JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A capacitor \(C\) is fully charged with voltage \(V _{0}\) After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance \(\frac{ C }{2} .\) The energy loss in the process after the charge is distributed between the two capacitors is\(.........\)\(CV _{0}^{2}\)
- A \(0.166\)
- B \(0.5\)
- C \(0.33\)
- D \(0.25\)
Answer & Solution
Correct Answer
(A) \(0.166\)
Step-by-step Solution
Detailed explanation
\(\frac{ CV _{0}- q }{ C }=\frac{ q }{ C / 2}=\frac{2 q }{ C }\) \(V _{0}=\frac{3 q }{ C } \Rightarrow q =\frac{ CV _{0}}{3}\) \(U _{ i }=\frac{1}{2} CV _{0}^{2}\)…
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