Electrical calculators

Transformer calculator

Find primary and secondary full-load amps and the turns ratio, size a transformer for a load, or estimate the secondary fault current from %Z.

Calculated with electrician.wiki · https://electrician.wiki/calculators/transformer/

Result

90.21 A primary · 208.2 A secondary

A 75 kVA three-phase transformer from 480 V to 208 V carries 90.21 A on the primary and 208.2 A on the secondary at full load.

Primary full-load current: 90.21 A
Secondary full-load current: 208.2 A
Turns ratio (N1 : N2): 3.997 : 1
Line voltage ratio: 2.308 : 1

Overcurrent protection for the primary and secondary is set by NEC 450.3, and secondary conductor protection by 240.21(C). This calculator gives full-load current only; it does not size the protection.

Show the work

1. Primary full-load current
$I_{1} = \frac{S}{\sqrt{3} \cdot V_{1}} = \frac{75,000 VA}{\sqrt{3} \cdot 480 V} = 90.21 A$
2. Secondary full-load current
$I_{2} = \frac{S}{\sqrt{3} \cdot V_{2}} = \frac{75,000 VA}{\sqrt{3} \cdot 208 V} = 208.2 A$
3. Turns ratio (delta primary, wye secondary)
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{\frac{V_{2}}{\sqrt{3}}} = \frac{480 V}{120.1 V} = 3.997$
The delta primary winding sees 480 V; the wye secondary winding sees 208 V ÷ √3 = 120.1 V. The line-to-line voltage ratio is 2.308 : 1, but a turns-ratio test reads 3.997 : 1.

References: NEC 450.3 (2023): transformer overcurrent protection (not calculated here)

Full-load current from the nameplate kVA at rated voltage; three-phase voltages are line-to-line.
The real primary current is slightly higher, by the transformer’s losses and magnetizing current.

New to this? How a transformer works

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How to use the transformer calculator

This transformer calculator has two modes. Full-load current takes the kVA rating, the number of phases and the primary and secondary voltages, and returns the full-load amps on each side and the turns ratio. Switch on the short-circuit option and enter the nameplate %Z to get the largest current a bolted fault at the secondary terminals can draw. Size for a load takes the load in kVA or in amps at the secondary voltage, adds a growth margin (25 % by default) and picks the next standard rating.

For three-phase units, enter line-to-line voltages: 480 and 208 for a 480 V delta to 208Y/120 V transformer. For a single-phase 120/240 V secondary, enter 240. The winding connection only affects the turns ratio, not the currents.

Transformer full-load current formulas

Full-load current is the kVA rating divided by the voltage, with √3 for three-phase because the rating is shared across three phases. The same formula applies to each side; the current is higher on the lower-voltage side.

Single-phase and three-phase full-load current

I = \frac{S}{V}

I = \frac{S}{\sqrt{3} \cdot V_{LL}}

Turns ratio: single-phase, delta–delta and wye–wye

\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}}

Turns ratio: delta primary, wye secondary

\frac{N_{1}}{N_{2}} = \frac{V_{1}}{\frac{V_{2}}{\sqrt{3}}} = \frac{\sqrt{3} \cdot V_{1}}{V_{2}}

Maximum secondary short-circuit current, infinite source

I_{sc} = \frac{I_{FLA} \cdot 100}{%Z}

Size with growth margin

S_{req} = S_{load} \cdot (1 + \frac{margin}{100})

Worked examples

75 kVA, 480 V delta to 208Y/120 V

A workhorse dry-type transformer in US commercial buildings, feeding a 120/208 V lighting and receptacle panel.

1. Primary full-load current
$I_{1} = \frac{S}{\sqrt{3} \cdot V_{1}} = \frac{75,000 VA}{\sqrt{3} \cdot 480 V} = 90.21 A$
2. Secondary full-load current
$I_{2} = \frac{S}{\sqrt{3} \cdot V_{2}} = \frac{75,000 VA}{\sqrt{3} \cdot 208 V} = 208.2 A$
3. Turns ratio (delta primary, wye secondary)
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{\frac{V_{2}}{\sqrt{3}}} = \frac{480 V}{120.1 V} = 3.997$

A 75 kVA three-phase transformer from 480 V to 208 V carries 90.21 A on the primary and 208.2 A on the secondary at full load.

