Two capacitors of capacities $2 \mu \mathrm{F}$ and $4 \mu \mathrm{F}$ are connected in parallel. A third…

1. $12 \mu C$
2. $16 \mu C$
3. $14 \mu C$
4. $11 \mu C$

Solution

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 $C_{e q}=2+4=6 \mu \mathrm{F}$ 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., $V_{1}=V_{2}=6 V$ 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. $\frac{Q_{1}}{Q_{2}}=\frac{C_{1}}{C_{2}}$ $\therefore \frac{Q_{1}}{Q_{1}+Q_{2}}=\frac{C_{1}}{C_{1}+C_{2}}$ $\therefore \frac{Q_{1}}{Q}=\frac{C_{1}}{C_{1}+C_{2}}=\frac{2}{6}=\frac{1}{3}$ $\therefore Q_{1}=\frac{Q}{3}=\frac{36}{3}=12 \mu C$

Asked in: MHT CET 2020 (19 Oct Shift 2)

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