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(B) The circuit consists of two parallel branches connected between points $A$ and $B$.In the upper branch,two capacitors of $20 \mu F$ each are conne

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$15$

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(B) The circuit consists of two parallel branches connected between points $A$ and $B$.In the upper branch,two capacitors of $20 \mu F$ each are connected in series. Their equivalent capacitance $C_1$ is given by:$\frac{1}{C_1} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10} \implies C_1 = 10 \mu F$.In the lower branch,two capacitors of $10 \mu F$ each are connected in series. Their equivalent capacitance $C_2$ is given by:$\frac{1}{C_2} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5} \implies C_2 = 5 \mu F$.Since the two branches are in parallel,the total effective capacitance $C_{	ext{eq}}$ is:$C_{	ext{eq}} = C_1 + C_2 = 10 \mu F + 5 \mu F = 15 \mu F$.

The effective capacitance between points $A$ and $B$ shown in the figure is (in $\mu F$)

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