The capacities and connection of five capacit | vedclass

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(D) $1$. By analyzing the circuit,we can see that the $10\;\mu F$,$5\;\mu F$,and $9\;\mu F$ capacitors are connected in parallel.$2$. The equivalent c

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$4\;\mu F;\;50\;\mu C$

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(D) $1$. By analyzing the circuit,we can see that the $10\;\mu F$,$5\;\mu F$,and $9\;\mu F$ capacitors are connected in parallel.$2$. The equivalent capacitance of this parallel combination is $C_p = 10 + 5 + 9 = 24\;\mu F$.$3$. Now,the circuit simplifies to three capacitors in series: $12\;\mu F$,$24\;\mu F$,and $8\;\mu F$.$4$. The equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{12} + \frac{1}{24} + \frac{1}{8} = \frac{2 + 1 + 3}{24} = \frac{6}{24} = \frac{1}{4}$. Thus,$C_{eq} = 4\;\mu F$.$5$. The total charge $Q$ flowing through the circuit is $Q = C_{eq} 	imes V = 4\;\mu F 	imes 60\;V = 240\;\mu C$.$6$. This charge $Q$ passes through the $12\;\mu F$ and $8\;\mu F$ capacitors,and splits among the parallel combination.$7$. The charge $Q'$ on the $5\;\mu F$ capacitor is determined by the ratio of its capacitance to the total parallel capacitance: $Q' = Q 	imes \frac{5}{10 + 5 + 9} = 240 	imes \frac{5}{24} = 50\;\mu C$.

The capacities and connection of five capacitors are shown in the adjoining figure. The potential difference between the points $A$ and $B$ is $60\;V$. Then the equivalent capacity between $A$ and $B$ and the charge on $5\;\mu F$ capacitor will be respectively:

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