A capacitor, \(C_1=6 \mu \mathrm{~F}\) is charged to a potential difference of \(\mathrm{V}_0=5 \mathrm{~V}\) using a 5 V battery. The battery is removed and another capacitor, \(\mathrm{C}_2=12 \mu \mathrm{~F}\) 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 \(\left(q_1\right.\) and \(\left.q_2\right)\) on the capacitors \(C_1\) and \(C_2\) when equilibrium condition is reached.

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Correct Answer

(A) \(q_1=10 \mu \mathrm{C}, \mathrm{q}_2=20 \mu \mathrm{C}\)

Step-by-step Solution

Detailed explanation

\(\mathrm{q}_1^{\prime}=6 \times 5=30 \mu \mathrm{C}\)
Finally

\(\begin{aligned} & 6 \mathrm{~V}_{\mathrm{C}}+12 \mathrm{~V}_{\mathrm{c}}=30+0 \\ & 18 \mathrm{~V}_{\mathrm{C}}=30 \\ & \mathrm{~V}_{\mathrm{C}}=\frac{30}{18}=\frac{5}{3} \mathrm{Volt} \\ & \Rightarrow \mathrm{q}_1=\frac{6 \times 5}{3}=10 \mu \mathrm{C} \\ & \Rightarrow \mathrm{q}_2=\frac{12 \times 5}{3}=20 \mu \mathrm{C}\end{aligned}\)

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