Physics - Capacitor Question with Solution | TestHub

The simplified circuit of the circuit given in question as follows:
The equivalent capacitance between C \& D capacitors of $2\mu \mathrm{F},5\mu \mathrm{F}$ and $5\mu \mathrm{F}$ are in parallel. $\therefore {\mathrm{C}}_{\mathrm{CD}}=2+5+5=12\mu \mathrm{F}(\because$ In parallel grouping ${\mathrm{C}}_{\mathrm{eq}}={\mathrm{C}}_1+{\mathrm{C}}_2+\dots .+{\mathrm{C}}_{\mathrm{n}}$ ) Similarly equivalent capacitance between $\mathrm{E}$ $\&{\mathrm{BC}}_{\mathrm{EB}}$ $=4+2=6\mu \mathrm{F}$ Now equivalent capacitance between A \& B $\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\frac{1}{6}+\frac{1}{12}+\frac{1}{6}=\frac{5}{12}$ $\Rightarrow {\mathrm{C}}_{\mathrm{eq}}=\frac{12}{5}=2.4\mu \mathrm{F}(\because$ In series grouping, $\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_2}+\dots \dots \dots +\frac{1}{{\mathrm{C}}_{\mathrm{n}}}$ )