simulate this circuit – Schematic created using CircuitLab
Above is the set-up of my simple experiment to determine the inductance of an inductor. The voltage across resistor R (5 ohms) and inductor L (20mH) are measured using a oscilloscope and labelled VR and VL respectively. Therefore, the inductor coil resistance RL is:
$$R_L=\frac{V_L}{V_R}(R)=2\pi fL$$
The experimental value of VR and VL are 1.9V and 2.4V:
$$\frac{2.4}{1.9}(5)=2\pi (50)(20\cdot 10^{-3})$$ $$6.32\approx 2\pi$$
Hence I can conclude that both the resistor and inductor used are in good condition.
What's troubling me is that when I'm asked to measure the inductor coil's resistance using a analog multimeter and compare the value of RL obtained from the experiment. The multimeter shows a non-zero reading of about 3 ohms.
From my understanding, ohmmeter provided a small battery to apply a voltage to a resistance to measure the current through the resistance, so shouldn't the value measured using the multimeter be 0 ohms since the battery provide DC current? Even if the multimeter does measure the resistance of the inductor coil, why it's significantly less than (by about 3 ohms) the experimental value?
SOLVED:
VR and VL are actually 1.4V and 1.9V, not 1.9V and 2.4V. My mistake. Now it makes sense. The RL should be:
$$R_L=\frac{V_L}{V_R}(R)=\sqrt{(2\pi fL)^{2}+R_l^{2}}$$ $$ where Rl is pure ohm resistance of inductor coil. Inserting the value:
$$\frac{1.9}{1.4}(5)\: ;\: \sqrt{(2\pi (50)(20\cdot 10^{-3}))^{2}+3^{2}}$$ $$6.78\approx 6.96$$ (Difference by 0.18 due to limitation of the precision of oscilloscope)