Use consistent time units and changing operating intervals
Add energy from separately stated steady intervals instead of assuming an initial reading or unrelated averages hold throughout.
Core and Extended; Physics 0625, 2026–2028. Use the stated source, operating conditions and meter ranges. Written practice is self-checked.
Before you begin
- Use E = IVt at constant conditions.
- Interpret power as a rate.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the total duration of 20 s followed by 10 s?
What you will learn
- Sum electrical energy across changing stages.
- Identify when E = IVt lacks the required interval data.
Split the interval where the conditions change
If a component operates at 6 V and 0.5 A for 20 s, then at 6 V and 1 A for 10 s, calculate each stage. The first supplies 60 J and the second 60 J, giving 120 J over 30 s. The average power is 4 W, although neither stage has a 4 W input.
A time-weighted average is needed when stages have unequal lengths. The simple average of the two stage powers, (3 + 6)/2 = 4.5 W, would incorrectly give 135 J over 30 s. Record the durations before averaging or simply sum the stage energy amounts.
Use simultaneous data
When both current and voltage vary, average power depends on their simultaneous products. Suppose equal 10 s stages have 1 A at 2 V and 3 A at 4 V: powers are 2 W and 12 W, so total energy is 140 J. Multiplying their separate averages gives 2 A × 3 V × 20 s = 120 J, which is different.
A meter showing 2 A once while voltage is 6 V does not prove the 12 W rate persisted for a whole minute. Without the later readings or a constant-condition statement, 720 J is only a conditional estimate. State what the data establishes and what still needs measurement.
Pause and explain
Equal 10 s stages have powers 2 W and 12 W. What is the energy?
Worked example
Find energy for 6 V at 0.5 A over 20 s followed by 6 V at 1 A over 10 s.
Show the worked solution
- First stage: 6 × 0.5 × 20 = 60 J.
- Second stage: 6 × 1 × 10 = 60 J.
- Total: 120 J over 30 s.
Answer 120 J; average power 4 W.
Try a topic test
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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 energy for 6 V at 0.5 A over 20 s, then 6 V at 1 A over 10 s.
Give me a hint
Calculate each stage with its own duration before adding.
Compare my reasoning
- First stage gives 60 J.
- Second stage gives 60 J.
- Total is 120 J.
120 J.
Look for these in your work
- Retain both durations.
- Add energy amounts.
Equal 10 s stages have 1 A at 2 V and 3 A at 4 V. Find total energy and average power.
Give me a hint
Use simultaneous products in each stage rather than separate averages.
Compare my reasoning
- Stage powers are 2 W and 12 W.
- Energy is 20 + 120 = 140 J.
- Average power is 140/20 = 7 W.
140 J; average power 7 W.
Look for these in your work
- Multiply matching values.
- Use the complete duration.
One reading shows 6 V and 2 A during a one-minute interval. Is 720 J proved?
Give me a hint
Ask whether the rate persisted for all sixty seconds.
Compare my reasoning
- That reading gives an instantaneous 12 W input.
- If it persisted for 60 s, energy would be 720 J.
- The one reading does not establish the full interval's energy.
No; 720 J requires constant values or equivalent interval data.
Look for these in your work
- State the conditional estimate.
- Identify missing interval evidence.
Common mistakes
- Weight unequal intervals by time.
- Use simultaneous current and voltage.
- A single reading does not prove a full-interval rate.
What can you explain now?
How should separate steady-stage energy amounts be combined?
Compare with the explanation
Add the stage amounts.
Each stage's power acts for its own specified duration before the energies are totalled.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Add stage energies and explain why unrelated averages or a single reading may not determine total energy.
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.