The equivalent capacitance between the points | vedclass

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(D) From the given circuit diagram, we can identify the arrangement of capacitors:$1$. Capacitors $C_2$ $(150 	ext{ F})$ and $C_3$ $(150 	ext{ F})$

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$60$

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(D) From the given circuit diagram, we can identify the arrangement of capacitors:$1$. Capacitors $C_2$ $(150 	ext{ F})$ and $C_3$ $(150 	ext{ F})$ are connected in series. Their equivalent capacitance $C_{23}$ is given by:$\frac{1}{C_{23}} = \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{150} + \frac{1}{150} = \frac{2}{150} = \frac{1}{75} \implies C_{23} = 75 	ext{ F}$.$2$. This combination $C_{23}$ is in parallel with capacitor $C_1$ $(75 	ext{ F})$. Their equivalent capacitance $C_{123}$ is:$C_{123} = C_1 + C_{23} = 75 + 75 = 150 	ext{ F}$.$3$. Finally, this combination $C_{123}$ is in series with capacitor $C_4$ $(100 	ext{ F})$ connected to terminals $A$ and $B$. The total equivalent capacitance $C_{AB}$ is:$\frac{1}{C_{AB}} = \frac{1}{C_{123}} + \frac{1}{C_4} = \frac{1}{150} + \frac{1}{100} = \frac{2 + 3}{300} = \frac{5}{300} = \frac{1}{60}$.Therefore, $C_{AB} = 60 	ext{ F}$.

The equivalent capacitance between the points $A$ and $B$ of the network shown in the figure is (in $\text{ F}$)

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