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

Options

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

Answer

A

Explanation

When key is closed at t = 0, the current through inductance L will increase with time, resulting in induced emf across L, which will oppose the current through the resistance `R_(1)`. Therefore, the current from battery will flow through `R_(2)` . This
current is `I = (V)/(R^(2)) at t = 0`
When `t = oo`, the current will reach to steady state in the circuit. The presence of L will become ineffective.Now effective resistance of circuit
`= (R_(1) R_(2))/(R_(1) + R_(2))`
current through battery at `t = oo`
`I' = (V)/(R_(1) R_(2) // (R_(1) + R_(2))) + (V(R_(1) + R_(2)))/(R_(1) R_(2))`