| Figure shows a network of capacitors where the number indicates capacitances in micro Farad. The value of capacitance C if the equivalent capacitance between point A and B is to be \[1\mu F\] is: [JEE ONLINE 10-04-2016] |
 |
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 :
| [c] \[\frac{8\times 12}{8}=4\mu F\] |
| \[4\mu F+4\mu F=8\mu E\] |
| \[\frac{8\times 1}{8+1}=\frac{8}{9}\mu F\] |
| \[\frac{8\times 1}{8+11}=\frac{32}{12}=\frac{8}{3}\mu F\] |
| \[\frac{8}{3}+\frac{8}{9}=\frac{24+8}{9}=\frac{32}{9}\mu C\] |
| \[\frac{\frac{32}{9}\times C}{\frac{32}{9}+C}=1\Rightarrow \frac{32}{9}\times C=\frac{32}{9}+C\] |
| \[\Rightarrow C\left( \frac{32-9}{9} \right)=\frac{32}{9}\] |
| \[C=\frac{32}{23}\mu F\] |
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