-2
\$\begingroup\$

I recently watched this video from EEVblog showing a method of measuring inductance. I'm probably missing something obvious, but why does this work? Mathematical proof would be great.

Thanks.

https://www.youtube.com/watch?v=UrS5ezesA9s

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Please include all the information needed to answer the question in the question post. The external link might not work for readers a year or more from now. \$\endgroup\$
    – The Photon
    Apr 3 at 17:12
  • 3
    \$\begingroup\$ I am am bit hesitant to give this as an answer, but it works due to the April Fools effect. \$\endgroup\$
    – Justme
    Apr 3 at 17:20

1 Answer 1

2
\$\begingroup\$

It is explained in the video but perhaps not as clear as it might be. The video shows how a multimeter can be used to measure inductance when used on its capacitance ranges. The assumption is that the multimeter is measuring the impedance of the device under test and then calculating the capacitance based on a known test frequency. Thus if the frequency is f, the measured impedance is Z, the capacitance (C) is calculated as C = 1/(2πfZ) based on the capacitive reactance (impedance) being given by Z =1/(2πfC). If one then measures an inductance, L, with the same multimeter on a capacitance range, the meter is actually measuring the inductive reactance which is 2πfL. So Z = 2πfL. Thus L is equal to Z/2πf. We see that to calculate the L from the multimeter's measurement of C, we need to take the reciprocal. This is what he did in the video.

\$\endgroup\$

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