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(C) By analyzing the circuit,we can see that the first three capacitors are connected in parallel between the input and point $P$. Each capacitor has

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$6$

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(C) By analyzing the circuit,we can see that the first three capacitors are connected in parallel between the input and point $P$. Each capacitor has a capacitance of $8 \mu F$. Therefore,the equivalent capacitance of these three parallel capacitors is $C_p = 8 \mu F + 8 \mu F + 8 \mu F = 24 \mu F$. This combination is then in series with the final $8 \mu F$ capacitor connected to point $B$. Thus,the total equivalent capacitance $C_{eq}$ is given by: $C_{eq} = \frac{C_p 	imes 8 \mu F}{C_p + 8 \mu F} = \frac{24 \mu F 	imes 8 \mu F}{24 \mu F + 8 \mu F} = \frac{192}{32} \mu F = 6 \mu F$.

The equivalent capacitance between points $A$ and $B$ in the figure shown below is .......... $\mu F$.

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