QJEEM16Ph88b9: JEE Main 2016 Physics - Class 12 JEEM16Ph88b9 Solution: Figure shows a network of capacitors where the numbers…

Equivalent capacitance of 6 $\mu$ F and 12 $\mu$ F is = ${{6 \times 12} \over {12 + 6}}$ = 4 $\mu$ F

Equivalent capacitance of 4 $\mu$ F and 4 $\mu$ F

is = 4 + 4 = 8 $\mu$ F

New circuit is $\to$



equivalent capacitance of 1 $\mu$ F and 8 $\mu$ F is

   = ${{1 \times 8} \over {8 + 1}}$ = ${8 \over 9}$ $\mu$ F

Equivalent capacitance of 8 $\mu$ F and 4 $\mu$ F is

   = ${{8 \times 4} \over {12}}$ = ${8 \over 3}$ $\mu$ F

New circuit is $\to$



Equation capacitance of ${8 \over 9}$ $\mu$ F and ${8 \over 3}$ $\mu$ F is

   = ${8 \over 9}$ + ${8 \over 3}$ = ${32 \over 9}$ $\mu$ F

$\therefore$    Equivalent capacitance of AB is

C_AB = ${{C \times {{32} \over 9}} \over {C + {{32} \over 9}}}$

Given that,

C_AB = 1 $\mu$ F

$\therefore$    1 = ${{C \times {{32} \over 9}} \over {C + {{32} \over 9}}}$

$\Rightarrow$    C + ${{{32} \over 9}}$ = ${{32C} \over 9}$

$\Rightarrow$     ${{23C} \over 9} = {{32} \over 9}$

$\Rightarrow$    C = ${{32} \over {23}}$ $\mu$ F