In the circuit shown below, the key K is closed at t =0. The current through the battery is

Options

A. `(V R_(1)R_(2))/(sqrt(R_(1)^(2)+R_(2)^(2))) "at t = 0"` and `(V)/(R_(2)) "at t" = oo`
B. `(V)/(R_(2)) "at t = 0"` and `(V(R_(1)+R_(2)))/(R_(1)R_(2)) "at t" = oo`
C. `(V)/(R_2) "at t = 0"` and `(VR_(1)R_(2))/(sqrt(R_(1)^(2)+R_(2)^(2))) "at t" = oo`
D. `(V(R_(1)+R_2))/(R_(1)R_(2)) "at t = 0"` and `(V)/(R_(2)) "at t" = oo`

Answer

B

Explanation

Just after closing hthe key (say at time `t = 0`) the inductor coil offers infinite resistance, hence it acts as an open circuit. The corresponding equivalent circuit diagram is as shown in the figure (i).

The current through battery is `I = (V)/(R_(2))`
At time when energy stored equally inelectric and magnetic field, at this time
Energy of a capacitor `= (1)/(2)` Total energy
`(1)/(2)(q^(2))/(C ) = (1)/(2)((1)/(2)(q_(0)^(2))/(C )) rArr q = (q_(0))/(sqrt(2))`
From equation (i) `" " (q_(0))/(sqrt(2)) = q_(0) cos omega t`
`cos omega t = (1)/(sqrt(2)) rArr omega t = cos^(-1)((1)/(sqrt(2))) = (pi)/(4)`
`t = (pi)/(4 omega) = (pi)/(4) sqrt(LC) (because omega = (1)/(sqrt(LC)))`