Question

In the following figure the resultant capacitance between A and Bis $1\mu F$. The capacitance C is

• A. $\frac{32}{11}\mu F$
• B. $\frac{11}{32}\mu F$
• C. $\frac{23}{32}\mu F$
• D. $\frac{32}{23}\mu F$

Answer 1

Answer: (D)

$\frac{32}{23}\mu F$

Answer 2

Given network can be simplified as follows

Given that equivalent capacitance between A and B i.e., C_AB $= 1\mu F$

But $C_{AB} = \frac{C \times \frac{32}{9}}{C + \frac{32}{9}}$ hence $\frac{C \times \frac{32}{9}}{C + \frac{32}{9}} = 1 \Rightarrow C = \frac{32}{23}\mu F.$