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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2020
  • Hard
  • MCQ

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\(.........\)\(CV _{0}^{2}\)

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Answer & Solution

Correct Answer

(A) \(0.166\)

Step-by-step Solution

Detailed explanation

\(\frac{ CV _{0}- q }{ C }=\frac{ q }{ C / 2}=\frac{2 q }{ C }\)

\(V _{0}=\frac{3 q }{ C } \Rightarrow q =\frac{ CV _{0}}{3}\)

\(U _{ i }=\frac{1}{2} CV _{0}^{2}\)

\(U _{ f }=\frac{\left(\frac{2 CV _{0}}{3}\right)^{2}}{2 C }+\frac{\left(\frac{ CV _{0}}{3}\right)^{2}}{2\left(\frac{ C }{2}\right)}\)

\(=\frac{1}{2} CV _{0}^{2}\left[\frac{4}{9}+\frac{2}{9}\right]=\frac{1}{2} CV _{0}^{2}\left(\frac{2}{3}\right)\)

Heat loss \(=\frac{1}{2} CV _{0}^{2}-\left(\frac{2}{3}\right)\left(\frac{1}{2} CV _{0}^{2}\right)\)

\(=\frac{1}{6} CV _{0}^{2}\)

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