In the circuit the key (K) is closed at t = 0. Find the current in ampere through the key (K) at the instant  $t={10}^{-3}\ell n2\sec$. (Assume energy stored in inductor and capacitor before t = 0 is zero)

$i_1$  is current in the first loop
$i_1=\frac{E}{R_1}(1-e^{-\frac{t}{\raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{$R_1$}\right.}})=\frac{40}{20}(1-e^{-\frac{{10}^{-3}\ln 2}{5\times {10}^{-4}}})=\frac{3}{2}A$ 
$i_2$  is current in the second loop
$i_2=\frac{E}{R_2}{e}^{-\frac{t}{R_{2}C}}=\frac{40}{40}{e}^{-\frac{{10}^{-3}\ln 2}{{10}^{-3}}}=\frac{1}{2}A$ 
So, total current through key (K)
$i=i_1+i_2\text{}=\frac{3}{2}+\frac{1}{2}=2A$