A circuit containing capacitors `C_(1)` and `C_(2)` as shown in the figure are in steady state with key `K_(1)` closed.At the instant `t=0`, if `K_(1)` is opened and `K_(2)` is closed then the maximum current in the circuit will be:

Options

A. `1 A`
B. `1/2 A`
C. `2 A`
D. None of these

Answer

A

Explanation

When swith `K_(1)` is and `K_(2)` is closed it becomes `I-C` circuit so applying energy conservation `Q_(0)^(2)/(2C)=1/2Li^(2).Q_(0)=C_(eq)V=(C_(1)C_(2))/(C_(1)+C_(2)) V=(20xx10^(-6))`
`((20xx10^(-6))^(2))/(2xx2xx10^(-6))=1/2(0.2xx10^(-3))i^(2) =i=1A`