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Q.

To form a composite 16μF,1000V capacitor from a supply of identical capacitors marked 8 μF and 250 V , we require a minimum number of capacitors

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a

40

b

32

c

2

d

8

answer is B.

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Detailed Solution

Suppose C = 8μF, C' = 16μF and V = 250 V, V' = 1000V                  

Suppose m rows of given capacitors are connected in parallel and each row contains n capacitors then potential difference across each capacitor 

V = \frac{{V'}}{n}

 and equivalent capacitance of network 

C' = \frac{{mC}}{n}

  on putting the values we get n = 4 and m = 8

 

 Total capacitors = n x m = 4 x 8 = 32

Short Trick : For such type of problems number of capacitors

= \frac{{C'}}{C} \times {\left( {\frac{{V'}}{V}} \right)^2} = \frac{{16}}{8}{\left( {\frac{{1000}}{{250}}} \right)^2} = 32

  

     

      

      

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