Transformer Calculations Practice

Turns ratios, full-load current on both windings, kVA sizing, and why the same kVA gives wildly different currents on each side.

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What it is, and why electricians need it

A transformer changes the ratio of voltage to current without creating energy, so volt-amperes in equals volt-amperes out apart from small losses. Everything else in this topic follows from that one sentence.

Voltage divides in the same ratio as the turns, and current moves the opposite way. Halve the voltage and you double the available current. This is why the same kVA rating produces very different currents on the two sides: a 75 kVA three-phase transformer draws about 90 amperes at 480 volts on its primary and delivers about 208 amperes at 208 volts on its secondary. Both represent 75 kVA. A useful check on any transformer calculation is that the volt-amperes balance on both sides — if they do not, you have inverted a ratio.

The kVA rather than kW question is worth understanding rather than memorising. What limits a transformer is winding current, because that is what generates heat. Winding current follows apparent power, not real power. So a 100 kVA transformer is fully loaded thermally when delivering 100 kVA, whether that represents 100 kW into a resistive load or 80 kW into a load at 0.8 power factor. Correcting the power factor is how you free up that lost capacity.

Two further points that appear on master-level exams. A transformer with separate windings creates a separately derived system: its secondary has no reference to ground until one is deliberately established with a system bonding jumper and a grounding electrode conductor. And an autotransformer, sharing one winding, does not create a separately derived system and offers no galvanic isolation — which is both why buck-boost transformers are so compact and why they need care.

Formulas and principles

  • Vp ÷ Vs = Np ÷ Ns

    Voltage divides in the same ratio as the turns.

    Rearranged: Vs = Vp × Ns ÷ Np.

  • VA in = VA out

    Ignoring losses. The basis of every transformer current calculation.

    So Ip = Is × Vs ÷ Vp — current moves opposite to voltage.

  • Single phase: I = VA ÷ V

    Full-load current on either winding of a single-phase transformer.

  • Three phase: I = VA ÷ (1.732 × V)

    Full-load current on either winding of a three-phase transformer.

    Omitting 1.732 overstates the current by about 73 percent.

  • kVA = V × I ÷ 1,000, or 1.732 × V × I ÷ 1,000 for three phase

    Sizing a transformer from a known load current.

    Then round up to the next standard size and allow for growth.

  • Three-phase bank: each unit is one third of the total kVA

    A 45 kVA bank built from single-phase units needs three 15 kVA transformers.

Now try it on 14 questions

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Worked examples

Scenario: A 75 kVA three-phase transformer has a 480-volt primary and a 208-volt secondary. Find the full-load current on each winding.

  1. Primary: I = VA ÷ (1.732 × V) = 75,000 ÷ (1.732 × 480) = 75,000 ÷ 831.4.
  2. Primary full-load current = 90.2 amperes.
  3. Secondary: I = 75,000 ÷ (1.732 × 208) = 75,000 ÷ 360.3.
  4. Secondary full-load current = 208.2 amperes.
  5. Check: the current ratio 208.2 ÷ 90.2 = 2.31, which matches the inverse voltage ratio 480 ÷ 208 = 2.31. Consistent.

Answer: 90.2 A primary, 208.2 A secondary

Scenario: A transformer has a 480-volt primary and a 240-volt secondary. The secondary is delivering 100 amperes. What current flows in the primary?

  1. Secondary volt-amperes: 240 × 100 = 24,000 VA.
  2. Primary volt-amperes are the same, 24,000 VA.
  3. Primary current: 24,000 ÷ 480 = 50 amperes.
  4. Note the inverse relationship: double the voltage, half the current.

Answer: 50 amperes

Scenario: A three-phase 208-volt load draws 100 amperes per line. What transformer size is required?

  1. kVA = 1.732 × V × I ÷ 1,000.
  2. kVA = 1.732 × 208 × 100 ÷ 1,000 = 36,032 ÷ 1,000.
  3. Calculated requirement = 36 kVA.
  4. In practice you would specify the next standard size, typically 45 kVA, to avoid running at the limit and to leave room for growth.

