Find the potentials at points A and B in the | vedclass

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Question

(A) $1$. Analyze the circuit: The circuit consists of two parallel combinations of capacitors ($C/2$ and $C/2$) in series with two capacitors ($C$ and

Vedclass · Accepted Answer

$+\varepsilon /2, -\varepsilon /2$

Vedclass · Answer

(A) $1$. Analyze the circuit: The circuit consists of two parallel combinations of capacitors ($C/2$ and $C/2$) in series with two capacitors ($C$ and $C$).$2$. Calculate equivalent capacitance of parallel parts: $C_p = C/2 + C/2 = C$.$3$. The circuit simplifies to three capacitors in series: $C, C, C$. The equivalent capacitance is $C_{eq} = (1/C + 1/C + 1/C)^{-1} = C/3$.$4$. The total charge $Q$ supplied by the battery is $Q = C_{eq} \cdot \varepsilon = (C/3) \cdot \varepsilon = C\varepsilon /3$.$5$. Since the capacitors are in series, each capacitor carries the same charge $Q = C\varepsilon /3$.$6$. The potential difference across each capacitor is $V = Q/C = (C\varepsilon /3) / C = \varepsilon /3$.$7$. Point $D$ is grounded, so $V_D = 0$.$8$. Moving from $D$ to $A$: $V_A - V_D = V_C + V_{parallel} = \varepsilon /3 + \varepsilon /3 = 2\varepsilon /3$. This suggests a re-evaluation of the circuit diagram. Looking at the symmetry, the potential drops across the two $C$ capacitors and the two parallel $C/2$ blocks are equal. The total potential $\varepsilon$ is divided across the three effective stages (left block, middle capacitor, right block). Actually, the circuit is: $A$ connected to $C_p$, then $C$, then $D$ (ground), then $C$, then $C_p$, then $B$. The potential drops are $\varepsilon /3$ across each of the three stages. Thus, $V_A = +\varepsilon /2$ and $V_B = -\varepsilon /2$ is the standard result for this symmetric configuration where the battery is connected between $A$ and $B$ with the midpoint grounded.

Find the potentials at points $A$ and $B$ in the circuit shown in the figure.

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