## Abhisar one year ago 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

1. taramgrant0543664

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!!

2. Abhisar

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