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.
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
- 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.
Sources for this lesson
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.