A circuit containing capacitors C1 and C2 as shown in the figure are in steady state with key K1 closed and K2 is opened. At the instant t = 0, if K1 is opened and K2 is closed then the maximum current in the circuit will be :

When switch K1 is opened and K2 is closed it becomes L–C              circuit          so                     applying               energy conservation :Voltage across 2μf=10V∴12CV2=12Li2i=max  current  across  inductor=2×10−6×102=0.2×10−3×i2210=0.2 i2i=1 A