Correct Answer: Option D

Explanation

For capacitors in series

C\(_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2}\)

C\(_{\text{eq}} = \frac{5 \times 15}{5 + 15} = \frac{75}{20}\) = 3.75μF

Charge on the series branch

Q = Ceq × V =3.75 × 24 = 90 μC.

Potential difference across the 5 µF capacitor

V\(_{5} = \frac{Q}{C} = \frac{90}{5}\) = 18 V

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