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IGCSE · Cambridge (CIE) · 0652

Calculate ideal transformer input and output currents

Use IpVp = IsVs for a stated 100%-efficient transformer with matched power quantities and suitable load assumptions.

Transformers · Cambridge IGCSE Physical Science 0652 · P4.6.3

Cross-referenced from the related Cambridge IGCSE Chemistry 0620 or Physics 0625 teaching to the 0652 syllabus. Exact 0652 Core/Supplement objective coverage and this final-series course’s examiner fit still need specialist review. Written practice is self-checked and does not predict a Cambridge grade or demonstrate supervised laboratory competence.

Before you begin

  • Use P = IV under stated conditions.
  • Rearrange a product equality.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

Which equality is required by ideal efficiency?

What you will learn

  • Use IpVp = IsVs for a stated 100%-efficient transformer with matched power quantities and suitable load assumptions.
  • Calculate ideal transformer currents from matched powers and retain real-load and no-load limits.
Original 100%-efficient loaded model has primary 240 V, secondary 12 V and secondary current 2 A. Both matched power products are 24 W and primary current is 0.10 A. Suitable matched a.c. magnitudes and the supplied resistive-load account are assumed; real no-load current is not inferred.100%-efficient loaded modelVp = 240 V · Vs = 12 VIs = 2 AIp = 0.10 APin = Pout = 24 WCorresponding a.c. magnitudesSupplied resistive-load accountReal no-load current not inferred
Original 100%-efficient loaded model has primary 240 V, secondary 12 V and secondary current 2 A. Both matched power products are 24 W and primary current is 0.10 A. Suitable matched a.c. magnitudes and the supplied resistive-load account are assumed; real no-load current is not inferred.

Apply the stated ideal power balance

For a transformer with 100% efficiency in the supplied simple power model, input power equals output power: IpVp = IsVs. The primary and secondary quantities must refer to the same operating state and use consistent voltage and current units. The exercise uses suitable matched a.c. magnitudes and a resistive-load power account; do not apply it unqualified to arbitrary phase-shifted loads.

For Vp = 240 V, Vs = 12 V and Is = 2 A, output power is 24 W. Input current is Ip = 24/240 = 0.10 A. Lower output voltage can accompany greater output current while power is conserved. The primary and secondary currents are not equal merely because the transformer is ideal.

Keep real and no-load cases separate

Real transformer losses mean the ideal equality does not establish actual efficiency. With an open secondary, Is can be zero while a real primary still draws a magnetising and loss-related current. Do not use the ideal load calculation to claim every unloaded real primary current must be zero.

The current equation requires load information. A voltage-turn ratio alone cannot give a unique actual current in both circuits. State the 100%-efficiency assumption before rearranging; output power cannot exceed input power in that passive ideal account.

Sources for this lesson

Pause and explain

Vp = 120 V, Vs = 24 V, Is = 0.50 A. What is ideal Ip?

Put the idea to work

Worked example

A supplied ideal transformer has Vp = 240 V, Vs = 12 V and Is = 2 A. Find Ip and both powers.

Show the worked solution
  1. Output power is 12 × 2 = 24 W.
  2. Ideal input power is also 24 W.
  3. Ip = 24/240 = 0.10 A.

Answer Ip = 0.10 A; input and output powers 24 W.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

Find Ip for Vp = 240 V, Vs = 12 V and Is = 2 A.

Give me a hint

Compute output power before dividing by primary voltage.

Compare my reasoning
  1. Pout = 24 W.
  2. Pin = 24 W under the ideal assumption.
  3. Ip = 24/240 = 0.10 A.

0.10 A.

Look for these in your work

  • State the ideal condition.
  • Retain amperes.
2 · Independent

Find Ip for Vp = 120 V, Vs = 24 V and Is = 0.50 A.

Give me a hint

Use the same-state power products.

Compare my reasoning
  1. Pout = 24 × 0.50 = 12 W.
  2. Ideal Pin = 12 W.
  3. Ip = 12/120 = 0.10 A.

0.10 A.

Look for these in your work

  • Use the output load current.
  • Check equal powers.
3 · Transfer

A supplied ideal step-up has Vp = 12 V, Ip = 2 A and Vs = 120 V. Find Is and explain why higher output voltage is not greater output power.

Give me a hint

Conserve the 24 W input product.

Compare my reasoning
  1. Input power is 12 × 2 = 24 W.
  2. Is = 24/120 = 0.20 A.
  3. Greater output voltage accompanies smaller current at unchanged ideal power.

Is = 0.20 A; both powers 24 W.

Look for these in your work

  • Use inverse current change.
  • Reject energy creation.

Common mistakes

  • State ideal efficiency.
  • Use matching operating quantities.
  • Do not set primary and secondary current equal.
Recall without your notes

What can you explain now?

What does 100% efficiency imply for the supplied transformer power account?

Compare with the explanation

Input power equals output power

The ideal account equates the primary and secondary power rather than their voltages or currents individually.

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Make it stick

Calculate ideal transformer currents from matched powers and retain real-load and no-load limits.

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

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Course scope and assessment

This page maps the 2025–2026 Cambridge IGCSE Physical Science 0652 syllabus: 38 Chemistry sections and 46 Physics sections. Every section links to original 0620/0625 teaching or one of five new 0652 lessons, with guided, independent and transfer practice and explained answers. Cambridge limits entry to Administrative Zone 3 and identifies November 2026 as the final exam series. The section crosswalk is complete as a first coverage map; a numbered-objective and Core/Supplement audit, independent subject-teacher review and cumulative assessment are still needed. Written practice does not award a Cambridge grade. Online activities cannot demonstrate supervised laboratory competence or complete Paper 5/6.

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