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 the inductor when the potential difference across the condenser reduces to $${V_2}$$ is

Solution :
$$\eqalign{ & q = C{V_1}\cos \omega t \Rightarrow i = \frac{{dq}}{{dt}} = - \omega C{v_1}\sin \omega t \cr & {\text{Also,}}\,{\omega ^2} = \frac{1}{{LC}}\,{\text{and}}\,V = {V_1}\cos \omega t \cr & {\text{At}}\,t = {t_1},V = {V_2}\,{\text{and}}\,i = - \omega C{V_1}\sin \omega {t_1} \cr & \therefore \cos \omega {t_1} = \frac{{{V_2}}}{{{V_1}}}\left( { - ve{\text{ sign gives direction}}} \right) \cr & {\text{Hence,}}\,i = {V_1}\sqrt {\frac{C}{L}} {\left( {1 - \frac{{V_2^2}}{{V_1^2}}} \right)^{\frac{1}{2}}} = {\left( {C\frac{{\left( {V_1^2 - V_2^2} \right)}}{L}} \right)^{\frac{1}{2}}} \cr} $$