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 P4.6.3 · Transformers
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.
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?
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
- Output power is 12 × 2 = 24 W.
- Ideal input power is also 24 W.
- Ip = 24/240 = 0.10 A.
Answer Ip = 0.10 A; input and output powers 24 W.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →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.
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
- Pout = 24 W.
- Pin = 24 W under the ideal assumption.
- Ip = 24/240 = 0.10 A.
0.10 A.
Look for these in your work
- State the ideal condition.
- Retain amperes.
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
- Pout = 24 × 0.50 = 12 W.
- Ideal Pin = 12 W.
- Ip = 12/120 = 0.10 A.
0.10 A.
Look for these in your work
- Use the output load current.
- Check equal powers.
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
- Input power is 12 × 2 = 24 W.
- Is = 24/120 = 0.20 A.
- 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.
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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