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?


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)


1 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.

  • \$\begingroup\$ How can I be so dumb to not realize this. BTW I typed the wrong result in my question and now it make sense. I'm stuck while doing my lab report and you just saved my day. \$\endgroup\$
    – khtan
    Commented Mar 29, 2015 at 12:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.