2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 10 April 2016)

  • question_answer
    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 Main Online Paper (Held On 10 April 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 :

                 \[\frac{8\times 12}{8}=4\mu F\]              \[4\mu F+4\mu F=8\mu E\]              \[\frac{8\times 1}{8+1}=\frac{8}{9}\mu F\]              \[\frac{8\times 1}{8+11}=\frac{32}{12}=\frac{8}{3}\mu F\]              \[\frac{8}{3}+\frac{8}{9}=\frac{24+8}{9}=\frac{32}{9}\mu C\]              \[\frac{\frac{32}{9}\times C}{\frac{32}{9}+C}=1\Rightarrow \frac{32}{9}\times C=\frac{32}{9}+C\]              \[\Rightarrow C\left( \frac{32-9}{9} \right)=\frac{32}{9}\]              \[C=\frac{32}{23}\mu F\]                


You need to login to perform this action.
You will be redirected in 3 sec spinner