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1. Electric Charges and Fields
normal
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 capacitors will be
A
$Q, 2Q$
B
$\frac{Q}{2},\frac{{2Q}}{3}$
C
$\frac{{3Q}}{2},3Q$
D
$\frac{{2Q}}{3},\frac{{4Q}}{3}$
Solution
In steady state, both the capacitors are at the same potential,
i.e., $ \quad \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}$
Also, $\mathrm{Q}_{1} +\mathrm{Q}_{2}=\mathrm{Q}$
$\therefore $ $\mathrm{Q}_{1}=\frac{\mathrm{Q}}{3}, \mathrm{Q}_{2}=\frac{2 \mathrm{Q}}{3}$
Standard 12
Physics
