lIndex":33,"initialQuestionWindow":[{"id":"bc650fd0-f058-48ea-b25c-04ba34e37b32","revisionId":"6fbcce1e-6635-41f9-9c61-6d8d765a6ff1","questionNumber":null,"questionType":"LA","specification":"An industrial ventilation system contains an electric fan motor with a continuous electrical power input of $5.6\\text{ kW}$.","difficulty":4,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"6a34cddc-87fc-4157-a29d-01ace8a93cb1","label":null,"content":"Show that the fan motor uses $30.24\\text{ MJ}$ of electrical energy when running continuously for $1\\text{ hour}$ and $30\\text{ minutes}$.","markscheme":"- Converts time to seconds: $t = 90 \\times 60 = 5400\\text{ s}$ M1\n- Recalls and applies $E = P \\times t$ with power in Watts ($P = 5600\\text{ W}$): $E = 5600 \\times 5400$ M1\n- Obtains $30,240,000\\text{ J}$ and divides by $1,000,000$ to show $30.24\\text{ MJ}$ A1","marks":3,"order":0,"markingRubric":null,"explanation":{"methods":[{"id":"power_energy","label":"Use power and time","steps":[{"type":"text","order":0,"title":"Choose the energy equation","content":"Electrical energy transferred is found using\n\n$$E=Pt$$\n\nwhere $E$ is energy in joules (J), $P$ is power in watts (W), and $t$ is time in seconds (s). So both given quantities must be converted into these units first.","highlights":[{"text":"continuous electrical power input of $5.6\\text{ kW}$","location":"specification"},{"text":"$$1\\text{ hour}$ and $30\\text{ minutes}$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"t=1\\text{ hour}+30\\text{ minutes}=90\\text{ minutes}"},{"latex":"t=90\\times60=5400\\text{ s}"}]},"order":1,"title":"Convert the time","description":"Write the duration in minutes first, then convert minutes to seconds. Converting the duration to seconds earns the first method mark (M1)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"P=5.6\\text{ kW}"},{"latex":"P=5.6\\times1000=5600\\text{ W}"}]},"order":2,"title":"Convert the power","description":"The equation needs power in watts. The prefix kilo means one thousand."},{"type":"hyper_latex","block":{"nodes":[{"latex":"E=Pt"},{"latex":"E=(5600\\text{ W})(5400\\text{ s})"},{"latex":"E=30\\,240\\,000\\text{ J}"}]},"order":3,"title":"Calculate the energy","description":"Substitute the converted power and time into the energy equation. This earns the second method mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"1\\text{ MJ}=1\\,000\\,000\\text{ J}"},{"latex":"E=\\frac{30\\,240\\,000}{1\\,000\\,000}\\text{ MJ}"},{"latex":"E=30.24\\text{ MJ}"}]},"order":4,"title":"State the answer in MJ","description":"One megajoule equals one million joules, so divide the energy in joules by one million. This completes the accuracy mark."}]}],"generatedAt":"2026-07-22T09:39:31.780Z","modelVersion":"gpt-5.6-terra+diagrams-1","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"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_elect

Electrical power

Question 9

Electrical power Questions

Electrical power

Question 9

Electrical power Questions