In the following diagram, the charge and potential difference across $8\,\mu F$ capacitance will be nearly
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equivalent circuit between$\mathrm{A}$ and $\mathrm{B}$

potential at $8\, \mu \mathrm{F}=\mathrm{V}_{\mathrm{AC}}$

$\mathrm{V} \propto \frac{1}{\mathrm{C}}$

$\mathrm{V}_{\mathrm{AC}}=\frac{36}{36+18} \times 40$

$\mathrm{V}_{\mathrm{AC}}=\frac{36}{54} \times 40$

$\mathrm{V}_{\mathrm{AC}} \approx 27 \mathrm{\,V}$

charge at $8 \mu \mathrm{F}$ capacitor $=27 \times 8\, \mu \mathrm{F}$

$=214\, \mu C$

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