Minimum number of capacitors each of 8μF and 250V used to make a composite capacitor of 16μF and 1000V are
- a)8
- b)32
- c)16
- d)24
Correct answer is option 'B'. Can you explain this answer?
Related Test
Answers
|
Shiksha Academy
Feb 11, 2022 |
|
Minimum number of capacitors in each row = 1000/250 = 4
Therefore, 4 capacitors connected in series.
If there are m such rows, that total capacity
= m × 2 = 16 ∴ m = 16/28
∴ minimum number of capacitors = 4 × 8 = 32
Therefore, 4 capacitors connected in series.
If there are m such rows, that total capacity
= m × 2 = 16 ∴ m = 16/28
∴ minimum number of capacitors = 4 × 8 = 32
Minimum number of capacitors in each row = 1000/250 = 4Therefore, 4 capacitors connected in series.If there are m such rows, that total capacity= m × 2 = 16 ∴ m = 16/28∴ minimum number of capacitors = 4 × 8 = 32
Top Courses for NEET
Related Content
Do you know? How Toppers prepare for NEET Exam
With help of the best NEET teachers & toppers, We have prepared a guide for student who are
preparing for NEET : 15 Steps to clear NEET Exam
Download free EduRev App
Track your progress, build streaks, highlight & save important lessons and more!
Minimum number of capacitors in each row = 1000/250 = 4Therefore, 4 capacitors connected in series.If there are m such rows, that total capacity= m × 2 = 16 ∴ m = 16/28∴ minimum number of capacitors = 4 × 8 = 32
|
© EduRev
|
Follow Us
|
