Compare transmission losses at matched load power
Use stated receiving voltages and load power to compare current and cable losses without equating source and load power when losses are present.
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 for a supplied resistive load.
- Calculate cable I²R losses.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
With non-zero cable heating, is source power equal to load power in the steady stated account?
What you will learn
- Use stated receiving voltages and load power to compare current and cable losses without equating source and load power when losses are present.
- Compare inverse-current and square-loss scaling with a consistent receiving-point power and source-loss account.
Use the voltage at the stated transfer point
A supplied resistive receiving load takes 200 kW. Compare receiving voltages 2000 V and 10000 V, with total transmission-loop resistance 5 Ω in each case. Load current is P_load/V_load: 100 A and 20 A respectively. Cable losses are 100² × 5 = 50000 W and 20² × 5 = 2000 W.
At steady transfer with no other stated losses or net storage, source powers are 250 kW and 202 kW respectively: load power plus cable heating. The given receiving voltages are not silently treated as identical source voltages. This keeps the lower-current heating comparison consistent while delivering the same load power.
Separate loss reduction from ideal transformation
For matched load power and fixed cable resistance, multiplying the receiving voltage by k divides current by k and cable heating by k². Raising receiving voltage fivefold in this exercise reduces loss to 1/25. The reduction is not merely 1/5 because current is squared.
The passive voltage transformation itself does not create the load energy. If cable resistance, load power or operating conditions change too, the simple ratio needs those changes included. The supplied calculations are a model comparison, not real-grid ratings or a safe high-voltage procedure.
Pause and explain
Matched power and resistance, receiving voltage doubled: what happens to cable heating?
Worked example
Compare I and cable loss for a 200 kW resistive receiving load at 2000 V and 10000 V, with total loop resistance 5 Ω.
Show the worked solution
- Currents are 200000/2000 = 100 A and 200000/10000 = 20 A.
- Losses are 100² × 5 = 50 kW and 20² × 5 = 2 kW.
- The higher receiving voltage reduces loss to 1/25.
Answer 100 A/50 kW and 20 A/2 kW; higher-voltage loss is 1/25.
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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.
Calculate currents for 200 kW load power at 2000 V and 10000 V.
Give me a hint
Convert kilowatts to watts before using P/V.
Compare my reasoning
- 200 kW is 200000 W.
- At 2000 V current is 100 A.
- At 10000 V current is 20 A.
100 A and 20 A.
Look for these in your work
- Use receiving voltages.
- Keep the load power fixed.
Calculate each cable loss at total R = 5 Ω and those currents.
Give me a hint
Apply I²R separately to each case.
Compare my reasoning
- At 100 A, loss is 50000 W.
- At 20 A, loss is 2000 W.
- Loss ratio is 2000/50000 = 1/25.
50 kW and 2 kW; ratio 1/25.
Look for these in your work
- Square both currents.
- Use total resistance once.
Find source powers for those steady cases with no other losses or net storage, then explain why 200 kW cannot be both source and load power.
Give me a hint
Include the cable heating in the same system boundary.
Compare my reasoning
- Load power is 200 kW in both cases.
- Add losses 50 kW and 2 kW.
- Source powers are 250 kW and 202 kW.
250 kW and 202 kW; non-zero cable loss requires additional source power.
Look for these in your work
- Keep the load boundary.
- Include every stated loss.
Common mistakes
- Use receiving-point voltage.
- Keep load power and loop resistance matched.
- Source supplies load plus loss.
What can you explain now?
Why does fivefold voltage reduce matched-power cable loss by twenty-fivefold?
Compare with the explanation
Current is divided by five and loss depends on its square
At fixed transfer power and resistance, the inverse current change is squared in I²R.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Compare inverse-current and square-loss scaling with a consistent receiving-point power and source-loss account.
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.