From the circuit given below, the capacitance | vedclass

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(D) Let the potential at $A$ be $V_A$ and at $B$ be $V_B$. By analyzing the circuit, we see that the wire connecting the node between $C_1$ and $C_2$

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$5/2$

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(D) Let the potential at $A$ be $V_A$ and at $B$ be $V_B$. By analyzing the circuit, we see that the wire connecting the node between $C_1$ and $C_2$ to the node between $C_2$ and $C_3$ effectively short-circuits $C_2$. Thus, the circuit simplifies to $C_1$ and $C_3$ in series, which are in parallel with $C_4$. However, looking closely at the diagram, $C_1$ and $C_2$ are in series, and this combination is in parallel with $C_4$. Actually, the circuit shows $C_1$ and $C_2$ in series, and this combination is in parallel with $C_4$. The capacitor $C_3$ is in series with the parallel combination of $(C_1, C_2)$ and $C_4$. Let $C_{12} = \frac{C_1 C_2}{C_1 + C_2} = \frac{1 	imes 1}{1 + 1} = 0.5	ext{ }\mu	ext{F}$. Now, $C_{12}$ is in parallel with $C_4$, so $C_{p} = C_{12} + C_4 = 0.5 + 2 = 2.5	ext{ }\mu	ext{F}$. Finally, $C_p$ is in series with $C_3$, so $C_{eq} = \frac{C_p C_3}{C_p + C_3} = \frac{2.5 	imes 1}{2.5 + 1} = \frac{2.5}{3.5} = \frac{25}{35} = \frac{5}{7}	ext{ }\mu	ext{F}$. Wait, re-evaluating the diagram: The wire connects the node after $C_1$ to the node after $C_2$. This means $C_2$ is shorted. The circuit is $C_1$ in series with $C_3$, and $C_4$ is in parallel with the combination of $C_1$ and $C_3$. $C_{13} = \frac{1 	imes 1}{1 + 1} = 0.5	ext{ }\mu	ext{F}$. $C_{eq} = C_{13} + C_4 = 0.5 + 2 = 2.5 = 5/2	ext{ }\mu	ext{F}$.

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

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