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

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

1.	In the circuit shown below, the key K is closed at t =0. The current through the battery is 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» Correct Answer - B 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))`

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