Question

In the given circuit, find the current through the $5mH$ inductor in steady state is $I\times {10}^{-2}$

Answer 1

In steady state, main current from battery is
$i_0=\frac{E}{R}=\frac{20}{5}$
$=4A$
Now this current distributes in inverse ratio of inductor
$\therefore$ $i_5=\left(\frac{10}{10+5}\right)\left(4A\right)=2.67A$