Three capacitors of values 6, mu F, 3, mu F a | vedclass

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(B) $1$. The capacitors $6\, \mu F$ and $3\, \mu F$ are connected in parallel. Their equivalent capacitance is $C_p = 6\, \mu F + 3\, \mu F = 9\, \mu

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$5$

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(B) $1$. The capacitors $6\, \mu F$ and $3\, \mu F$ are connected in parallel. Their equivalent capacitance is $C_p = 6\, \mu F + 3\, \mu F = 9\, \mu F$.$2$. Now,this equivalent capacitor $C_p = 9\, \mu F$ is in series with the $9\, \mu F$ capacitor.$3$. The total capacitance $C_{eq}$ of the circuit is given by $\frac{1}{C_{eq}} = \frac{1}{9} + \frac{1}{9} = \frac{2}{9}$,so $C_{eq} = 4.5\, \mu F$.$4$. The total charge $Q$ supplied by the battery is $Q = C_{eq} 	imes V = 4.5\, \mu F 	imes 10\, V = 45\, \mu C$.$5$. Since the $9\, \mu F$ capacitor is in series with the parallel combination,the same charge $Q = 45\, \mu C$ flows through it.$6$. The potential difference $V_9$ across the $9\, \mu F$ capacitor is $V_9 = \frac{Q}{C} = \frac{45\, \mu C}{9\, \mu F} = 5\, V$.

Three capacitors of values $6\, \mu F, 3\, \mu F$ and $9\, \mu F$ are connected as shown in the figure. This combination is connected to a $10\, V$ battery. What is the potential difference across the plates of the $9\, \mu F$ capacitor in $V$?

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