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(B) The given circuit consists of two groups of capacitors connected in series between points $A$ and $B$.Each group consists of three identical capac

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$400$

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(B) The given circuit consists of two groups of capacitors connected in series between points $A$ and $B$.Each group consists of three identical capacitors connected in parallel.Let $C$ be the capacitance of each individual capacitor.For a parallel combination of three capacitors,the equivalent capacitance is $C_{eq} = C + C + C = 3C$.Since each capacitor can withstand a maximum of $200\, V$,the entire parallel group of three capacitors can also withstand a maximum of $200\, V$.The circuit effectively becomes two equivalent capacitors of $3C$ connected in series.Let $V_1$ and $V_2$ be the voltages across the two groups. Since the capacitors in each group are identical,the total voltage $V_{AB}$ is divided equally across the two groups.$V_{AB} = V_1 + V_2 = 200\, V + 200\, V = 400\, V$.Thus,the maximum voltage that can be safely applied between $A$ and $B$ is $400\, V$.

All six capacitors shown are identical. Each can withstand a maximum of $200\, V$ between its terminals. The maximum voltage that can be safely applied between $A$ and $B$ is.....$V$

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