Equivalent capacitance between $A$ and $B$ in the figure is

• A

  $20 \mu \mathrm{F}$
• Correct answer
  B

  $8 \mu \mathrm{F}$
• C

  $12 \mu \mathrm{F}$
• D

  $16 \mu \mathrm{F}$

Solution

$\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}$ $C=2 \mu F$ $2 \mu \mathrm{F}$ will be parallel to $4 \mu \mathrm{F}$. so the resultant capacitance $\begin{aligned} \frac{1}{C^{\prime \prime}} &=\frac{1}{4 \mu F}+\frac{1}{4 \mu F}=\frac{2}{4} \mu F \\ C^{-} &=2 \mu F \\ C_{A Q} &=2 \mu F+6 \mu F \\ &=8 \mu F \end{aligned}$

Hence, the correct answer is option: 'B'.