In the given circuit, let `i_(1)` be the current drawn battery at time `t=0` and `i_(2)` be steady current at `t=oo` then the ratio `(i_(1))/(i_(2))` is

Options

A. `1.0`
B. `0.8`
C. `1.2`
D. `1.5`

Answer

B

Explanation

At `t=0`, inductor behave as open circuit so `i_(1)` `=(10)/(10)=1 A`
At `t=oo`, inductor behave as short circuit, so `i_(2)` `=(10)(8)=(5)/(4)A`
Hence, `(i_(1))/(i_(2))=(1)/(5//4)=(4)/(5)=0.8`