I'm having a hard time understanding why one couldn't measure the inductance of a coil using a simple multimeter capable of measuring AC voltages and currents:


simulate this circuit – Schematic created using CircuitLab

The total series impedance could be expressed as:

$${Z=R+R_L+\omega L}$$

EDIT: given the 90 degrees phase difference, the correct formula is:

$$Z=\sqrt{(R+R_L)^2+(\omega L)^2}$$

Now, if I'm not mistaking, the same impedance can be expressed as the ratio between the measured effective AC voltage and current, V and I (as complex values):


We can then obtain L as:

$$L=\frac{V/I-(R+R_L)}{2\pi f}$$

EDIT: the correct formula is:

$$L=\frac{\sqrt{(V/I)^2-(R+R_L)^2}}{2\pi f}$$

In conclusion, by measuring the AC voltage, current, coil DC resistance and by choosing a well known external resistor, we could find out the coil inductance.

Am I mistaking? If yes, where? If no, why all google searches for "measuring inductance" present more complex methods involving function generators, oscilloscopes and phase-shifting?

  • 2
    \$\begingroup\$ If the inductance is large enough to give meaningful voltages at 50/60 Hz, then no problem. For uH, you need higher frequency to get enough amplitude. \$\endgroup\$
    – winny
    Commented Mar 8, 2017 at 21:32
  • 1
    \$\begingroup\$ But you ARE using a function generator \$\endgroup\$
    – Trevor_G
    Commented Mar 8, 2017 at 21:32
  • 1
    \$\begingroup\$ readme electronics.stackexchange.com/questions/290327/… \$\endgroup\$ Commented Mar 8, 2017 at 21:41
  • 1
    \$\begingroup\$ If your reference is a 50 Ohm generator and the nomograph indicates a frequency for 50 Ohms in your range , you can use directly across generator for 50%Voc. Other method is use a known Cap and sweep for resonant frequency and use nomograph or compute. \$\endgroup\$ Commented Mar 8, 2017 at 21:53
  • 1
    \$\begingroup\$ While for instantaneous voltage: \$ V(t) = V_{R+R_L}(t) + V_L(t) \$. A voltmeter does not give instantaneous measurements, it may give RMS measurements. For that, \$ V_{RMS} = \sqrt{{V_{R+R_L.RMS}}^2 + {V_{L.RMS}}^2} \$. So you need to adjust your formula accordingly. Or just measure the current and the voltage across the inductor directly. \$\endgroup\$
    – rioraxe
    Commented Mar 8, 2017 at 22:26

1 Answer 1


Your calculation is wrong - sorry! The inductance causes a phase lag to the current. The impedance of the inductor for that reason can't add directly to the resistances The proper formula for the total impedance is

$$Z = \frac{V}{I} = \sqrt{(R+RL)^2 + (\omega L)^2}$$

[Offtopic : This formula is easily derived by using the complex phasor calculus. That calculus is developed for AC circuit analysis nearly 125 years ago]

L must be solved from this equation. The result is

$$L= \frac{\sqrt{Z^2-(R+RL)^2}}{\omega}$$

This method is not practical if the inductance is small, say 1 mH or less.

For sub-mH inductors you must have quite high frequency to have well measurable U and I. You run out of the frequency range of your multimeter. Additionally: The stray capacitances make disturbing currents that bypass the coil and the skin effect makes the resistance of the coil much higher than its DC resistance.

Thus in uH range you should start to use for example LC resonance as the grounding of the measurements. Add a well known series C or a parallel C. Have an external diode rectifier for your voltmeter. You can find the resonant frequency and solve the L from the equation of the reaonant frequency. Even the stray capacitances can be solved by adding 1...2 new known capacitors.

  • \$\begingroup\$ it may be advantageous if you plan to stick around to change your profile name to set you apart from the other users that don't intend to stick around. Its hard to tell users that say user followed by 6 digits apart. \$\endgroup\$
    – Voltage Spike
    Commented Mar 9, 2017 at 6:32
  • \$\begingroup\$ Thanks for your answer. I corrected my formulas according to your remarks. I also selected this as the answer since you gave the actual reason why this method is impractical (high frequencies not playing well with AC multimeters). \$\endgroup\$
    – ccrisan
    Commented Mar 9, 2017 at 7:44

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.