• # question_answer In the Q.26, the current ${{I}_{1}}$ long after pressing the switch is A) 2.5 amp B) 2 amp C) 5/6 amp D) 5/3 amp

 After long time induced emf in L become zero so, L is replaced with simple conducting wire. So new equivalent circuit is shown. ${{I}_{1}}=\frac{E}{{{R}_{eq}}}$ ${{R}_{eq}}=2+{{R}_{eq}}$ of 6 &$3\Omega =2+\frac{6\times 3}{6+3}=2+2=4$ ${{I}_{1}}=\frac{10}{4}=2.5\,\,amp$