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

Options

A. `(VR_(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

At `t = 0`, inductor behaves like an infinite resistance so at `t = 0, i = (V)/(R_(2))`
and at `t = oo`, inductor behaves line an a conducting wire `i = (V)/(R_("eq")) = (V(R_(1)+R_(2)))/(R_(1)R_(2))`