question_answer

Minimum number of 8|nF and 250 V capacitors are used to make a combination of \[\text{16 }\!\!\mu\!\!\text{ F}\]and 1000 V are

A)  4              

B)  32

C)  8              

D)  3

Correct Answer: B

Solution :

 Let m rows of n series capacitor be taken then  minimum number of capacitors required is \[N=m\times n\] Also effective voltage is              \[V=1000=n\times 250\] \[\Rightarrow \] \[n=\frac{1000}{250}=4\] Also these four capacitors are connected in series then effective capacitance is \[\frac{1}{C}=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{4}{8}\] \[\Rightarrow \] \[C=2\,\mu F\] \[\therefore \] \[C=16=2\times m\] \[\Rightarrow \] \[m=\frac{16}{2}=8\] Hence, \[N=m\times n=8\times 4=32\]