Physics - Capacitor Question with Solution | TestHub

Two capacitors of capacities $2\mu \mathrm{F}$ and $4\mu \mathrm{F}$ are connected in parallel. A third capacitor of $6\mu \mathrm{F}$ capacity is connected in series with this combination. A battery of $12\mathrm{V}$ is connected across this combination. The charge on $2\mu \mathrm{F}$ capacitor is $12\mu \mathrm{C}$. Explanation: The circuit diagram of given situation is as shown below
The equivalent capacitance of $2\mu \mathrm{F}$ and $4\mu \mathrm{F}$ capacitors connected in parallel Is $C_{eq}=2+4=6\mu \mathrm{F}$ The circuit now becomes
As, both the capacitors are of the same capacitance, so the potential of $12\mathrm{V}$ is equally divided in them i.e., $V_1=V_2=6V$ In parallel combination, the potential remains the same. $\therefore$ Charge on $2\mu \mathrm{F},\mathrm{Q}=2\times {\mathrm{V}}_1=2\times 6=12\mu \mathrm{C}$ OR
Let $C_1=2\mu F_1{C}_2=4\mu F$ and $C_3=6\mu F$ Equivalent of $C_1$ and $C_2=C_4=6\mu F$ Equivalent of $C_3$ and $C_4=C_5=3\mu F$ Charge on $C_5=Q=C_{5}V=3\times 12=36\mu C$. $C_3$ and $C_4$ are in series. Hence charge on them is also $36\mu C$. The charge of $36\mu C$ is divided between $C_1$ and $C_2$ in proportion to their capacitance. $\frac{Q_1}{Q_2}=\frac{C_1}{C_2}$ $\therefore \frac{Q_1}{Q_1+Q_2}=\frac{C_1}{C_1+C_2}$ $\therefore \frac{Q_1}{Q}=\frac{C_1}{C_1+C_2}=\frac{2}{6}=\frac{1}{3}$ $\therefore Q_1=\frac{Q}{3}=\frac{36}{3}=12\mu C$