A) \[\frac{33}{23}\mu F\]
B) \[\frac{31}{23}\mu F\]
C) \[\frac{32}{23}\mu F\]
D) \[\frac{34}{23}\mu F\]
Correct Answer: C
Solution :
\[\frac{8\times 12}{8}=4\mu F\] \[4\mu F+4\mu F=8\mu E\] \[\frac{8\times 1}{8+1}=\frac{8}{9}\mu F\] \[\frac{8\times 1}{8+11}=\frac{32}{12}=\frac{8}{3}\mu F\] \[\frac{8}{3}+\frac{8}{9}=\frac{24+8}{9}=\frac{32}{9}\mu C\] \[\frac{\frac{32}{9}\times C}{\frac{32}{9}+C}=1\Rightarrow \frac{32}{9}\times C=\frac{32}{9}+C\] \[\Rightarrow C\left( \frac{32-9}{9} \right)=\frac{32}{9}\] \[C=\frac{32}{23}\mu F\]You need to login to perform this action.
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