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

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.
\(\frac{Q_{1}}{Q_{2}}=\frac{C_{1}}{C_{2}}\)
\(\therefore \frac{Q_{1}}{Q_{1}+Q_{2}}=\frac{C_{1}}{C_{1}+C_{2}}\)
\(\therefore \frac{Q_{1}}{Q}=\frac{C_{1}}{C_{1}+C_{2}}=\frac{2}{6}=\frac{1}{3}\)
\(\therefore Q_{1}=\frac{Q}{3}=\frac{36}{3}=12 \mu C\)