lIndex":18,"initialQuestionWindow":[{"id":"3e0a1b73-d369-4bd4-84dc-443b5400a4fd","revisionId":"29d71b25-54c1-4678-bdb5-845c39b6d6fb","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"3dc47f95-7d00-4ee9-bd7a-e0850cb39603","label":null,"content":"An electric scooter motor is connected to a $36\\text{ V}$ battery pack. Over a climb lasting $8\\text{ minutes}$, it transfers $129\\,600\\text{ J}$ of electrical energy.\n\nCalculate the current in the motor during this climb.\n\nUse the equation:\n$$I = \\frac{E}{V \\times t}$$","markscheme":"- Converts time from minutes to seconds: $8 \\times 60 = 480\\text{ s}$ M1\n- Substitutes values into the given equation: $I = \\dfrac{129\\,600}{36 \\times 480}$ M1\n- Calculates the correct current of $7.5\\text{ A}$ A1","marks":3,"order":0,"markingRubric":null,"explanation":{"methods":[{"id":"substitution","label":"Convert then substitute","steps":[{"type":"text","order":1,"title":"Convert the time","content":"The equation needs $t$ in seconds, so convert the given $8\\text{ minutes}$:\n\n$$8 \\times 60 = 480\\text{ s}$$\n\nWriting this conversion earns **M1**.","highlights":[{"text":"$$8\\text{ minutes}$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"I=\\frac{E}{V\\times t}"},{"latex":"I=\\frac{\\htmlData{anchor=energy}{\\color{@sky}{129\\,600}}}{\\htmlData{anchor=denominator}{\\color{@amber}{36\\times480}}}"},{"latex":"I=\\frac{129\\,600}{17\\,280}"}],"highlights":[{"color":"sky","style":"text-color","anchor":"energy"},{"color":"amber","style":"text-color","anchor":"denominator"}]},"order":2,"title":"Substitute the values","description":"Put the energy, voltage and converted time into the equation provided. This correct substitution earns M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"I=\\frac{129\\,600}{17\\,280}"},{"latex":"I=\\htmlData{anchor=answer}{\\color{@green}{7.5\\text{ A}}}"}],"highlights":[{"color":"green","style":"text-color-with-background","anchor":"answer"}]},"order":3,"title":"Calculate the current","description":"Now divide the energy by the voltage-time product. Obtaining I = 7.5 A earns A1."}]}],"generatedAt":"2026-07-21T18:03:14.968Z","modelVersion":"gpt-5.6-terra+diagrams-1","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"5ec0b684-eb52-400d-87e8-e5b2fd34df63","revisionId":"cf5b8d15-5b64-4bdf-9ba7-8baecb14a108","questionNumber":null,"questionType":"LA","specification":"","difficulty":6,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"238d8dc2-dec7-467f-bb00-37a4a0fbc4f5","label":null,"content":"An electric delivery van's battery is recharged at a depot charging station.\n\nThe charging station provides an average current of 40.0 A to the battery, at a potential difference of 400 V.\n\nIt is claimed that 115.2 MJ of energy can be transferred to the battery in less than 1.8 hours.\n\nComment on this claim.\n\nUse this equation to support your answer:\n\n$$t = \\frac{E}{I \\times V}$$","markscheme":"- Converts energy to joules: $115.2\\text{ MJ} = 1.152 \\times 10^8\\text{ J}$ OR converts time to seconds: $1.8\\text{ hours} = 6480\\text{ s}$ B1\n- Calculates the time required as $7200\\text{ s}$ (or $2.0\\text{ hours}$) OR calculates the maximum energy transferred in $1.8\\text{ hours}$ as $103.7\\text{ MJ}$ (or $1.04 \\times 10^8\\text{ J}$) M1\n- Concludes that the claim is incorrect since $2.0\\text{ hours}$ is greater than $1.8\\text{ hours}$ (or $104\\text{ MJ}$ is less than $115.2\\text{ MJ}$) A1","marks":3,"order":0,"markingRubric":null,"explanation":{"methods":[{"id":"calculate_charging_time","label":"Calculate the charging time","steps":[{"type":"text","order":1,"title":"Convert the energy","content":"The equation uses energy in joules, so convert megajoules first:\n\n$$115.2\\text{ MJ}=1.152\\time

Electrical power

Question 6

Electrical power Questions

Electrical power

Question 6

Electrical power Questions