
The resultant capacity between points $\mathrm{A}$ and $\mathrm{B}$ in the given circuit is

- $\mathrm{C}$
- $\frac{\mathrm{C}}{3}$
- $3 \mathrm{C}$
- $2 \mathrm{C}$
Solution
$\mathrm{C}_1$ and $\mathrm{C}_2$ are in parallel, their equivalent capacitance $\mathrm{C}_7=2 \mathrm{C}$.
$\mathrm{C}_7$ and $\mathrm{C}_3$ are in series, their equivalent capacitance is $\mathrm{C}_8=\mathrm{C}$.
$\mathrm{C}_8$ and $\mathrm{C}_4$ are in parallel, their equivalent capacitance is $\mathrm{C}_9=2 \mathrm{C}$.
$\mathrm{C}_9$ and $\mathrm{C}_5$ are in series, their equivalent capacitance $\mathrm{C}_{10}=\mathrm{C}$.
$\mathrm{C}_{10}$ and $\mathrm{C}_6$ are in parallel, their equivalent capacitance is $3 \mathrm{C}$
Hence equivalent capacitance between $\mathrm{A}$ and $\mathrm{B}$ is $3 \mathrm{C}$Asked in: MHT CET 2021 (21 Sep Shift 1)