In the circuit Fig the final current through `30 Omega` resistance when circuit is completed is

Options

A. `3 A`
B. `0.1 A`
C. `5 A`
D. `0.5 A`

Answer

B

Explanation

The current is short circuited through inductor. In steady state, current will pass through inductor of `20 Omega`.
Then current through `30 Omega` is:
`I = (3)/(30) = 0.1 A`