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An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given by:a)v = 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v = 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v = 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v = 622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given by:a)v = 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v = 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v = 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v = 622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given by:a)v = 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v = 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v = 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v = 622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer?.

Solutions for An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given by:a)v = 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v = 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v = 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v = 622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE). Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.

Here you can find the meaning of An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given by:a)v = 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v = 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v = 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v = 622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given by:a)v = 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v = 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v = 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v = 622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given by:a)v = 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v = 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v = 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v = 622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given by:a)v = 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v = 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v = 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v = 622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given by:a)v = 622 sin (314t) Voltsi = 7.07 cos (314t-90°) Ampsb)v = 311 sin (314t) Voltsi = 14 .14 sin (314t- 90°) Ampsc)v = 311 sin (314t)Voltsi =7.07 sin (314t - 90°) Ampsd)v = 622 sin (314t) Voltsi = 14.14 cos (314 t - 90°) AmpsCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Electrical Engineering (EE) tests.