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[
{
"slide": 1,
"fragments": [
{
"fragment_index": -1,
"text_description": "Why Bulbs Glow Bright\nVoltage, current, resistance—lighting up the story.",
"image_description": ""
}
]
},
{
"slide": 2,
"fragments": [
{
"fragment_index": -1,
"text_description": "Ohm’s Law",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Ohm’s Law",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Ohm’s Law: Current through a conductor is\ndirectly proportional\nto the voltage across it, if temperature stays constant.",
"image_description": ""
}
]
},
{
"slide": 3,
"fragments": [
{
"fragment_index": -1,
"text_description": "Key Equation",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "\\[V = I \\cdot R\\]",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Variable Definitions\nV\nPotential difference (volt, V)\nI\nCurrent (ampere, A)\nR\nResistance (ohm, Ω)",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Applications\nOhm’s Triangle\nCover the unknown in the triangle to reveal the correct rearrangement.\nFind Current of a Bulb\nWith supply \\(V\\) and bulb resistance \\(R\\), compute \\(I = \\tfrac{V}{R}\\).\nSet Resistor Values\nChoose \\(R\\) so that a device draws safe current from a given voltage.",
"image_description": ""
}
]
},
{
"slide": 4,
"fragments": [
{
"fragment_index": -1,
"text_description": "Water Pipe Analogy",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Electricity ↔ Water Flow\nPicture a tank pushing water through a pipe: that is how voltage pushes charge.\nCurrent is the water flow rate, while resistance is how narrow the pipe is.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Key Points:\nHigher pressure (V) → greater flow (I) if pipe width stays same.\nWider pipe means lower resistance, so more current for a given voltage.\nNarrow pipe raises resistance and limits current, matching Ohm’s Law.",
"image_description": ""
}
]
},
{
"slide": 5,
"fragments": [
{
"fragment_index": -1,
"text_description": "Straight-Line Proof",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "V–I graph: slope = 2 Ω",
"image_description": "https://asset.sparkl.ac/pb/sparkl-vector-images/img_ncert/bzi30DWFlAl8YUIjMtuDJTbx0WquzFvN7r5jUNBz.png"
},
{
"fragment_index": 2,
"text_description": "Straight V–I line for a 2 Ω resistor\nThe graph confirms voltage varies linearly with current for a fixed resistor.\nKey Points:\nLine passes through origin ⇒ \\(V \\propto I\\).\nSlope \\(= \\frac{V}{I} = 2\\,\\Omega\\) (resistance).\nDoubling \\(I\\) doubles \\(V\\); e.g., \\(I = 2\\text{ A}\\) gives \\(V = 4\\text{ V}\\).",
"image_description": ""
}
]
},
{
"slide": 6,
"fragments": []
},
{
"slide": 7,
"fragments": [
{
"fragment_index": -1,
"text_description": "Worked Example",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "1\nWrite Given Data\nVoltage \\(V = 12 \\text{ V}\\); Resistance \\(R = 4 \\, \\Omega\\).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "2\nRecall Formula\nOhm’s Law gives current: \\(I = \\frac{V}{R}\\).",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "3\nSubstitute & Calculate\n\\(I = \\frac{12\\text{ V}}{4\\,\\Omega} = 3 \\text{ A}\\).",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "4\nState Answer\nThe resistor carries a current of \\(3 \\text{ A}\\).",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Pro Tip:\nAlways convert to SI units before using \\(I = V/R\\).",
"image_description": ""
}
]
},
{
"slide": 8,
"fragments": [
{
"fragment_index": -1,
"text_description": "Multiple Choice Question\nSubmit Answer\nCorrect!\nRight! Doubling resistance halves current because \\(I\\) varies inversely with \\(R\\).\nIncorrect\nCheck Ohm's law \\(I = \\frac{V}{R}\\); current decreases when resistance increases.\nconst correctOption = 0;\n const answerCards = document.querySelectorAll('.answer-card');\n const submitBtn = document.getElementById('submitBtn');\n const feedbackCorrect = document.getElementById('feedbackCorrect');\n const feedbackIncorrect = document.getElementById('feedbackIncorrect');\n \n let selectedOption = null;\n \n answerCards.forEach((card, index) => {\n card.addEventListener('click', () => {\n answerCards.forEach(c => c.classList.remove('border-blue-500', 'bg-blue-50'));\n card.classList.add('border-blue-500', 'bg-blue-50');\n selectedOption = index;\n });\n });\n \n submitBtn.addEventListener('click', () => {\n if (selectedOption === null) return;\n \n if (selectedOption === correctOption) {\n feedbackCorrect.classList.remove('hidden');\n feedbackIncorrect.classList.add('hidden');\n } else {\n feedbackIncorrect.classList.remove('hidden');\n feedbackCorrect.classList.add('hidden');\n }\n });",
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},
{
"fragment_index": 1,
"text_description": "Question\nIf resistance doubles while voltage remains constant, what happens to the current?",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "1\nHalves",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "2\nDoubles",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "3\nStays the same",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "4\nBecomes zero",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Hint:\nRecall Ohm's law \\(I = \\frac{V}{R}\\).",
"image_description": ""
}
]
},
{
"slide": 9,
"fragments": [
{
"fragment_index": -1,
"text_description": "Key Takeaways",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Ohm’s Law\nVoltage equals current times resistance: \\(V = I R\\).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Voltage (V)\nActs as the push that drives charges through the circuit.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Current (I)\nMeasures how fast charge flows; more push gives more current.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Resistance (R)\nOpposes current; higher resistance means lower current for the same voltage.",
"image_description": ""
}
]
}
]