Two capacitors of capacitances \(C\) and \(2\, C\) are charged to potential differences \(V\) and \(2\, V\), 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\(.....CV^2\)

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Answer & Solution

Correct Answer

(D) \(1.5\)

Detailed explanation

\(Q _{1}= CV \quad Q _{2}=2 C \times 2 V =4 CV\)

\(\Rightarrow\) By conservation of charge

\(q_{i}=q_{f}\)

\(Q_{1}+Q_{2}=q_{1}+q_{2}\)

\(4 CV - CV =( C +2 C ) V _{ C }\)

\(v _{ C }=\frac{3 CV }{3 C } \Rightarrow V\)

\(\Rightarrow \frac{1}{2} \times(3 C) \times V_{ c }^{2}\)

\(=\frac{1}{2} \times 3 C \times V ^{2}=\frac{3}{2} CV ^{2}\)

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