Check voltage-loss scaling and changed cable conditions
Explain the squared loss reduction for a matched transfer and account for simultaneous resistance changes and different transfer boundaries.
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 inverse current changes.
- Distinguish IR from I²R.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Receiving voltage triples at fixed load power. What happens to current?
What you will learn
- Explain the squared loss reduction for a matched transfer and account for simultaneous resistance changes and different transfer boundaries.
- Apply square-current loss scaling with explicit transfer boundaries and include changed resistance rather than hiding it.
Build a controlled ratio
For the same stated receiving-load power and matched resistive conditions, doubling receiving voltage halves current. If total loop resistance is fixed, cable heating becomes one quarter. For a 6000 W load at receiving voltages 3000 V and 6000 V, currents are 2 A and 1 A; with total R = 10 Ω, losses are 40 W and 10 W.
Do not claim the whole system is 100%-efficient because a transformer sub-model is ideal. Cables with finite resistance still transfer energy to heating. Identify the voltage and power boundary and retain the supplied operating conditions when comparing cases.
Include a second changed factor
If receiving voltage triples at the same load power but loop resistance doubles, current becomes one third while loss becomes (1/3)² × 2 = 2/9 of its original value. It is not 1/9 because resistance was not held fixed. This example deliberately tests a changed cable condition.
Voltage drop IR and heating power I²R are different quantities with different units. Multiplying a voltage drop by current gives its associated heating power under the stated resistive model. A smaller loss claim needs matching current and resistance, not only a high voltage label.
Sources for this lesson
Pause and explain
Voltage triples at fixed load power, but loop resistance doubles. What is the loss ratio?
Worked example
A 6000 W resistive receiving load uses 3000 V or 6000 V, with total R = 10 Ω. Calculate currents and cable losses.
Show the worked solution
- Currents are 2 A and 1 A.
- Losses are 2² × 10 = 40 W and 1² × 10 = 10 W.
- The doubled-voltage loss is one quarter.
Answer 2 A/40 W and 1 A/10 W; ratio 1/4.
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 currents and losses for 6000 W at 3000 V and 6000 V with total R = 10 Ω.
Give me a hint
Use each receiving voltage, then square its current.
Compare my reasoning
- Currents are 6000/3000 = 2 A and 6000/6000 = 1 A.
- Losses are 40 W and 10 W.
- The ratio is 1/4.
2 A/40 W and 1 A/10 W.
Look for these in your work
- Keep load power fixed.
- Use the common total resistance.
At fixed load power, receiving voltage triples and total cable resistance doubles. Find the loss ratio.
Give me a hint
Multiply the squared current ratio by the resistance ratio.
Compare my reasoning
- Current ratio is 1/3.
- Squared ratio is 1/9.
- Multiply by 2 for the changed resistance.
2/9 of the original loss.
Look for these in your work
- Include both changes.
- State a dimensionless ratio.
For I = 2 A and R = 10 Ω, a report calls IR = 20 W the cable loss. Correct quantity, units and heating rate.
Give me a hint
Use the appropriate expression for each physical quantity.
Compare my reasoning
- IR = 20 V is a voltage drop.
- I²R = 4 × 10 = 40 W is heating power.
- The power also equals current times that 20 V drop.
Voltage drop 20 V; heating power 40 W.
Look for these in your work
- Distinguish volts and watts.
- Use the same cable boundary.
Common mistakes
- Fix the stated transfer boundary.
- Include resistance changes when present.
- IR and I²R have different units.
What can you explain now?
What units distinguish cable voltage drop and heating power?
Compare with the explanation
Volts for IR; watts for I²R
The resistive voltage drop and its associated power rate are different quantities.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Apply square-current loss scaling with explicit transfer boundaries and include changed resistance rather than hiding it.
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.