In the circuit shown, the switch $S$ has been kept closed for a long time and then opened. Just after the switch is opened, what is the voltage across the inductor $\left(V_L\right)$ and which labeled point $\left(AorB\right)$ of the inductor is at a higher potential? Take $R_1=4.0\Omega ,R_2=8.0\Omega$ and $L=2.5H.$

When the switch is closed, effective resistance is parallel combination of  $R_1$ and  $R_2$

$R_0=\frac{R_1{R}_2}{R_1+R_2}=\frac{4\times 8}{4+8}=\frac{8}{3}\Omega$

Current through  $L$ is $I=\frac{12V}{\frac{8}{3}\Omega }=\frac{9}{2}A$

Immediately after the switch is opened, the current in the inductor remains  $I=\frac{9}{2}A$ and the only 

resistance in the circuit is  $R_1=4\Omega .$

Drop of potential in $R_1=\frac{9}{2}\times 4=18V$

Obviously, induced emf in the inductor is $6V$ with $B$ at higher potential.