Physics - EMI/AC Question with Solution | TestHub

At $t=0$ the inductor acts as infinite resistance (open branch)

Battery current $I=\frac{9}{6}=1.5\mathrm{A}$ (P$\to$4); current in $3\Omega$ is zero (R$\to$3)

At $t=\infty$ the inductor acts as a wire: $R_{eq}=\frac{3\times 6}{6+3}=2\Omega$, so battery current $I=\frac{9}{2}=4.5\mathrm{A}$ (Q$\to$5)

$6I_1=3I_2$ and $I_1+I_2=4.5$ give $I_2=3\mathrm{A}$ through $3\Omega$ (S$\to$1)

Answer: (A)