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Eshan
2 years ago
Dear student,

Since charge oncapacitor is$\dpi{80} 24\mu C$, the voltage across it is$\dpi{80} 12V$. Since it is in parallel to capapcitor C, C also has 12V across it.

Since the battery provides 18V, the voltage on$\dpi{80} 6\mu F$capacitor is$\dpi{80} 18V-12V=6V$.
Hence the charge stored on it is$\dpi{80} 36\mu C$. This current must have divided into the two capacitors in series. Hence charge stored on C is$\dpi{80} 36\mu C-24\mu C=12\mu C$. But potential difference across it is$\dpi{80} 12V$. Hence the capacitance C is$\dpi{80} \dfrac{12\mu C}{12V}=1\mu F$