Question

The capacity of each capacitor is $9\mu \text{F}$ in the following figure.Then the equivalent capacitance(in $\mu \text{F}$) across A and B is

• A. 12
• B. 12.00
• C. 12.0

Answer 1

Answer: (A), (B), (C)

After rearranging the circuit we get,
Considering the two series combinations and then a net parallel combination we get the required capacitance between A and B as,

$C_{\text{AB}}=\frac{2C\times C}{3C}+\frac{2C\times C}{3C}=\frac{4C}{3}$

$=\frac{4\times 9}{3}=12\mu \text{F}$