Question

Challenge Question:
The minimum number of 8\(\mu\)F and 250 V capacitors required to make a combination of 16\(\mu\)F and 1000 V are

Answer 1

since you require 1000V, there should be atleast 4 capacitors in the series.
The capacitance of this combination becomes 8/4=2μF
Now since the capacitance of the combination should be 16μF there should be 8 rows in parallel.
total number of capacitors = 4*8=32
I don't know if that makes any sense but I hope it does!!

Answer 2

That's correct!
For others I am explaining it in detail.

Since we need 1000V the capacitors must be added in series. Let the number of capacitors needed be n then,

250n=1000
=> n=4

capacitance of each row will be 8/4=2\(\mu\)F
Now, we also need the capacitance be increased from 8\(\sf \mu\)F to 16\(\sf \mu\), m such rows must be added in parallel (in parallel connection of capacitors, capacitance increases). Let m such rows be added then,

m2\(\mu\)F=16
=> m=8

\(\therefore\) Total number of capacitors = 8\(\times\)4=32