Measure inductor coil's resistance using multimeter

Question

^{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 V_R and V_L respectively. Therefore, the inductor coil resistance R_L is:

$$R_L=\frac{V_L}{V_R}(R)=2\pi fL$$

The experimental value of V_R and V_L 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 R_L 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?

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SOLVED:

V_R and V_L are actually 1.4V and 1.9V, not 1.9V and 2.4V. My mistake. Now it makes sense. The R_L should be:

$$R_L=\frac{V_L}{V_R}(R)=\sqrt{(2\pi fL)^{2}+R_l^{2}}$$ $$ where R_l 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)

Answer 1

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.