In the circuit shown, calculate the potential drop across the capacitor

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Solution

When capacitor is fully charged, no currents draws from the source. So we can remove the capacitor and the circuit becomes open as shown in figure.
Applying KVL , $18 - 9 = 4 I + 2 I$ or $I = 9 / 6 = 3 / 2 A$
here, $V_{A} - V_{B} = 9 + 2 I = 9 + 2 (3 / 2) = 12 V$
The potential across capacitance $12 - 9 = 3 V$

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