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\(.........\)\(CV _{0}^{2}\)
Pick an option to check your answer.
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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