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IGCSE · Cambridge (CIE) · 0625

Calculate ideal transformer input and output currents

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

Section 4.5.6 · The transformer

Transformer and transmission pathway · 4.5.6

Extended only (Supplement); Physics 0625, 2026–2028. Use the stated source, operating conditions and meter ranges. Written practice is self-checked.

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.

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.

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Course scope and assessment

491 sequenced Physics lessons provide 1473 guided, independent and application problems. 58 of 58 syllabus sections have full linked teaching sequences; 0 sections remain pending. Three earlier starter overviews remain available but are not counted as full sequences. The map covers all six subject areas for 2026–2028 and keeps review and assessment limits visible. This is available content, not full programme completion, an exam grade or certified mastery. Written practice is self-checked. Supervised practical experience, cumulative assessment and human subject-teacher review remain pending. Paper method exercises do not demonstrate practical competence or complete either practical-assessment route.

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