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, \[N=m\times n=8\times 4=32\]You need to login to perform this action.
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