lIndex":34,"initialQuestionWindow":[{"id":"d78fc9fc-c52b-455b-b1a8-f790c1560688","revisionId":"05086bfc-1089-49f3-a703-c58d442322d8","questionNumber":null,"questionType":"LA","specification":"An engineer tests a miniature heating element designed for a microfluidic thermal cycler.\n\nWhen connected to a $12.0\\text{ V}$ DC power supply, the current in the heating element is $450\\text{ mA}$.","difficulty":5,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"a8a6d5bf-abe3-4f52-8c33-e1559b752fc9","label":null,"content":"Calculate the energy transferred to the heating element in $8\\text{ minutes}$.\n\nUse an equation selected from the physics equations sheet.","markscheme":"- Converts current to $0.45\\text{ A}$ or time to $480\\text{ s}$ M1\n- Substitutes into electrical energy and power equations, e.g., $E = 12.0 \\times 0.45 \\times 480$ M1\n- Calculates the final energy as $2592\\text{ J}$ (or $2.592\\text{ kJ}$) A1","marks":3,"order":0,"markingRubric":null,"explanation":{"methods":[{"id":"direct_electrical_energy","label":"Use electrical energy","steps":[{"type":"text","order":1,"title":"Convert the quantities","content":"Use SI units before substituting into an equation: current in amperes and time in seconds.\n\n$$450\\text{ mA}=0.45\\text{ A}$$\n\n$$8\\text{ min}=480\\text{ s}$$\n\nEither conversion earns the first mark (M1). Converting both makes the later substitution clear.","highlights":[{"text":"the current in the heating element is $450\\text{ mA}$","location":"specification"},{"text":"in $8\\text{ minutes}$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"P=\\htmlData{anchor=power_formula}{VI}"},{"latex":"P=\\htmlData{anchor=power_sub}{12.0\\times0.45}"},{"latex":"P=\\htmlData{anchor=power_value}{5.4\\text{ W}}"},{"latex":"E=\\htmlData{anchor=energy_formula}{Pt}"},{"latex":"E=\\htmlData{anchor=energy_sub}{5.4\\times480}"},{"latex":"E=\\htmlData{anchor=final_energy}{2592\\text{ J}}\\quad\\text{(A1)}"}],"highlights":[{"color":"green","style":"box","anchor":"final_energy"}],"connections":[{"to":"power_sub","from":"power_formula","color":"sky","label":"substitute V and I"},{"to":"energy_sub","from":"power_value","color":"green","label":"use P = 5.4 W"},{"to":"energy_sub","from":"energy_formula","color":"amber","label":"substitute P and t"}]},"order":2,"title":"Use the equation sheet","description":"Use P = VI to find the electrical power, then E = Pt to find the energy transferred. Substituting the values in SI units earns the second M1; the final calculation earns A1."}]}],"generatedAt":"2026-07-22T11:52:09.820Z","modelVersion":"gpt-5.6-terra+diagrams-1","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"e0912ec6-192b-4606-bc13-8c6e28c65629","revisionId":"155c17b6-a732-44d9-a333-04faa68ad25f","questionNumber":null,"questionType":"LA","specification":"An engineer records the electrical energy transferred and the operating time for two different industrial induction furnaces.\n\n| Furnace | Electrical energy transferred (MJ) | Operating time (minutes) |\n| :--- | :---: | :---: |\n| Furnace A | 180 | 5.0 |\n| Furnace B | 240 | 8.0 |","difficulty":5,"difficultyBand":"medium","totalMarks":2,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"561b1f5d-7443-426e-b95c-a4dc29e29f10","label":null,"content":"Compare the average electrical power of the two furnaces, using calculations to support your answer.","markscheme":"- Calculates the average power of at least one furnace using $P = \\frac{E}{t}$ with consistent units, for example, Furnace A: $\\frac{180 \\times 10^6 \\text{ J}}{300 \\text{ s}} = 600,000 \\text{ W}$ or Furnace B: $\\frac{240 \\times 10^6 \\text{ J}}{480 \\text{ s}} = 500,000 \\text{ W}$ M1\n- Correctly calculates both average power values ($600 \\text{ kW}$ and $500 \\text{ kW}$) and concludes that Furnace A has the higher average power A1","marks":2,"order":0,"markingRubric":null,"explan

Electrical power

Question 10

Electrical power Questions

Electrical power

Question 10

Electrical power Questions