Two capacitors $C_1$ and $C_2=$ $2 C _1$ are connected in a circuit with a switch between them as shown in the figure.Initially the switch is open and $C _1$ holds charge $Q$. The switch is closed. At steady state, the charge on each capacitor will be
A$Q , 2 Q$
B$\frac{Q}{3}, \frac{2 Q}{3}$
C$\frac{3 Q}{2}, 3 Q$
D$\frac{2 Q }{3}, \frac{4 Q }{3}$
Medium
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B$\frac{Q}{3}, \frac{2 Q}{3}$
b (b)
In steady state, both the capacitors are at the same potential, i.e., $\frac{ Q _1}{ C _1}=\frac{ Q _2}{ C _2}$ or $\frac{ Q _1}{ C }=\frac{ Q _2}{2 C }$ or $Q _2=2 Q _1$
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