Fault current at a 150 kVA transformer, %Z 4.5

The same voltages, a larger unit, and the nameplate impedance, to check the interrupting rating needed in the secondary panel.

1. Primary full-load current
$I_{1} = \frac{S}{\sqrt{3} \cdot V_{1}} = \frac{150,000 VA}{\sqrt{3} \cdot 480 V} = 180.4 A$
2. Secondary full-load current
$I_{2} = \frac{S}{\sqrt{3} \cdot V_{2}} = \frac{150,000 VA}{\sqrt{3} \cdot 208 V} = 416.4 A$
3. Turns ratio (delta primary, wye secondary)
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{\frac{V_{2}}{\sqrt{3}}} = \frac{480 V}{120.1 V} = 3.997$
4. Maximum secondary fault current (infinite source)
$I_{sc} = \frac{I_{2} \cdot 100}{%Z} = \frac{416.4 A \cdot 100}{4.5} = 9,252 A$

A 150 kVA three-phase transformer from 480 V to 208 V carries 180.4 A on the primary and 416.4 A on the secondary at full load; with 4.5 % impedance and an infinite source, a bolted fault at the secondary terminals can reach 9.252 kA.

25 kVA single-phase pole transformer, 7,200 V to 120/240 V

A utility transformer serving a few houses from a 12.47Y/7.2 kV distribution line.

1. Primary full-load current
$I_{1} = \frac{S}{V_{1}} = \frac{25,000 VA}{7,200 V} = 3.472 A$
2. Secondary full-load current
$I_{2} = \frac{S}{V_{2}} = \frac{25,000 VA}{240 V} = 104.2 A$
3. Turns ratio
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}} = \frac{7,200 V}{240 V} = 30$

A 25 kVA single-phase transformer from 7,200 V to 240 V carries 3.472 A on the primary and 104.2 A on the secondary at full load.

Sizing for a 150 A load at 208 V

A tenant space whose calculated demand is 150 A per phase at 208 V three-phase, with a 25 % allowance for growth.

1. Load in kVA
$S_{load} = \sqrt{3} \cdot V_{2} \cdot I = \sqrt{3} \cdot 208 V \cdot 150 A = 54.04 kVA$
2. Add the growth margin
$S_{req} = S_{load} \cdot 1.25 = 54.04 kVA \cdot 1.25 = 67.55 kVA$
3. Next standard three-phase size
$67.55 kVA \leq 75 kVA$
4. Secondary full-load current of that size
$I_{2} = \frac{S}{\sqrt{3} \cdot V_{2}} = \frac{75,000 VA}{\sqrt{3} \cdot 208 V} = 208.2 A$

A 150 A load at 208 V (54.04 kVA) plus a 25 % growth margin needs 67.55 kVA: the next standard three-phase size is 75 kVA, with a full-load current of 208.2 A at 208 V.

630 kVA, 11 kV to 400 V (Dyn11), %Z 4

A typical European distribution substation transformer feeding a 230/400 V switchboard.

1. Primary full-load current
$I_{1} = \frac{S}{\sqrt{3} \cdot V_{1}} = \frac{630,000 VA}{\sqrt{3} \cdot 11,000 V} = 33.07 A$
2. Secondary full-load current
$I_{2} = \frac{S}{\sqrt{3} \cdot V_{2}} = \frac{630,000 VA}{\sqrt{3} \cdot 400 V} = 909.3 A$
3. Turns ratio (delta primary, wye secondary)
$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{\frac{V_{2}}{\sqrt{3}}} = \frac{11,000 V}{230.9 V} = 47.63$
4. Maximum secondary fault current (infinite source)
$I_{sc} = \frac{I_{2} \cdot 100}{%Z} = \frac{909.3 A \cdot 100}{4} = 22,733 A$

A 630 kVA three-phase transformer from 11,000 V to 400 V carries 33.07 A on the primary and 909.3 A on the secondary at full load; with 4 % impedance and an infinite source, a bolted fault at the secondary terminals can reach 22.73 kA.

