Thank you for reporting, we will resolve it shortly
Q.
In the circuit shown in figure, key $K$ is closed at $t=0$, the current through the key at the instant $t=10^{-3} \ln 2 s$ is
Electromagnetic Induction
Solution:
Current in branches containing $L$ and $R$ will flow independently
$I_{1}=\frac{20}{10}\left(1-e^{-\frac{t}{5 \times 10^{-4}}}\right)$
$=\frac{3}{2}=1.5\, A$
$I_{2}=\frac{20}{10} e^{-\frac{t}{10^{-3}}}=1.0\, A$
$I=I_{1}+I_{2}=2.5\, A$
