Calculate cable heating losses with current squared resistance
Use P_loss = I²R with the stated cable-loop resistance and distinguish heating power from transmitted load power.
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028. Use the stated source, operating conditions and meter ranges. Written practice is self-checked. Human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Square numerical values.
- Distinguish power and energy.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
If R already includes outward and return cables, should it be doubled again?
What you will learn
- Use P_loss = I²R with the stated cable-loop resistance and distinguish heating power from transmitted load power.
- Calculate same-path I²R losses and keep loop boundaries, rates and energy intervals explicit.
Identify the current and resistance boundary
The electrical heating power in a resistive cable is P_loss = I²R. Use the current through the named cable path and the resistance of that same path. If R is stated as total outward-and-return loop resistance, use it once rather than doubling it again. Squaring current makes current reduction especially effective.
For a supplied loop with I = 20 A and total R = 5 Ω, the heating loss is 20² × 5 = 2000 W, or 2 kW. This is a loss rate in the cable, not automatically the entire source power or the useful power delivered to a load.
Keep power and energy distinct
For an interval at constant current and resistance, cable heating energy is P_loss × time. A power loss in watts is a rate; it is not an energy in joules until a time is included. Resistance changes with conditions in real cables, so use the stated matched or fixed model.
At steady transfer with no net stored-energy change, source power supplies load power plus stated losses. A finite loss means source and load powers are different. The exercise does not certify a cable rating or provide a real transmission installation design.
Pause and explain
A cable loop has I = 4 A and R = 3 Ω. What is its heating rate?
Worked example
A supplied transmission loop has I = 20 A and total R = 5 Ω. Calculate cable heating power.
Show the worked solution
- Square the loop current: 20² = 400.
- Multiply by the stated total resistance 5 Ω once.
- The heating rate is 2000 W.
Answer 2000 W or 2 kW.
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.
Calculate heating power for I = 20 A and total loop resistance 5 Ω.
Give me a hint
Square the current before multiplying by the loop resistance.
Compare my reasoning
- 20² = 400.
- 400 × 5 = 2000.
- The result is 2000 W or 2 kW.
2000 W.
Look for these in your work
- Use the stated total boundary.
- State power units.
Calculate heating power for I = 4 A and R = 3 Ω.
Give me a hint
Keep current squared and resistance to its first power.
Compare my reasoning
- 4² = 16.
- 16 × 3 = 48.
- The heating power is 48 W.
48 W.
Look for these in your work
- Use I²R.
- Distinguish watts from volts.
A loop with constant I = 20 A and total R = 5 Ω operates for 60 s. Find cable heating energy and explain why 2 kW is not already an energy.
Give me a hint
Find the rate, then multiply by the interval.
Compare my reasoning
- Heating rate is 2000 W.
- Energy is 2000 × 60 = 120000 J.
- Watts express joules per second, not total joules.
120000 J; 2 kW is a power rate.
Look for these in your work
- Include the stated time.
- Separate energy and power.
Common mistakes
- Square current, not resistance.
- Use total loop resistance once.
- Loss power is not automatically load power.
What can you explain now?
Which equation describes the stated resistive cable heating rate?
Compare with the explanation
P_loss = I²R
The supplied loss rate is current squared times resistance of the same cable path.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Calculate same-path I²R losses and keep loop boundaries, rates and energy intervals explicit.
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.