The equivalent capacity between terminal A and $B$ is
- $\frac{\mathrm{C}}{4}$
- $\frac{3 \mathrm{C}}{4}$
- $\frac{\mathrm{C}}{3}$
- $\frac{4 \mathrm{C}}{3}$
Solution
$\begin{aligned}
\frac{1}{C_5} & =\frac{1}{C}+\frac{1}{C}+\frac{1}{C} \\
\therefore \quad C_5 & =\frac{C}{3}
\end{aligned}$
Now, $\mathrm{C}_5$ and $\mathrm{C}$ are connected in parallel,
$\therefore \quad \mathrm{C}_{\text {net }}=\mathrm{C}_{\mathrm{s}}+\mathrm{C}=\frac{\mathrm{C}}{3}+\mathrm{C}=\frac{4 \mathrm{C}}{3}$Asked in: MHT CET 2023 (12 May Shift 1)