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In a centre tapped full wave rectifier, RL=1KΩ and for diode Rf=10Ω. The primary voltage is 800sinωt with transformer turns ratio=2. The ripple factor will be _________​
  • a)
    54%
  • b)
    48%
  • c)
    26%
  • d)
    81%
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
In a centre tapped full wave rectifier, RL=1KΩ and for diode Rf=...
The ripple factor ϒ= [(IRMS/IAVG)2 – 1]1/2. IRMS =Im /√2=Vm/(Rf+RL)√2=200/1.01=198.
(Secondary line to line voltage is 800/2=400. Due to centre tap Vm=400/2=200)
IRMS=198/√2=140mA, IAVG=2*198/π=126mA. ϒ=[(140/126)2-1]1/2=0.48. So, ϒ=48%.
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In a centre tapped full wave rectifier, RL=1KΩ and for diode Rf=10Ω. The primary voltage is 800sinωt with transformer turns ratio=2. The ripple factor will be _________​ a)54%b)48%c)26%d)81%Correct answer is option 'B'. Can you explain this answer? for Electrical Engineering (EE) 2026 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about In a centre tapped full wave rectifier, RL=1KΩ and for diode Rf=10Ω. The primary voltage is 800sinωt with transformer turns ratio=2. The ripple factor will be _________​ a)54%b)48%c)26%d)81%Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a centre tapped full wave rectifier, RL=1KΩ and for diode Rf=10Ω. The primary voltage is 800sinωt with transformer turns ratio=2. The ripple factor will be _________​ a)54%b)48%c)26%d)81%Correct answer is option 'B'. Can you explain this answer?.
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