There are a couple of nonlinear magnetic material effects that might be at play here, although this answer must be described as speculation without more detail. Both effects are more pronounced if your inductor is ungapped.
1) At very low current levels (corresponding to very low levels of magnetic field H), the inductance can be lower than nominal. (The B-H characteristic of the magnetic core material has a lower slope right at the origin.) As current increases from these low levels, the calculated inductor impedance $Z_L =2 \pi f L$ would increase and then stabilize at the nominal value.
2) As current continues to increase, eventually the inductor starts to saturate. (The B-H characteristic flattens at high fields.) Inductance then becomes a decreasing function of current, so the calculated inductor impedance would decrease.
You can see some B-H curves illustrating these effects in the wikipedia article on "Saturation (magnetic)".
Introducing a gap in the magnetic core reduces the component's inductance but stabilizes it against these effects.
First of all, don't mix up voltage with current. In your examples 1 and 2 it is certainly true that the voltages across the resistor and inductor are the same w.r.t. the source voltage. This is just Kirchhoff's voltage law. However, this still results in a current lag in the inductor compared to the resistor. Say the source voltage is
$\Delta V_S = V_0 \sin(\omega t)$
So the resistor voltage is
$\Delta V_R = - V_0 \sin(\omega t)$
so that $\sum \Delta V = 0$ as Kirchhoff's law requires. The same goes for the voltage across the inductor, $\Delta V_L$.
But for the resistor we have Ohm's Law
$\Delta V_R = IR$ so the current through the resistor is just
$I_R = -\frac{V_0}{R} \sin(\omega t)$
But for the inductor we have
$\Delta V_L = - L \frac{dI}{dt}$.
So to get the current, $I_L$, you need to integrate $\Delta V_L$ w.r.t. time so in this example you will get a cosine instead of a sine. Thus, we see a phase shift in the current (but not the voltage). It is worth going through the integral yourself, but it is also in most elementary circuits textbooks.
Best Answer
The basic formula for finding the potential drop is $V=X_{L}I$ ----(1);
Where $X_L$ is the inductive reactance
and I is current
Step 1: To find $X_L$
its simple $X_L$=$\omega$L
Step 2: To find current I
In LCR circuit we find net reactance which is similar to net resistance
${Z}^{2}$=$R^2$+${X_{L}}^2$
Then we have I=V/Z.
Now you can substitute the values of I and $X_L$ and substitute in eqn (1).