Question

From a supply of identical capacitors rated `8 mu F, 250 V`, the minimum number of capacitors required to form a composite `16 mu F, 1000 V` is? (A) 2 (B) 4 (C) 16 (D) 32

Answer 1

32

Explanation:

Recall

`1/C_"eq"= 1/C_1+1/C_2+1/C_3...` in series

`C_"eq" = C_1+C_2+ C_3+...` in parallel

To have 1000 V tolerance, you need to stack up four 8`muF` capacitor in series. The equivalent capacitance of 4 such capacitors is:

`1/C_"eq"= 1/(8muF)+1/(8muF)+1/(8muF) + 1/(8muF) = 1/(2muF)`

`rArr C_(eq)= 2 muF`

To make a total of 16 muF that tolerates 1000V, you need 8 set of them arranged in parallel such that

`2muF+2muF+2muF+... =16muF`

`rArr` total number of capacitors = 8 x 4 = 32

Can you prove that this is the minimum number of them needed?