16:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 17:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 15:T502,{"@context":"https://schema.org","@type":"Quiz","name":"Ap Eapcet 2024 Capacitor PYQ with Solution","url":"https://pyq-test-series-platform-alsjdcxu2-sameers-projects-41c5a71e.vercel.app/ap-eapcet/questions/2024/physics/capacitor/m8svz5uj","educationalLevel":"Entrance exam preparation","about":"Ap Eapcet / Physics / Capacitor","hasPart":{"@type":"Question","name":"Ap Eapcet 2024 Capacitor PYQ with Solution","text":"Three parallel plate capacitors of capacitances 4 \\mu \\mathrm{~F}, 6 \\mu \\mathrm{~F} and 12 \\mu \\mathrm{~F} are first connected in series and then in parallel. The ratio of the effective capacitances in the two cases is","eduQuestionType":"mcq","acceptedAnswer":{"@type":"Answer","text":"A"},"comment":{"@type":"Comment","text":"Given the capacitances: C_1 = 4 \\, \\mu \\mathrm{F} C_2 = 6 \\, \\mu \\mathrm{F} C_3 = 12 \\, \\mu \\mathrm{F} When connected in series: To find the equivalent capacitance for capacitors in series, we use the formula: C_{\\text{eq}} = \\frac{C_1 C_2 C_3}{C_1 C_2 + C_2 C_3 + C_3 C_1} Substituting the given values: C_{\\text{eq}} = \\frac{4 \\times 6 \\times 12}{4 \\times 6 + 6 \\times 12 + 12 \\times 4} = \\frac{288}{144} = 2 \\, \\mu \\mathrm{F} When connected in parallel: For capacitors in parallel, the equivalent..."}}}18:T42f,

Given the capacitances:

$ C_1 = 4 \, \mu \mathrm{F} $

$ C_2 = 6 \, \mu \mathrm{F} $

$ C_3 = 12 \, \mu \mathrm{F} $

When connected in series:

To find the equivalent capacitance for

Capacitor

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