Electronic – Determining inductance using a multimeter

Keep in mind that your coil is made of a long and thin wound wire, so it also has a resistance. Though it heavily depends on the gauge and length of the wire, I would say, 3 Ohm makes sense.

Finally, the total impedance is the sum of the (Ohm's) resistance of R and the coil, plus the impedance of the coil:

$$Z(f)=R_R+(\underbrace{\color{blue}{R_L+2\pi jLf}}_{coil})$$

First of all, this is a complex expression. If you measure the voltage across the coil, it is

$$V_L=Re\left(\frac{R_L+2\pi jLf}{R_R+R_L+2\pi jLf}\right)\cdot V_0$$

However, you can thrust your multimeter here, as it measures the pure ohm's resistance. Maybe, it can also measure the inductance. For correct results of the oscilloscope, you will have to do the math.

Assuming this equation is correct: $$Z = \frac{R}{1+i \omega cR} + i \omega L$$

We multiply the first term's nominator and the denominator by the complex conjugate of the denominator: $$.. = \frac{R(1-i\omega c R)}{(1+i \omega cR)(1-i\omega c R)} + i \omega L$$ and thus getting rid of imaginary stuff in the denominator: $$.. = \frac{R(1-i\omega c R)}{(1+ \omega^2 c^2R^2)} + i \omega L$$ And this thing is easily separated to real and imaginary parts: $$.. = \frac{R}{1+ \omega^2 c^2R^2} + i \left(\omega L - \frac{\omega c R^2}{1+ \omega^2 c^2R^2} \right)$$