MHT CET20225 Aug 2022Morning ShiftPhysicsAlternating CurrentActual
A condenser of capacity ' C ' is charged to a potential difference of ' V ₁ '. The plates of the condenser are then connected to an ideal inductor of inductance ' L '. The current through an inductor |when the potential difference across the condenser reduces to ' V ' is
Options
- AC ( V ₁^2- V ₂^2 ) L
- BC ( V ₁^2+ V ₂^2 ) L
- C[ C ( V ₁^2- V ₂^2 )^ 1 2 ~L ]
- D[ C ( V ₁- V ₂ )^ 1 2 ~L ]
Correct answer
C. [ C ( V ₁^2- V ₂^2 )^ 1 2 ~L ]
Step-by-step solution
The correct option is (C). Concept: For the LC circuit the potential drop across the circuit can be written as: q C + L di dt =0 On rewriting, q C + L ( dq dt ) di dt =0 Therefore, q C + Li di dq =0 On integrating, _ cv ₁ ^ cv ₂ q c dq =- ₀^ i Lidi On solving for I= [ c ( v ₁^2- v ₂^2 )^ 1 / 2 ~L ]