In the following diagram, the charge and potential difference across $8\, \mu F$ capacitance will be respectively
Diffcult
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(c) Given circuit can be redrawn as follows capacitors, $9\, \mu F, 9\, \mu F$ and $7 \,\mu F$ are short circuited. So they are deleted.
${V_1} + {V_2} = 40\,V$
and $\frac{{{V_1}}}{{{V_2}}} = \frac{{36}}{{18}} = 2$
Hence ${V_1} = \frac{{80}}{3}\,V$
${\rm{and }}\,{V_2} = \frac{{40}}{3}\,V$
Charge on $8 \,\mu F$ capacitor $ = 8 \times \frac{{80}}{3} = 213.3\,\mu F \approx 214\,\mu F$
${V_1} + {V_2} = 40\,V$
and $\frac{{{V_1}}}{{{V_2}}} = \frac{{36}}{{18}} = 2$
Hence ${V_1} = \frac{{80}}{3}\,V$
${\rm{and }}\,{V_2} = \frac{{40}}{3}\,V$
Charge on $8 \,\mu F$ capacitor $ = 8 \times \frac{{80}}{3} = 213.3\,\mu F \approx 214\,\mu F$
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