JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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\(.....CV^2\)
- A \(4.5\)
- B \(4.16\)
- C \(0\)
- D \(1.5\)
Answer & Solution
Correct Answer
(D) \(1.5\)
Step-by-step Solution
Detailed explanation
\(Q _{1}= CV \quad Q _{2}=2 C \times 2 V =4 CV\) \(\Rightarrow\) By conservation of charge \(q_{i}=q_{f}\) \(Q_{1}+Q_{2}=q_{1}+q_{2}\) \(4 CV - CV =( C +2 C ) V _{ C }\) \(v _{ C }=\frac{3 CV }{3 C } \Rightarrow V\) \(\Rightarrow \frac{1}{2} \times(3 C) \times V_{ c }^{2}\)…
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