Physics - EMI/AC Question with Solution | TestHub

Steady state with switch closed: $I_0=\frac{9}{3}=3A$ through the inductor. After opening, $\tau =L/R=9/9=1s$ and $I(t)=3e^{-t}$.

At $t=\ln 2$: $I=1.5A$.

$V_{3\Omega }=4.5V$ (Q$\to 2$); $V_{6\Omega }=9V$ (R$\to 1$); $V_L=I(6+3)=13.5V$ (P$\to 5$); $V_{bc}=V_{6\Omega }=9V$ (S$\to 1$). Option (A).