question_answer

Minimum number of \[8\mu F\] and \[250\,\,V\] capacitors are used to make a combination of \[16\mu F\] 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,\[N=m\times n=8\times 4=32\]