A capacitor, $C_{1} = 6 μ F$ is charged to a potential difference of $V_{0} = 5 V$ using a 5V battery. The battery is removed and another capacitor, $C_{2} = 12 μ F$ is inserted in place of the battery. When the switch ‘S’ is closed, the charge flows between the capacitors for some time until equilibrium condition is reached. What are the charges (q₁ and q₂) on the capacitors C₁ and C₂ when equilibrium condition is reached.

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$q_{1} = 20 μC, q_{2} = 10 μC$

$q_{1} = 30 μ C, q_{2} = 15 μ C$

$q_{1} = 15 μ C, q_{2} = 30 μ C$

$q_{1} = 10 μ C, q_{2} = 20 μ C$

answer is C.

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Detailed Solution

$q_{1}^{'} = 6 \times 5 = 30 μ C$
Finally
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$\begin{aligned} 6 V_{c} + 12 V_{c} = 30 + 0 \\ 18 V_{c} = 30 \\ V_{C} = \frac{30}{18} = \frac{5}{3} Volt \\ \Rightarrow q_{1} = \frac{6 \times 5}{3} = 10 μ C \\ \Rightarrow q_{2} = \frac{12 \times 5}{3} = 20 μ C \end{aligned}$