A capacitor of capacitance $C_{1}$ is charged by connecting it to a battery. The battery is now removed and this capacitor is connected to a second uncharged capacitor of capacitance $C_{2}$. If the charge distributes equally on the two capacitors, the ratio of the total energy stored in the…

Question

A capacitor of capacitance $C_{1}$ is charged by connecting it to a battery. The battery is now removed and this capacitor is connected to a second uncharged capacitor of capacitance $C_{2}$. If the charge distributes equally on the two capacitors, the ratio of the total energy stored in the capacitors after connection to the total energy stored in them before connection is

Aicharya · Accepted Answer

If $Q$ is the initial charge on capacitor $C_{1}$, the initial energy is given by $U_{i}=Q^{2} / 2 C_{1}$ When the two capacitors are connected together, and as the charge is distributed equally, the charge on each capacitor is $Q / 2$. Since the potential difference (in a parallel connection) across the two capacitors is also the same, it follows that their capacitances are equal (since $C=Q / V)$. Thus $C_{1}=C_{2}=C$ (say). Also, $Q_{1}=Q_{2}=Q / 2$ Therefore, final energy stored in the two capacitors is $U_{f}=\frac{Q_{1}^{2}}{2 C_{1}}+\frac{Q_{2}^{2}}{2 C_{2}}=\frac{(Q / 2)^{2}}{2 C}+\frac{(Q / 2)^{2}}{2 C}=\frac{Q^{2}}{4 C}$ But $U_{i}=\frac{Q^{2}}{2 C}$ $\therefore \quad \frac{U_{f}}{U_{i}}=\frac{1}{2}$, which is choice (b).

A capacitor of capacitance \(C_{1}\) is charged by connecting it to a battery. The battery is now removed…

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