Parallel Transformers Calculator

About this calculator

This calculator models two single-phase transformers operating in parallel — a common arrangement when one transformer can no longer supply a growing load, or when service continuity matters (if one unit fails, the other keeps supplying part of the load). From each transformer's nameplate and short-circuit test data it computes how the load is shared, the circulating current between the units, the maximum power the group can deliver without overloading either machine, and whether the parallel connection is acceptable at all.

---

How to use it

Enter each transformer's rated power (S_N), its high- and low-voltage ratings, and its short-circuit test data: U_cc in volts, referred to the high-voltage winding, and P_cc in watts. Then set the group's total load with the slider, the power factor (inductive) and the working voltage U₂. The catalog data is converted to percentage impedance (u_cc%, u_r%, u_x%) automatically, and that conversion is shown in its own table so you can compare it with the u_r% and u_x% given in your own problem. To see a circulating current appear, change the low-voltage rating of one transformer by a few volts.

---

Understanding the results

The two status banners check the conditions that decide whether the parallel connection makes sense: the transformation ratios must match within 0.5%, and the circulating current must stay below 10% of the smaller transformer's rated current.

The loading bars show each winding's apparent power (U₂ × I) as a percentage of its own rated power S_N. Each bar is scaled independently, since a small transformer can reach 100% long before a large one does. A bar turns red at 100%, and it turns yellow between 90% and 100% only when the two loadings differ by more than 5 percentage points; two well-matched units can safely run close to their rating together. The load-share chart shows what fraction of the delivered load each transformer supplies. The per-transformer table gives each winding's current and phase angle and, when a circulating current is present, an extra row with the part of each current that actually reaches the load.

The group results give the circulating current I_C (magnitude and angle), the maximum group power without overloading either machine, and which transformer is the limiting one (or a tie). In the phasor diagram, I₁ and I₂ are drawn against the reference voltage U₂, and I_C is the dashed segment joining the tips of I₁ and I₂: the difference the two currents would still have with the load removed.

---

How the load is shared

From catalog data to percentage impedance

u_cc% = U_cc / U_N,HV × 100 u_r% = P_cc / (10 · S_N[kVA]) u_x% = √(u_cc%² − u_r%²) φ_cc = arctan(u_x% / u_r%)

Equal transformation ratios

With the same ratio, the currents divide inversely to the equivalent impedances, and the loading d = S / S_N of each unit is inversely proportional to its u_cc%:

I₁ / I₂ = Z₂ / Z₁ d₁ · u_cc1 = d₂ · u_cc2

The machine with the lowest u_cc% carries the most load and reaches full load first. When rated powers and u_cc% both differ, it is better for the smaller transformer to be the one with the higher u_cc%, otherwise the smaller unit is the one that limits the group. The result does not depend on the size of the load, only on the impedances.

Different transformation ratios: circulating current

With k = U_HV / U_LV for each unit and Z the equivalent impedance referred to the secondary, the currents are:

I₁ = [ k₂·Z₂·I + (k₂ − k₁)·U₂ ] / ( k₁·Z₁ + k₂·Z₂ ) I₂ = [ k₁·Z₁·I − (k₂ − k₁)·U₂ ] / ( k₁·Z₁ + k₂·Z₂ ) I_C = (k₂ − k₁)·U₂ / ( k₁·Z₁ + k₂·Z₂ )

I_C is the value the currents take with no load (I = 0). It adds to one winding and subtracts from the other, and it disappears when k₁ = k₂. Because the impedances are small, even a slight ratio mismatch produces a large circulating current.

Maximum group power

With equal ratios and in-phase currents, the unit with the lowest u_cc% runs at d = 1 and the other at d = u_cc,limiting / u_cc,other:

S_max = S_N,limiting + S_N,other · (u_cc,limiting / u_cc,other)

The two currents are not exactly in phase when u_r%/u_x% differ, so the calculator solves the full phasor equations. Its value is therefore slightly lower than this shortcut.

---

Conditions for connecting transformers in parallel

Condition	Type	If it is not met
Same frequency	Mandatory	With primaries on lines of different frequency the waveforms are no longer sinusoidal.
Same transformation ratio (within 0.5%)	Mandatory	A circulating current flows between the secondaries even at no load.
Same polarity	Mandatory	A voltmeter across the open connection reads about twice the rated voltage instead of zero; closing it would short-circuit the units.
Same connection group (three-phase)	Mandatory	The phase displacement between secondaries differs and large short-circuit currents circulate.
Equal ucc%	Recommended	The load is not shared in proportion to the ratings, and the unit with the lowest ucc% reaches 100% first.
Equal ur% and ux%	Recommended	The two currents are out of phase, so the group can deliver less than the sum of the individual ratings.

A small ratio mismatch is tolerated in practice: a difference below 0.5%, or a circulating current not exceeding 10% of the rated current of either unit, is generally acceptable.

---

Worked example

---

Frequently asked questions