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Directions: For the regulator circuit shown below, various parameters areVz = 6.3 VIzT = 40 mARz = 2 ΩVs = 12 V to 18 VMinimum load current = 0 mAMinimum zener diode current = 1 mAMaximum zener diode current = 119 mAThe value of RS that limits the zener current Iz(max) (in Ω) will be (Answer up to one decimal place)Correct answer is '96.3'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared
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the Electronics and Communication Engineering (ECE) exam syllabus. Information about Directions: For the regulator circuit shown below, various parameters areVz = 6.3 VIzT = 40 mARz = 2 ΩVs = 12 V to 18 VMinimum load current = 0 mAMinimum zener diode current = 1 mAMaximum zener diode current = 119 mAThe value of RS that limits the zener current Iz(max) (in Ω) will be (Answer up to one decimal place)Correct answer is '96.3'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam.
Find important definitions, questions, meanings, examples, exercises and tests below for Directions: For the regulator circuit shown below, various parameters areVz = 6.3 VIzT = 40 mARz = 2 ΩVs = 12 V to 18 VMinimum load current = 0 mAMinimum zener diode current = 1 mAMaximum zener diode current = 119 mAThe value of RS that limits the zener current Iz(max) (in Ω) will be (Answer up to one decimal place)Correct answer is '96.3'. Can you explain this answer?.
Solutions for Directions: For the regulator circuit shown below, various parameters areVz = 6.3 VIzT = 40 mARz = 2 ΩVs = 12 V to 18 VMinimum load current = 0 mAMinimum zener diode current = 1 mAMaximum zener diode current = 119 mAThe value of RS that limits the zener current Iz(max) (in Ω) will be (Answer up to one decimal place)Correct answer is '96.3'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electronics and Communication Engineering (ECE).
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Here you can find the meaning of Directions: For the regulator circuit shown below, various parameters areVz = 6.3 VIzT = 40 mARz = 2 ΩVs = 12 V to 18 VMinimum load current = 0 mAMinimum zener diode current = 1 mAMaximum zener diode current = 119 mAThe value of RS that limits the zener current Iz(max) (in Ω) will be (Answer up to one decimal place)Correct answer is '96.3'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Directions: For the regulator circuit shown below, various parameters areVz = 6.3 VIzT = 40 mARz = 2 ΩVs = 12 V to 18 VMinimum load current = 0 mAMinimum zener diode current = 1 mAMaximum zener diode current = 119 mAThe value of RS that limits the zener current Iz(max) (in Ω) will be (Answer up to one decimal place)Correct answer is '96.3'. Can you explain this answer?, a detailed solution for Directions: For the regulator circuit shown below, various parameters areVz = 6.3 VIzT = 40 mARz = 2 ΩVs = 12 V to 18 VMinimum load current = 0 mAMinimum zener diode current = 1 mAMaximum zener diode current = 119 mAThe value of RS that limits the zener current Iz(max) (in Ω) will be (Answer up to one decimal place)Correct answer is '96.3'. Can you explain this answer? has been provided alongside types of Directions: For the regulator circuit shown below, various parameters areVz = 6.3 VIzT = 40 mARz = 2 ΩVs = 12 V to 18 VMinimum load current = 0 mAMinimum zener diode current = 1 mAMaximum zener diode current = 119 mAThe value of RS that limits the zener current Iz(max) (in Ω) will be (Answer up to one decimal place)Correct answer is '96.3'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Directions: For the regulator circuit shown below, various parameters areVz = 6.3 VIzT = 40 mARz = 2 ΩVs = 12 V to 18 VMinimum load current = 0 mAMinimum zener diode current = 1 mAMaximum zener diode current = 119 mAThe value of RS that limits the zener current Iz(max) (in Ω) will be (Answer up to one decimal place)Correct answer is '96.3'. Can you explain this answer? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.