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Two capacitors of capacitances C and 2C are charged to potential differences V and 2V, respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is :
Option: 1 \frac{25}{6}CV^{2}
Option: 2 \frac{3}{2}CV^{2}
Option: 3 zero
Option: 4 \frac{9}{2}CV^{2}

Answers (1)

best_answer

\mathrm{Q}_{1}=\mathrm{CV} \quad \mathrm{Q}_{2}=2 \mathrm{C} \times 2 \mathrm{~V}=4 \mathrm{CV}

\begin{aligned} &\Rightarrow \text { By conservation of charge }\\ &q_{i}=q_{f}\\ &\mathrm{Q}_{1}+\mathrm{Q}_{2}=\mathrm{q}_{1}+\mathrm{q}_{2}\\ &4 C V-C V=(C+2 C) V_{C}\\ &\mathrm{V}_{\mathrm{C}}=\frac{3 \mathrm{CV}}{3 \mathrm{C}} \Rightarrow \mathrm{V} \end{aligned}

\begin{array}{l} \Rightarrow \frac{1}{2} \times(3 \mathrm{C}) \times \mathrm{V}_{\mathrm{c}}^{2} \\ =\frac{1}{2} \times 3 \mathrm{C} \times \mathrm{V}^{2}=\frac{3}{2} \mathrm{CV}^{2} \end{array}

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Deependra Verma

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