In the given capacitive network the resultant capacitance between point $\mathrm{A}$ and $\mathrm{B}$ is
In the given capacitive network the resultant capacitance between point
$\mathrm{A}$ and $\mathrm{B}$ is
- $8 \mu \mathrm{F}$
- $4 \mu \mathrm{F}$
- $2 \mu \mathrm{F}$
- $16 \mu \mathrm{F}$
Solution
In the given circuit, $\mathrm{C}_3$ and $\mathrm{C}_4$ are in series and
$\begin{aligned}
\mathrm{C}_3 & =\mathrm{C}_4=8 \mu \mathrm{F} \\
\therefore \quad \frac{1}{\mathrm{C}_5} & =\frac{1}{\mathrm{C}_3}+\frac{1}{\mathrm{C}_4}
\end{aligned}$
$\begin{aligned}
\therefore \quad \mathrm{C}_{\mathrm{s}} & =\frac{\mathrm{C}_3^2}{2 \mathrm{C}_3} \\
\mathrm{C}_{\mathrm{s}} & =\frac{\mathrm{C}_3}{2} \\
\therefore \quad \mathrm{C}_{\mathrm{S}} & =4 \mu \mathrm{F}
\end{aligned}$
$\mathrm{C}_5$ and $\mathrm{C}_6$ are in parallel and $\mathrm{C}_5=\mathrm{C}_6=4 \mu \mathrm{F}$
$\begin{aligned}
\mathrm{C}_P & =\mathrm{C}_5+\mathrm{C}_6 \\
\therefore \quad \mathrm{C}_{\mathrm{P}} & =8 \mu \mathrm{F}
\end{aligned}$
$\therefore \quad$ Equivalent circuit is as shown in figure.
Now, $C_2$ and $C_P$ are in series and their combination in parallel with $\mathrm{C}_5$
$\begin{aligned}
\therefore \quad C_E & =\frac{C_2 C_P}{C_2 C_P}+C_5 \\
& C_E=\frac{(8)(8)}{16}+4 \\
\therefore \quad C_E & =8 \mu \mathrm{F}
\end{aligned}$
Now, $C_1$ and $C_E$ are in series,
$\begin{aligned}
& \therefore \quad \mathrm{C}=\frac{\mathrm{C}_1 \mathrm{C}_{\mathrm{E}}}{\mathrm{C}_1+\mathrm{C}_{\mathrm{E}}} \\
& \therefore \quad \mathrm{C}=\frac{(8)(8)}{16} \\
& \therefore \quad \mathrm{C}=4 \mu \mathrm{F}
\end{aligned}$
Asked in: MHT CET 2023 (12 May Shift 1)
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