Moderate
In a push - pull converter, the filter capacitor can be obtained as____________?
Correct answer: C. Cmin= ( 1 - 2 D ) V / 32 ( Vr L f2 )
- A. Cmin = V / ( Vr L f2 )
- B. Cmin= ( 1 - D ) V / ( Vr L f2 )
- C. Cmin= ( 1 - 2 D ) V / 32 ( Vr L f2 )
- D. Cmin= ( 1 - 2 D ) V / 42 ( Vr L f2 )
Explanation
In a push-pull converter, the effective ripple frequency is doubled, and the capacitor ripple relation introduces the triangular-current factor, giving Cmin = (1 − 2D)V/(32VrLf²). Thus option c matches the standard expression, assuming the printed fractions are formatted correctly.
Last updated
Practise Power Electronics
253 free Power Electronics MCQs from Electrical Engineering, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Electrical Engineering questions like this
Electrical Engineering is on 2 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.