Combine transformer turns voltage and ideal power balances
Use the voltage ratio first, then the 100%-efficiency power balance with explicit load conditions.
Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade.
Before you begin
- Use the voltage-turn ratio.
- Use the ideal power equality.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Can turns alone determine both actual winding currents?
What you will learn
- Use the voltage ratio first, then the 100%-efficiency power balance with explicit load conditions.
- Combine corresponding voltage and ideal loaded-current balances without converting ratios into safe equipment ratings.
Use two relationships for different unknowns
The turn ratio gives the simple voltage relationship. The ideal power equality gives the related currents once a load or current condition is supplied. For primary 120 V/600 turns and secondary 150 turns, the secondary voltage is 30 V. If Ip = 0.50 A in the stated ideal loaded model, Pin = 60 W and Is = 60/30 = 2 A.
Keep the voltage calculation and power calculation as separate steps. A lower secondary voltage does not force output current to one fixed value unless the operating load or input power is also specified. The ideal assumptions used here include corresponding a.c. magnitudes and the supplied resistive-load power account.
Check inverse current and role changes
At equal ideal power, current ratio is inverse to voltage ratio: Is/Ip = Vp/Vs. For step-down output, a greater secondary current can accompany the lower voltage. This relation describes the specified operating state rather than a universal safe current capacity for the winding.
If a suitable stated arrangement changes which winding is input, assign primary and secondary again before computing. Real insulation, heating and current ratings do not follow merely from the mathematical ratio. Do not infer no-load input current or actual 100% efficiency from these ideal exercises.
Sources for this lesson
Pause and explain
Ideal Vp = 240 V, Np = 1200, Ns = 60 and Ip = 0.10 A. What are Vs and Is?
Worked example
Find Vs, Is and power for primary 120 V/600 turns, secondary 150 turns and ideal loaded Ip = 0.50 A.
Show the worked solution
- Vs = 120 × 150/600 = 30 V.
- Pin = 120 × 0.50 = 60 W.
- Is = 60/30 = 2 A at ideal efficiency.
Answer Vs = 30 V; Is = 2 A; both powers 60 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.
Combine the supplied 120 V/600-turn input and 150-turn output with ideal Ip = 0.50 A.
Give me a hint
Find voltage, then apply power balance.
Compare my reasoning
- Secondary voltage is 30 V.
- Input power is 60 W.
- Output current is 2 A.
30 V, 2 A and 60 W.
Look for these in your work
- Show both relationships.
- Check output power.
Find Vs and Is for 240 V, 1200 input turns, 60 output turns and ideal Ip = 0.10 A.
Give me a hint
Keep the same operating state through both steps.
Compare my reasoning
- Vs = 240 × 60/1200 = 12 V.
- Pin = 24 W.
- Is = 24/12 = 2 A.
Vs = 12 V; Is = 2 A.
Look for these in your work
- Use correct turn roles.
- Conserve the ideal power.
A specified suitable arrangement uses primary 30 V/150 turns, secondary 600 turns and ideal Ip = 2 A. Find Vs and Is.
Give me a hint
Assign roles from the newly specified input.
Compare my reasoning
- Vs = 30 × 600/150 = 120 V.
- Input power is 60 W.
- Is = 60/120 = 0.50 A.
Vs = 120 V; Is = 0.50 A.
Look for these in your work
- Identify the new primary.
- Use inverse current change.
Common mistakes
- Use the voltage ratio before current.
- State the loaded ideal condition.
- Current capacity needs real specifications.
What can you explain now?
How does ideal output current compare when output voltage is smaller at equal power?
Compare with the explanation
It is greater
The current increases inversely to the voltage change when the ideal power is held equal.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Combine corresponding voltage and ideal loaded-current balances without converting ratios into safe equipment ratings.
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.