Power Stations WORLDWIDETHE GLOBAL EDITION 01
LESSON 6 OF 6 · ORIGINAL EDUCATIONAL EXAMPLE

Ideal transformer ratios

Calculate voltage and inverse current ratios from winding turns, and check apparent-power conservation.

Learning goal: Use corresponding winding quantities and recognize why a voltage step-down does not create power.

Prepared by Power Stations Worldwide. Sources checked on 30 September 2026; not independently engineering peer reviewed.

The essentials

Turns set the winding voltage ratio

For an ideal transformer, the winding-voltage ratio equals the turns ratio a = N₁/N₂ = V₁/V₂. More secondary turns step up voltage; fewer step it down. These are corresponding winding values, not arbitrary three-phase line-to-line nameplate voltages.

Current scales the other way

For an ideal lossless transformer supplying a load, current magnitudes follow I₂/I₁ = a. Stepping voltage down therefore steps current up at the same apparent power. Current signs depend on the chosen reference arrows; the example uses positive magnitudes.

Real transformers need more

Losses, excitation current, impedance, voltage regulation, taps, frequency and saturation are omitted here. Wye–delta connections introduce line-to-winding factors and phase shifts. A line-voltage nameplate ratio cannot always be treated as a winding turns ratio.

Ten-to-one voltage, one-to-ten current

Ten-to-one voltage, one-to-ten currentAn abstract transformer has a primary winding labelled 1,000 turns and 2,400 volts, and a secondary labelled 100 turns and 240 volts. Primary current 2.083 amperes enters the transformer; secondary current 20.83 amperes supplies the load. Both sides are labelled 5 kVA in the ideal model.Primary inputSecondary output1,000 turns2,400 V2.083 A100 turns240 V20.83 AIdeal: 5 kVA on both sides
Original ideal single-phase transformer abstraction; winding symbols are not to scale. No real core, insulation, grounding, fault protection or construction details are specified.
HYPOTHETICAL VALUES · NOT PLANT RECORDS

Worked example

An ideal single-phase 2,400 V / 240 V transformer

Given

  • Primary N₁ = 1,000 turns; secondary N₂ = 100 turns
  • Primary V₁ = 2,400 V RMS
  • Secondary load S₂ = 5 kVA = 5,000 VA

Formula toolbox

  • a = N₁ / N₂ = V₁ / V₂
  • V₂ = V₁ / a; I₂ / I₁ = a (magnitudes)
  • Ideal single-phase: S₁ = V₁I₁ = V₂I₂ = S₂

Calculate step by step

  1. Turns ratio: a = 1,000 / 100 = 10.
  2. Secondary voltage: V₂ = 2,400 / 10 = 240 V.
  3. Secondary current: I₂ = 5,000 / 240 ≈ 20.83 A.
  4. Primary current: I₁ = 5,000 / 2,400 ≈ 2.083 A; I₂/I₁ = 10.
  5. Using unrounded currents: V₁I₁ = V₂I₂ = 5,000 VA.

240 V secondary · 20.83 A secondary · 2.083 A primary

Apparent power is conserved in the ideal model. The 5 kVA load is not necessarily 5 kW: its real power also depends on load PF. The illustrative turns are not a transformer construction design.

Results are rounded only for display. Reproduce the calculations using unrounded intermediate values.

Quick quiz

Choose one answer for each question, then check your answers. This is a practice check, not a qualification. Answers stay in this page only; refreshing or leaving resets them.

1. An ideal transformer has N₁/N₂ = 4 and V₁ = 800 V. What is V₂?
2. For that ideal ratio of 4, if I₁ = 3 A, what is the secondary current magnitude?

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  1. 200 V. V₂ = V₁ / a = 800 / 4 = 200 V.
  2. 12 A. I₂ = aI₁ = 4 × 3 = 12 A; 800 × 3 = 200 × 12 = 2,400 VA.

Curriculum connection: MIT 6.061 (Spring 2011). Its readings include 2007 notes; some equations use peak rather than RMS amplitudes. This lesson states its own value conventions and uses original diagrams, numbers and quiz questions. No MIT affiliation or endorsement is claimed.

Textbook references

Suggested technical background for this topic. These books are further reading, not proof that every explanation was checked against every edition. See the full bibliography and review scope.

  1. Electrical Machines, Drives and Power Systems (6th ed., Pearson New International Edition) — Wildi, T., Pearson, 2013 · ISBN 9781292024585Ideal and practical transformers, ratios and three-phase arrangements.
  2. Electric Power Systems (5th ed.) — Weedy, B. M., Cory, B. J., Jenkins, N., Ekanayake, J. B. & Strbac, G., Wiley, 2012 · ISBN 9780470682685Transformer models in a wider power-system context.