A capacitor of capacity \(C\) is charged to a potential difference \(V\) and another capacitor of capacity…

A capacitor of capacity \(C\) is charged to a potential difference \(V\) and another capacitor of capacity \(2 \mathrm{C}\) is charged to a potential difference \(4 \mathrm{~V}\). The charging batteries are disconnected and the two capacitors are connected with reverse polarity (i.e. positive plate of first capacitor is connected to negative plate of the second capacitor and the negative plate of first capacitor is connected to positive plate of second capacitor). The heat produced during the redistribution of charge between the capacitor will be
  1. \(\frac{125 C V^{2}}{3}\)
  2. \(\frac{50 C V^{2}}{3}\)
  3. \(2 \mathrm{CV}^{2}\)
  4. \(\frac{25 C V^{2}}{3}\)

Solution

Initial energy of the system,
\(\mathrm{U}_{\mathrm{i}}=\frac{1}{2} C V^{2}+\frac{1}{2} \cdot 2 C(4 V)^{2}=\frac{33}{2} \mathrm{CV}^{2}\)
Final charge on the two capacitors after connections are \(\frac{7 C V}{3}\) and \(\frac{14 C V}{3}\) Final energy of system
\(\mathrm{U}_{\mathrm{f}}=\frac{1}{2 C}\left(\frac{7 C V}{3}\right)^{2}+\frac{1}{2.2 C}\left(\frac{14 C V}{3}\right)^{2}=\frac{49 C V^{2}}{6}\)
\(\therefore\) Heat produced \(=\mathrm{U}_{\mathrm{i}}-\mathrm{U}_{\mathrm{f}}=\frac{25}{3} \mathrm{CV}^{2}\) ^

Asked in: JEE Mains - Capacitance - Chapter Test

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