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(A) From the circuit diagram,capacitors $B$ and $C$ are in parallel,and their equivalent capacitance is $C_{BC} = C_B + C_C = 8 \mu	ext{F} + 12 \mu

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$5: 2: 3$

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(A) From the circuit diagram,capacitors $B$ and $C$ are in parallel,and their equivalent capacitance is $C_{BC} = C_B + C_C = 8 \mu	ext{F} + 12 \mu	ext{F} = 20 \mu	ext{F}$.This combination is in series with capacitor $A$ $(5 \mu	ext{F})$.The equivalent capacitance of the circuit is $C_{eq} = \frac{C_A 	imes C_{BC}}{C_A + C_{BC}} = \frac{5 	imes 20}{5 + 20} = \frac{100}{25} = 4 \mu	ext{F}$.The total charge drawn from the battery is $q_{	ext{net}} = C_{eq} V = 4 \mu	ext{F} 	imes 25 	ext{ V} = 100 \mu	ext{C}$.Since capacitor $A$ is in series with the combination of $B$ and $C$,the charge on capacitor $A$ is $q_A = q_{	ext{net}} = 100 \mu	ext{C}$.The voltage across the parallel combination of $B$ and $C$ is $V_{BC} = V - V_A = V - \frac{q_A}{C_A} = 25 - \frac{100}{5} = 25 - 20 = 5 	ext{ V}$.Now,the charges on $B$ and $C$ are:$q_B = C_B V_{BC} = 8 \mu	ext{F} 	imes 5 	ext{ V} = 40 \mu	ext{C}$.$q_C = C_C V_{BC} = 12 \mu	ext{F} 	imes 5 	ext{ V} = 60 \mu	ext{C}$.The ratio of charges is $q_A : q_B : q_C = 100 : 40 : 60 = 5 : 2 : 3$.

Three capacitors $A, B$ and $C$ are connected to a battery of $25 \text{ V}$ as shown in the figure. The ratio of charges on capacitors $A, B$ and $C$ will be $-$

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