Transformer amperage chart: full-load amps for standard sizes

Every value below is computed by the same engine as the calculator. They are rated currents, not breaker sizes: protection is chosen separately (see below).

Three-phase full-load current in amps, line-to-line voltage.
kVA	208 V	240 V	400 V	480 V	600 V
3	8.327	7.217	4.33	3.608	2.887
6	16.65	14.43	8.66	7.217	5.774
9	24.98	21.65	12.99	10.83	8.66
15	41.64	36.08	21.65	18.04	14.43
30	83.27	72.17	43.3	36.08	28.87
45	124.9	108.3	64.95	54.13	43.3
75	208.2	180.4	108.3	90.21	72.17
112.5	312.3	270.6	162.4	135.3	108.3
150	416.4	360.8	216.5	180.4	144.3
225	624.5	541.3	324.8	270.6	216.5
300	832.7	721.7	433	360.8	288.7
500	1,388	1,203	721.7	601.4	481.1
750	2,082	1,804	1,083	902.1	721.7
1,000	2,776	2,406	1,443	1,203	962.3
1,500	4,164	3,608	2,165	1,804	1,443
2,000	5,551	4,811	2,887	2,406	1,925
2,500	6,939	6,014	3,608	3,007	2,406

Single-phase full-load current in amps.
kVA	120 V	240 V	277 V	480 V
3	25	12.5	10.83	6.25
5	41.67	20.83	18.05	10.42
7.5	62.5	31.25	27.08	15.63
10	83.33	41.67	36.1	20.83
15	125	62.5	54.15	31.25
25	208.3	104.2	90.25	52.08
37.5	312.5	156.3	135.4	78.13
50	416.7	208.3	180.5	104.2
75	625	312.5	270.8	156.3
100	833.3	416.7	361	208.3
167	1,392	695.8	602.9	347.9
250	2,083	1,042	902.5	520.8
333	2,775	1,388	1,202	693.8
500	4,167	2,083	1,805	1,042

Sizing a transformer for a load

Start from the maximum demand, not the connected load: the result of a load calculation, or a measured peak. Add a margin for growth, then round up to a standard rating. With the default 25 %, a 50 kVA load needs 62.5 kVA and lands on a 75 kVA unit. If the load is known in amps, the calculator converts it first: 150 A at 208 V three-phase is 54.04 kVA.

Single-phase loads on a three-phase transformer. Each phase can supply only a third of the rating. Size from three times the most heavily loaded phase, not from the total.
Motors. Large motors started across the line pull several times their running current and dip the voltage for everything else on the transformer. Allow more margin, or a soft start or drive.
Nonlinear loads. Computers, LED lighting and drives produce harmonic currents that heat a standard transformer more than their kVA suggests. K-factor rated units are built for this.
Ambient and altitude. Ratings assume a normal ambient and altitude. A hot electrical room or a site at high altitude needs derating; the manufacturer’s data gives the factors.
Other markets. The standard sizes here are common North American ratings. IEC markets use other steps, such as 250, 400, 630 and 1,000 kVA; choose the next one up from your supplier’s range.

Short-circuit current from %Z

The nameplate impedance, %Z, is the percentage of rated voltage that pushes full-load current through the transformer with its secondary shorted. If 5 % of the voltage drives full-load current, full voltage drives 100 ÷ 5 = 20 times as much. For the 75 kVA example, 208.2 A × 20 = 4.164 kA. This assumes an infinite source with no impedance of its own, so it is the most the transformer can deliver, not what you will actually see.

Real fault current is lower once the utility source and the cable to the fault are included: the available fault current calculator (NEC) and the prospective fault current calculator (IEC) take those into account. Two adjustments push the other way. Nameplate %Z has a manufacturing tolerance, so worst-case studies often use 90 % of it. Running motors also feed a fault for the first few cycles, commonly estimated at about four times their full-load current. Equipment must be rated for the fault current where it is installed (NEC 110.9 and 110.10), and service equipment outside dwellings is marked with the available fault current (110.24).

