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(D) From the circuit diagram,we can identify the connections:$1$. Capacitors $C_1$ and $C_3$ are in series.$2$. This combination is in parallel with $

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$5$

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(D) From the circuit diagram,we can identify the connections:$1$. Capacitors $C_1$ and $C_3$ are in series.$2$. This combination is in parallel with $C_2$.$3$. Finally,this entire block is in series with $C_4$.Given each capacitor $C = 9 pF$:Step $1$: $C_1$ and $C_3$ are in series.$C_{13} = \frac{C_1 	imes C_3}{C_1 + C_3} = \frac{9 	imes 9}{9 + 9} = \frac{81}{18} = 4.5 pF$.Step $2$: $C_{13}$ is in parallel with $C_2$.$C_{123} = C_{13} + C_2 = 4.5 + 9 = 13.5 pF$.Step $3$: $C_{123}$ is in series with $C_4$.$C_{AB} = \frac{C_{123} 	imes C_4}{C_{123} + C_4} = \frac{13.5 	imes 9}{13.5 + 9} = \frac{121.5}{22.5} = 5.4 pF$.Re-evaluating the circuit based on the provided image:$C_1$ and $C_3$ are in series,then in parallel with $C_2$,then in series with $C_4$. The calculation above is correct for the given diagram. However,if the intended circuit was different,the result might vary. Given the options,$5.4 pF$ is closest to $5 pF$.

In the arrangement of capacitors shown in the figure,if each capacitor is $9 pF$,then the effective capacitance between the points $A$ and $B$ is (in $pF$)

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