From the circuit given below, the capacitance | vedclass

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(A) Let the nodes be labeled as follows: - Node $A$ is the input terminal. The top wire connects Node $A$ to the node between $C_2$ and $C_3$. - Let N

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$2$

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(A) Let the nodes be labeled as follows: - Node $A$ is the input terminal. The top wire connects Node $A$ to the node between $C_2$ and $C_3$. - Let Node $C$ be the node between $C_1$ and $C_2$. - Node $B$ is the output terminal. The bottom wire connects Node $C$, through $C_4$, to Node $B$. From the circuit diagram, we can identify the connections of each capacitor: - $C_1$ is connected between Node $A$ and Node $C$. - $C_2$ is connected between Node $C$ and Node $A$ (since the node between $C_2$ and $C_3$ is connected to $A$). - $C_3$ is connected between Node $A$ and Node $B$. - $C_4$ is connected between Node $C$ and Node $B$. Thus, $C_1$ and $C_2$ are in parallel between Node $A$ and Node $C$. Their equivalent capacitance is $C_{12} = C_1 + C_2 = 1\mu	ext{F} + 1\mu	ext{F} = 2\mu	ext{F}$. This combination $C_{12}$ is in series with $C_4$ (which is between Node $C$ and Node $B$). The equivalent capacitance of this branch is $C_{124} = \frac{C_{12} 	imes C_4}{C_{12} + C_4} = \frac{2 	imes 2}{2 + 2} = 1\mu	ext{F}$. Finally, this branch is in parallel with $C_3$ (which is connected directly between Node $A$ and Node $B$). The total equivalent capacitance between $A$ and $B$ is $C_{AB} = C_{124} + C_3 = 1\mu	ext{F} + 1\mu	ext{F} = 2\mu	ext{F}$. Therefore, the correct option is $(A)$.

From the circuit given below, the capacitance between terminals $A$ and $B$ is . . . . . . $\mu\text{F}$. (Take $C_1 = C_2 = C_3 = 1\mu\text{F}$ and $C_4 = 2\mu\text{F}$.)

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