizing branch is negligible — no-load losses and magnetizing current are ignored; (2) single-phase operation; (3) the load is linear and balanced; (4) skin effect and frequency dependence of Req are not modeled; (5) the turns ratio is referred to one winding (1:1 per-unit). For full analysis including core losses, three-phase systems, harmonic loads, or transient behavior, a multi-winding finite-element or per-unit model is required.
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