In the steady state,what is the potential dif | vedclass

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(D) In the steady state,no current flows through the capacitors. Therefore,the resistors in series with the capacitor network do not affect the potent

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$25$

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(D) In the steady state,no current flows through the capacitors. Therefore,the resistors in series with the capacitor network do not affect the potential distribution across the capacitors.The total potential difference across the capacitor network between points $A$ and $C$ is $100 \ V$.From the simplified circuit diagram (as shown in the solution image),the capacitor branch $AB$ has a capacitance of $C_1 = 6 \ \mu F$ and the branch $BC$ has a capacitance of $C_2 = 2 \ \mu F$.Since the capacitors are in series,the potential difference across each capacitor is inversely proportional to its capacitance: $V \propto \frac{1}{C}$.Thus,$\frac{V_{AB}}{V_{BC}} = \frac{C_2}{C_1} = \frac{2 \ \mu F}{6 \ \mu F} = \frac{1}{3}$.This implies $V_{BC} = 3 V_{AB}$.Since $V_{AC} = V_{AB} + V_{BC} = 100 \ V$,we have $V_{AB} + 3 V_{AB} = 100 \ V$.$4 V_{AB} = 100 \ V$,which gives $V_{AB} = 25 \ V$.

In the steady state,what is the potential difference between points $A$ and $B$ in volts? (All capacitors are in $\mu F$.)

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