In the circuit shown in the figure, the potential difference across the \(4 \mu\text{F}\) capacitor is

The equivalent capacitance of the parallel combination of \(9 \mu\text{F}\) and \(3 \mu\text{F}\) capacitors is:

\(C_p = 9 + 3 = 12 \mu\text{F}\)

The total equivalent capacitance of the circuit is:

\(C_{eq} = \dfrac{4 \times 12}{4 + 12} = \dfrac{48}{16} = 3 \mu\text{F}\)

The total charge supplied by the battery is:

\(Q = C_{eq} \times V = 3 \mu\text{F} \times 12 \text{V} = 36 \mu\text{C}\)

Since the \(4 \mu\text{F}\) capacitor is in series with the battery, the total charge \(Q\) flows through it.

The potential difference across the \(4 \mu\text{F}\) capacitor is:

\(V_4 = \dfrac{Q}{C_4} = \dfrac{36 \mu\text{C}}{4 \mu\text{F}} = 9 \text{V}\)

Answer: \(9\) V