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  • 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

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  1. taramgrant0543664
    • one year ago
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    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
    • one year ago
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    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

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