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 :
| \[12\mu F\]and \[6\mu F\]are in sense and again are in parallel with \[4\mu F\]. |
| 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 \[1\mu F\]. |
| Its equivalent capacitance \[=\frac{8\times 1}{8+1}=\frac{8}{9}\mu F\] ...(i) |
| Equivalent of \[8\mu F,2\mu F\] and \[2\mu F\] |
| \[=\frac{4\times 8}{4+8}=\frac{32}{12}=\frac{8}{3}\mu F\] .....(ii) |
| (i) and (ii) are in parallel and are in series with C |
| \[\therefore \,\,\frac{8}{9}+\frac{8}{3}=\frac{32}{9}\]and \[{{C}_{eq}}=1=\frac{\frac{32}{9}+C}{\frac{32}{9}+C}\Rightarrow \,C=\frac{32}{23}\mu F\] |
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