Physics - Capacitor Question with Solution | TestHub

$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}$ $C=2\mu F$ $2\mu \mathrm{F}$ will be parallel to $4\mu \mathrm{F}$. so the resultant capacitance$\begin{array}{rl}\frac{1}{C^{\prime \prime }}& =\frac{1}{4\mu F}+\frac{1}{4\mu F}=\frac{2}{4}\mu F\\ C^{-}& =2\mu F\\ C_{AQ}& =2\mu F+6\mu F\\ & =8\mu F\end{array}$