question_answer

To form a composite \[16\mu F,\ 1000V\] capacitor from a supply of identical capacitors marked \[8\mu F,\ 250V\], we require a minimum number of capacitors                                                                       [MP PET 1996; AIIMS 2000]

A)                    40                                      

B)            32

C)                    8                                        

D)            2

Correct Answer: B

Solution :

           Suppose \[C=\text{ 8}\mu F,C'=\text{ 16}\mu F\]                    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 ´ m = 4 ´ 8 = 32 Short Trick : For such type of problems number of capacitors = \[\frac{C'}{C}\times {{\left( \frac{V'}{V} \right)}^{2}}=\frac{16}{8}{{\left( \frac{1000}{250} \right)}^{2}}=32\]