A) \[\frac{32}{11}\mu F\]
B) \[\frac{11}{32}\mu F\]
C) \[\frac{23}{32}\mu F\]
D) \[\frac{32}{23}\mu F\]
Correct Answer: D
Solution :
12 mF and 6mF are in series and again are in parallel with 4mF. Therefore, resultant of these three will be \[=\frac{12\times 6}{12+6}+4=4+4=8\mu F\] This equivalent system is in series with 1 mF. Its equivalent capacitance \[=\frac{8\times 1}{8+1}=\frac{8}{9}\mu F\] ....(i) Equivalent of 8mF, 2mF and 2mF \[=\frac{4\times 8}{4+8}=\frac{32}{12}=\frac{8}{3}\mu F\] .....(ii) (i) and (ii) are in parallel and are in series with C \ \[\frac{8}{9}+\frac{8}{3}=\frac{32}{9}\] and \[{{C}_{eq}}=1=\frac{\frac{32}{9}\times C}{\frac{32}{9}+C}\]Þ \[C=\frac{32}{23}\mu F\]
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