A) \[3\mu F\]
B) \[2\text{ }\mu F\]
C) \[4\text{ }\mu F\]
D) 8 µF
Correct Answer: C
Solution :
\[2\mu F\] and \[2\mu F\] capacitors are in parallel, \[\therefore \] \[C'=2+2=4\mu F\] \[4\mu F\] and C' are in series \[\therefore \] \[\frac{1}{C''}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\] \[C''=2\mu F\] \[4\mu F\] and \[4\mu F\] capacitors are in series, \[\frac{1}{C'''}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\] \[\Rightarrow \] s\[C'''=2\mu F\] Now \[C'\,'\] and \[C'\,'\,'\] are in parallel \[\therefore \] \[C=C'\,'+C'\,'\,'\] \[=2+2=4\mu F\]
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