lIndex":22,"initialQuestionWindow":[{"id":"68313e29-0808-40e9-8959-11dd924fad6c","revisionId":"837ebdc1-4316-49e1-9087-07f2d90be1f7","questionNumber":null,"questionType":"LA","specification":"","difficulty":2,"difficultyBand":"easy","totalMarks":2,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"9761cff2-ac88-4216-bf8c-64d2dbe9ed82","label":null,"content":"A small toy motor is connected to a potential difference of $6.0\\text{ V}$.\n\nThe current in the motor is $0.25\\text{ A}$.\n\nCalculate the power of the motor.\nUse the equation:\n$$\\text{power} = \\text{current} \\times \\text{potential difference}$$","markscheme":"- Substitutes values into the power formula: $0.25 \\times 6.0$ M1\n- Calculates the power to be $1.5\\text{ W}$ A1","marks":2,"order":0,"markingRubric":null,"explanation":{"methods":[{"id":"power_formula","label":"Use the power equation","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"P=\\htmlData{anchor=formula}{IV}"},{"latex":"P=\\htmlData{anchor=substitution}{0.25\\ \\mathrm{A}\\times6.0\\ \\mathrm{V}}"},{"latex":"P=\\htmlData{anchor=answer}{1.5\\ \\mathrm{W}}"},{"text":"This earns A1."}],"highlights":[{"color":"green","style":"box","anchor":"answer"}],"connections":[{"to":"substitution","from":"formula","color":"sky","label":"substitute values"}]},"order":1,"title":"Substitute the values","description":"Use the supplied equation P = IV. Put the current and potential difference into it; this correct substitution earns M1."}]}],"generatedAt":"2026-07-22T11:54:01.084Z","modelVersion":"gpt-5.6-terra+diagrams-1","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"72b5c573-2fcd-42e5-a6bd-034a455b3088","revisionId":"c8e6913e-c0bb-4d00-900d-3b2e1d79dc13","questionNumber":null,"questionType":"LA","specification":"","difficulty":4,"difficultyBand":"easy","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"308351a4-f3b9-4c95-9ebc-dc1dafe92ac6","label":null,"content":"An LED desk lamp is connected to the $230\\text{ V}$ mains supply. Over a period of $4\\text{ minutes}$, it transfers $11\\,040\\text{ J}$ of energy.\n\nCalculate the current in the LED desk lamp.\n\nUse the equation:\n$$I = \\frac{E}{V \\times t}$$","markscheme":"- Converts time to seconds: $t = 4 \\times 60 = 240\\text{ s}$ B1\n- Substitutes values into the formula: $I = \\frac{11\\,040}{230 \\times 240}$ M1\n- Calculates the correct current: $I = 0.2\\text{ A}$ A1","marks":3,"order":0,"markingRubric":null,"explanation":{"methods":[{"id":"energy_current","label":"Use energy and time","steps":[{"type":"text","order":0,"title":"Convert the time","content":"The equation needs time in seconds, so convert the given minutes:\n\n$$t=4\\times60=240\\text{ s}$$\n\nThis conversion earns the **B1** mark.","highlights":[{"text":"Over a period of $4\\text{ minutes}$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"I=\\frac{E}{V\\times t}"},{"latex":"I=\\frac{11\\,040}{230\\times240}"}]},"order":1,"title":"Substitute the values","description":"Use the formula with energy in joules, voltage in volts, and the converted time in seconds. Correct substitution earns the M1 mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"I=\\frac{11\\,040}{\\htmlData{anchor=factors}{230\\times240}}"},{"latex":"I=\\frac{11\\,040}{\\htmlData{anchor=denominator}{55\\,200}}"},{"latex":"I=\\htmlData{anchor=current}{0.2\\text{ A}}"}],"highlights":[{"color":"green","style":"box","anchor":"current"}],"connections":[{"to":"denominator","from":"factors","color":"teal","label":"multiply"}]},"order":2,"title":"Calculate the current","description":"Evaluate the denominator, then divide the energy by it. Therefore, I = 0.2 A; this earns A1."}]}],"generatedAt":"2026-07-22T11:31:40.255Z","modelVersion":"gpt-5.6-terra+diagrams-1","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"7ab18dcf-2c28-462f-914b-1a15d79bd3ac","revisionId":

Electrical power

Question 16

Electrical power Questions

Electrical power

Question 16

Electrical power Questions