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?

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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

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
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