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…

Question

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.

Aicharya · Accepted Answer

$\mathrm{q}_1^{\prime}=6 	imes 5=30 \mu \mathrm{C}$ Finally $\begin{aligned} & 6 \mathrm{~V}_{\mathrm{C}}+12 \mathrm{~V}_{\mathrm{c}}=30+0 \ & 18 \mathrm{~V}_{\mathrm{C}}=30 \ & \mathrm{~V}_{\mathrm{C}}=\frac{30}{18}=\frac{5}{3} \mathrm{Volt} \ & \Rightarrow \mathrm{q}_1=\frac{6 	imes 5}{3}=10 \mu \mathrm{C} \ & \Rightarrow \mathrm{q}_2=\frac{12 	imes 5}{3}=20 \mu \mathrm{C}\end{aligned}$

A capacitor, $C_1=6 \mu \mathrm{~F}$ is charged to a potential difference of $\mathrm{V}_0=5 \mathrm{~V}$…

Solution