The equivalent capacitance between $A$ and $B$ in the given circuit is

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As we know that, Series equivalent capacitance, $ \frac{1}{C_{\mathrm{cq}}}=\frac{1}{C_a}+\frac{1}{C_b}+\frac{1}{C_c}+\ldots $ and parallel equivalent capacitance, $ C_{\mathrm{cq}}=C_a+C_b+C_c+\ldots $ Now, since $C_4$ and $C_5$ are in parallel. So, their equivalent capacitance, $ \therefore \quad C_a=2+1=3 \mu \mathrm{F} $ Now $C_a, C_3$ and $C_6$ are in series will give equivalent capacitance, $ \begin{aligned} & \therefore \\ & \frac{1}{C_b}=\frac{1}{C_a}+\frac{1}{C_3}+\frac{1}{C_6}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{3}{3}=1 \\ & \Rightarrow \quad C_b=1 \mu \mathrm{F} \\ & \end{aligned} $ Again $C_C$ and $C_8$ in parallel will give, $ \begin{aligned} C_c & =C_b+C_8 \\ & =1+2=3 \mu \mathrm{F} \end{aligned} $ Again $C_6, C_2$ and $C_7$ are in series will give, $ \begin{aligned} & \therefore \quad \frac{1}{C_d}=\frac{1}{C_2}+\frac{1}{C_7}+\frac{1}{C_c} \\ & =\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{3}{3} \\ & \Rightarrow \quad C_d=1 \mu \mathrm{F} \\ & \end{aligned} $ Again $C_d$ is in parallel with $C_9$, and give, $ \begin{aligned} C_f & =C_d+C_9 \\ & =1+2=3 \mu \mathrm{F} \end{aligned} $ and finally $C_f$ is in series with $C_1$ and $C_{10}$ and give $ \begin{aligned} & \frac{1}{C_{\mathrm{cq}}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{3}{3}=1 \\ & \Rightarrow \quad C_{\mathrm{cq}}=1 \mu \mathrm{F} \\ & \end{aligned} $

The equivalent capacitance between $A$ and $B$ in the given circuit is

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