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:T525,{"@context":"https://schema.org","@type":"Quiz","name":"Mht Cet 2025 Alternating Current PYQ with Solution","url":"https://pyq-test-series-platform-alsjdcxu2-sameers-projects-41c5a71e.vercel.app/mht-cet/questions/2025/physics/alternating-current/mf18er8u","educationalLevel":"Entrance exam preparation","about":"Mht Cet / Physics / Alternating Current","hasPart":{"@type":"Question","name":"Mht Cet 2025 Alternating Current PYQ with Solution","text":"An alternating e.m.f. is given by e=e_0 \\sin \\omega t. In how much time the e.m.f. will have half its maximum value, if e starts from zero ? \\left(\\mathrm{T}=\\text { Time Period, } \\sin 30^{\\circ}=\\frac{1}{2}\\right)","eduQuestionType":"mcq","acceptedAnswer":{"@type":"Answer","text":"C"},"comment":{"@type":"Comment","text":"We are asked: An alternating emf is given by e = e_0 \\sin (\\omega t). We want the time when emf becomes half its maximum value for the first time starting from zero . Step 1: Condition for half maximum. Maximum value = e_0. Half of that = \\tfrac{e_0}{2}. So: e_0 \\sin (\\omega t) = \\tfrac{e_0}{2} \\quad \\Rightarrow \\quad \\sin (\\omega t) = \\tfrac{1}{2}. Step 2: Solve for \\omega t. \\sin (\\omega t) = \\tfrac{1}{2} \\quad \\Rightarrow \\quad \\omega t = \\frac{\\pi}{6}, \\, \\frac{5\\pi}{6}, \\dots The first tim..."}}}18:T450,

We are asked:

An alternating emf is given by

$ e = e_0 \sin (\omega t). $

We want the time when emf becomes half its maximum value for the first time starting from zero

Alternating Current

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