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All capacitors used in the diagram are identical and each is of capacitance C. Then the effective capacitance between the point $A$ and $B$ is
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Verified Answer
The correct answer is:
$1.5 \mathrm{C}$
The equivalent diagram of the given circuit is
$\mathrm{C}_{1}, \mathrm{C}_{2}$ and $\mathrm{C}_{3}$ are in parallel
$$
\begin{aligned}
\therefore \quad \mathrm{C}_{123} &=\mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3} \\
&=\mathrm{C}+\mathrm{C}+\mathrm{C}=3 \mathrm{C}
\end{aligned}
$$
Similarly, $\quad C_{456}=3 C$
$\mathrm{C}_{123}$ and $\mathrm{C}_{456}$ are in series
$$
\therefore \quad C_{\text {eq }}=\frac{(3 C)(3 C)}{3 C+3 C}=1.5 C
$$
$\mathrm{C}_{1}, \mathrm{C}_{2}$ and $\mathrm{C}_{3}$ are in parallel
$$
\begin{aligned}
\therefore \quad \mathrm{C}_{123} &=\mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3} \\
&=\mathrm{C}+\mathrm{C}+\mathrm{C}=3 \mathrm{C}
\end{aligned}
$$
Similarly, $\quad C_{456}=3 C$
$\mathrm{C}_{123}$ and $\mathrm{C}_{456}$ are in series
$$
\therefore \quad C_{\text {eq }}=\frac{(3 C)(3 C)}{3 C+3 C}=1.5 C
$$
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