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As shown in the figure, an inductor of inductance 200 mH is connected to an AC source of emf 220 V and frequency 50 Hz. The instantaneous voltage of the source is 0 V when the peak value of current is √a/πA. The value of a is ___.(In Integers)Correct answer is '242'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about As shown in the figure, an inductor of inductance 200 mH is connected to an AC source of emf 220 V and frequency 50 Hz. The instantaneous voltage of the source is 0 V when the peak value of current is √a/πA. The value of a is ___.(In Integers)Correct answer is '242'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for As shown in the figure, an inductor of inductance 200 mH is connected to an AC source of emf 220 V and frequency 50 Hz. The instantaneous voltage of the source is 0 V when the peak value of current is √a/πA. The value of a is ___.(In Integers)Correct answer is '242'. Can you explain this answer?.

Solutions for As shown in the figure, an inductor of inductance 200 mH is connected to an AC source of emf 220 V and frequency 50 Hz. The instantaneous voltage of the source is 0 V when the peak value of current is √a/πA. The value of a is ___.(In Integers)Correct answer is '242'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

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