• # question_answer In the circuit shown in the following figure E = 10 V, ${{R}_{1}}=2\,ohm,$ ${{R}_{2}}=3\,ohm$ and  ${{R}_{3}}=\text{6}\,ohm$and L = 5 henry. The current ${{I}_{1}}$ just after pressing the switch S is A) 2.5 amp B) 2 amp C) 5/6 amp D) 5/3 amp

 just after closing of switch inductor behave as open circuit equivalent circuit is shown ${{I}_{1}}=\frac{E}{{{R}_{1}}+{{R}_{2}}}=\frac{10}{2+3}=\frac{10}{5}=2mp$