In the arrangement of the capacitors as shown in figure, each capacitor is of $6 \mu \mathrm{F}$, then…
In the arrangement of the capacitors as shown in figure, each capacitor is of $6 \mu \mathrm{F}$, then equivalent capacity between points $\mathrm{A}$ and $B$ is
$12 \mu \mathrm{F}$
$6 \mu \mathrm{F}$
$4 \mu \mathrm{F}$
$10 \mu \mathrm{F}$
Solution
$\mathrm{C}_1$ and $\mathrm{C}_3$ are in parallel.
Hence their equivalent capacitance $\mathrm{C}_5=2 \times 6=12 \mu \mathrm{F}$
$\mathrm{C}_5$ and $\mathrm{C}_2$ in series. Hence their equivalent capacitance $\mathrm{C}_6$ is given by
$\frac{1}{\mathrm{C}_6}=\frac{1}{12}+\frac{1}{6}=\frac{3}{12}=\frac{1}{4}$
$\therefore \mathrm{C}_6$ and $\mathrm{C}_4$ are in parallel.
Hence their equivalent capacitance is $\mathrm{C}=4+6=10 \mu \mathrm{F}$