Answer: 36 kVA calculated; a 45 kVA unit in practice

Common mistakes

  • Omitting 1.732 on a three-phase transformer. Read the question for the phase configuration. Using the single-phase formula on a three-phase transformer overstates the current by about 73 percent and leads to badly oversized conductors.
  • Reading a ratio in the wrong direction. Ratios are written primary-to-secondary, so 10-to-1 steps down. Reading it backwards turns a 48-volt answer into 4,800 volts.
  • Multiplying current in the same direction as voltage. They move oppositely. If voltage doubles, current halves. Checking that volt-amperes balance on both sides catches this immediately.
  • Confusing the kVA rating with deliverable kilowatts. A 100 kVA transformer delivers 100 kW only at unity power factor. At 0.8 power factor it delivers 80 kW while being just as fully loaded thermally.
  • Ordering a full-rated single-phase unit for a three-phase bank. Each unit in a three-phase bank carries one third of the total. A 45 kVA bank takes three 15 kVA transformers, not three 45 kVA ones.
  • Treating an autotransformer as an isolation transformer. An autotransformer shares a single winding, so there is no galvanic isolation and it does not create a separately derived system. A fault or an open neutral can put primary voltage onto the secondary.

Related practice topics

Topics that come up alongside this one on the exam.

Questions about this topic

Why are transformers rated in kVA rather than kW?

Because what limits a transformer is how much current its windings can carry without overheating, and that current is set by volt-amperes regardless of the load's power factor. A 100 kVA transformer delivers 100 kW into a unity power factor load but only 80 kW at 0.8 power factor — and the winding current, and therefore the heating, is identical in both cases.

How do I find full-load current from a kVA rating?

Divide the volt-amperes by the winding voltage for single phase, or by 1.732 times the voltage for three phase. A 75 kVA three-phase transformer with a 480-volt primary draws 75,000 ÷ (1.732 × 480) = 90.2 amperes on the primary, and with a 208-volt secondary delivers 75,000 ÷ (1.732 × 208) = 208.2 amperes. The kVA is the same on both sides; only the voltage and current differ.

How does current relate between primary and secondary?

Inversely to voltage, because volt-amperes in equals volt-amperes out apart from small losses. A transformer with a 480-volt primary and a 240-volt secondary delivering 100 amperes draws 50 amperes on the primary. Checking that the volt-amperes balance on both sides, 24,000 in that example, is the fastest way to catch an inverted ratio.

What is the difference between delta and wye on a transformer?

In a wye winding, line-to-line voltage is 1.732 times the line-to-neutral voltage, and line current equals winding current. In a delta winding it is the reverse: line voltage equals winding voltage, but line current is 1.732 times winding current. A delta primary with a wye secondary at 120 volts to neutral gives 208 volts line-to-line, which is where the 208Y/120 designation comes from.

What is an autotransformer, and when would I use one?

An autotransformer has a single winding tapped to give the output voltage, rather than separate primary and secondary windings. That makes it smaller and cheaper for a given capacity, but because the windings share a conductive path there is no galvanic isolation. The common application is a buck-boost transformer, field-connected to add or subtract a modest voltage — useful for feeding 240-volt equipment from a 208-volt supply, because the unit only handles the difference rather than the whole load.

Why does a transformer secondary form a separately derived system?

Because with separate windings the secondary has no electrical connection to the supply conductors, so it has no reference to ground until one is deliberately established. A system bonding jumper and a grounding electrode conductor create that reference and give secondary-side faults a low-impedance path back to the secondary source. An autotransformer, sharing a winding, does not create a separately derived system.

Why does an energized transformer with no load still get warm?

Core losses — hysteresis from repeatedly remagnetising the steel, and eddy currents circulating within it — plus the magnetizing current needed to establish the flux. These are roughly constant whenever the transformer is energized. Copper losses, by contrast, vary with the square of load current and are nearly zero at no load.

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