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(B) The circuit consists of four identical capacitors $C$ in series. The three capacitors between $A$ and $X$ can be replaced by an equivalent capacit

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$7.5$

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(B) The circuit consists of four identical capacitors $C$ in series. The three capacitors between $A$ and $X$ can be replaced by an equivalent capacitor $C_{eq1} = C/3$. The single capacitor between $X$ and $B$ is $C_{eq2} = C$.Since point $X$ is grounded,its potential $V_X = 0 \ V$.The total potential difference across the series combination is $10 \ V$. Since the battery is connected such that $A$ is at a higher potential than $B$,the potential difference is divided inversely proportional to the capacitance: $V_1 / V_2 = C_{eq2} / C_{eq1} = C / (C/3) = 3 / 1$.Given $V_1 + V_2 = 10 \ V$,we have $3V_2 + V_2 = 10 \ V \implies 4V_2 = 10 \ V \implies V_2 = 2.5 \ V$.Thus,$V_1 = 7.5 \ V$.Since $V_X = 0 \ V$ and $A$ is at a higher potential relative to $X$ (as the positive terminal of the battery is towards $A$),the potential at $A$ is $V_A = V_X + V_1 = 0 + 7.5 = 7.5 \ V$.

Four identical capacitors are connected in series with a battery of $emf$ $10 \ V$. Point $X$ is grounded. Find the potential at point $A$ in $V$.

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