Turns ratio, voltage ratio and taps

Why the ratio of turns sets the ratio of voltages, and where the losses and the %Z come from, is explained in what a transformer is and how it works.

For single-phase, delta–delta and wye–wye units, the turns ratio equals the voltage ratio. A delta–wye unit is different: the 480 V delta winding is matched to a 120.1 V wye winding (208 V ÷ √3), so the turns ratio is 3.997 : 1 while the line voltages are in the ratio 2.308 : 1. A turns-ratio test set reads the winding ratio, so expect 3.997, not 2.308.

Primary taps, typically in 2.5 % steps above and below nominal, adjust the ratio to suit the actual supply. If a 480 V primary measures 504 V, the matching +5 % tap brings the secondary back to nominal; left on the 480 V tap, a 208 V secondary would read about 218.4 V.

Protection and conductors are separate steps

Full-load current is the starting point for protection, not the answer. In the US, the overcurrent protection for a transformer’s primary and secondary is governed by NEC 450.3, and the secondary conductors by 240.21(C). This calculator does not size either. Take the currents to the wire size calculator or the IEC cable size calculator, and check long secondary runs with the voltage drop calculator. For kW, kVA and power factor on the load side, use the three-phase power calculator.

Method and verification

Full-load current is kVA × 1,000 ÷ V for single-phase and kVA × 1,000 ÷ (√3 × V) for three-phase, with line-to-line volts. The turns ratio divides the primary winding voltage by the secondary winding voltage, where a delta winding sees the line-to-line voltage and a wye winding the line-to-neutral voltage. Fault current is the secondary full-load current × 100 ÷ %Z. Sizing multiplies the load by (1 + margin ÷ 100) and takes the smallest standard rating at or above the result. Values are computed in double precision and displayed to four significant figures.

The tests check ten full-load currents worked by hand, among them 75 kVA at 480 V (90.21 A) and at 208 V (208.2 A), and the single-phase 75 kVA unit at 240 V (312.5 A). They also check turns ratios for all four winding connections, fault currents from %Z, that every standard size maps to itself while a size just above it moves up one rating, that kVA to amps and back is exact, and that zero, negative and out-of-range inputs are refused.

Assumes nameplate kVA at rated voltage and, for fault current, an infinite source. Updated 2026-10-14. Found a problem? Report it and we log every fix publicly.

Questions people ask

How do you calculate transformer full-load amps?

Divide the rating in VA by the voltage, and by √3 as well for three-phase. A 75 kVA three-phase transformer at 480 V carries 75,000 ÷ (1.732 × 480) = 90.21 A; at 208 V, 208.2 A.

How many amps is a 75 kVA transformer?

Three-phase: 90.21 A at 480 V and 208.2 A at 208 V. Single-phase: 312.5 A at 240 V. The full-load amps tables above list every standard size.

What size transformer do I need for my load?

Take the maximum demand in kVA, add a growth margin (25 % is common) and choose the next standard size up. A 50 kVA three-phase load becomes 62.5 kVA, so a 75 kVA transformer.

What does %Z on a transformer nameplate mean?

It is the percentage of rated voltage that drives full-load current through the windings with the secondary shorted. It limits fault current: a 5 % unit can deliver at most 20 times its full-load current into a bolted fault, and a lower %Z means more.

How do I calculate the fault current at a transformer secondary?

Multiply the secondary full-load current by 100 and divide by %Z. For 75 kVA at 208 V and 5 %, that is 4.164 kA. This assumes an infinite source; source and cable impedance make the real value lower.

Why doesn’t the turns ratio of my delta–wye transformer match 480 ÷ 208?

Because the windings see different voltages from the lines. The delta primary winding sees 480 V and the wye secondary winding sees 208 V ÷ √3 = 120.1 V, so the turns ratio is 3.997 : 1, while the line voltage ratio is 2.308 : 1.