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for the purpose of calculating the inductance of my unlabeled inductors I try to use equation for impedance of a series connection of inductor and resistor:

$$ Z = \sqrt{(R+R_l)^2 + (\omega l)^2}$$ $$ Z^2 = (R+R_{l})^2 + L^2\omega^2$$ $$ L = \sqrt{\frac{Z^2 - (R+R_{l})^2}{\omega^2}} $$ making it more usable in practice: $$ L = \sqrt{\frac{(\frac{U}{I})^2 - (R+R_{l})^2}{(2\pi f)^2}} $$ Ok so now, having the equation for L it's time do put values into it. Here is my test circuit:

enter image description here

Function generator creates 100kHz sin function with Voltage peak to peak equal to 4Vpp. Across the resistor I have scope probe for current measurement - I simply read the Vpp across resistor and divide it by known resistance to get current. Putting all of these vales into equation I get:

$$ L = \sqrt{\frac{(\frac{4}{\frac{2.55}{200}})^2 - (200+1.5)^2}{(2\pi 100000)^2} = 0.000395} $$

WolframAlpha link here

But this is wrong, my indycatnce meter says 0.987 uH which is close to its true inducatnce of 1mH.

So, what is wrong in my calculation?

Thanks for help :)

UPDATE

It looks like it works only if resistor voltage is half the inductor voltage. For example if resistor voltage is 2V and inductor voltage is 4V then putting all values into equation for L gives reasonable, quite accurate result. Any other way result is wrong. Can we explain it?

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  • $\begingroup$ The way you have connected your scope, you are creating a ground in the circuit. Can you confirm that there is no problem with grounding multiple points in your circuit? Usually you might want to use differential probes ("A-B" on the scope) to prevent this problem. $\endgroup$ – Floris Mar 17 '18 at 17:26
  • $\begingroup$ Also - is there any stray capacitance you need to worry about? At 100 kHz that may be significant. $\endgroup$ – Floris Mar 17 '18 at 17:26
  • $\begingroup$ Possible migration to Electrical Engineering SE as it seems more of a practical engineering problem than a theoretical one. $\endgroup$ – StephenG Mar 17 '18 at 18:21
  • $\begingroup$ In my judgment, this question is more or less equally at home at PSE and EESE. This kind of setup is often found in an undergraduate physics lab as well as an undergraduate EE lab. $\endgroup$ – Alfred Centauri Mar 17 '18 at 21:39
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For simplicity, set $R_L = 0$ and note that the (AC, phasor) voltage across the resistor is given by

$$V_R = V_s \frac{R}{R + j2\pi fL}$$

For $f = 100\,\mathrm{kHz},\quad L = 1\,\mathrm{mH},\quad R = 200\,\Omega$, the magnitude of the voltage across the resistor is then given by

$$|V_R| = |V_s|\frac{200}{\sqrt{200^2 + 4\cdot 10^4\cdot\pi^2}} \approx |V_s|\cdot 0.3033$$

If you insert this result into your formula, you find that

$$L = \sqrt{\frac{\left(\frac{200}{0.3033}\right)^2 - 200^2}{(2\pi\cdot10^5)^2}}\,\mathrm{H} = 1\,\mathrm{mH}$$

So your formula is correct. But look, if your source voltage is $4\,\mathrm{V_{pp}}$, then the resistor voltage should be less than a third of that or $1.213 \,\mathrm{V_{pp}}$. If I understand your post correctly, you're measuring more than double that at $2.55 \,\mathrm{V_{pp}}$.

Have you actually measured the voltage at the output of the signal generator?

If you haven't, then one possibility that comes to mind is that you have the signal generator set to $50\,\Omega$ output impedance. In that case, the voltage at the output of the generator will be higher than the output voltage setting since the load impedance is considerably larger than $50\,\Omega$.

In fact, if this is the case, and assuming the values of $f,L,R$ you've given, the output voltage would be greater than the output setting displayed by a factor of about 1.859 and the voltage across the resistor would be approximately $2.255\,\mathrm{V_{PP}}$ (which is in better agreement with the voltage I assume you measured).

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  • $\begingroup$ Thanks for answer @Alfred Centauri. Your suggestion to check source voltage was correct, it is not always what generator says. I managed to calculate two inductors inductance correctly however to get accurate result, the voltage on resistor must be half of the inductor volatge. Why is that? $\endgroup$ – DannyS Mar 17 '18 at 22:35
  • $\begingroup$ Also, you have an error in the numerator of equation for L - $$(\frac{200}{0.333})^2$$ is definitely wrong. Plus "the output setting displayed by a factor of about 1.859" where did yoy get this value? $\endgroup$ – DannyS Mar 17 '18 at 23:03
  • $\begingroup$ @DannyS, (1) it's $\left(\frac{200}{0.3033}\right)^2$ in the first term in the numerator (and not what you've written in your comment) and (2), there's no error in the numerator. $\endgroup$ – Alfred Centauri Mar 18 '18 at 2:15
  • $\begingroup$ how is that possible? Inside this parenthisis shall be Z, impedance or simpler V/I. You put there resistance over 0.3033 - what is that? Also could you please answer my first question? ;) $\endgroup$ – DannyS Mar 18 '18 at 8:38
  • $\begingroup$ @DannyS, well, for one thing, the unit of impedance is ohms and so you should expect a resistance over some dimensionless quantity. In my answer, I wrote that the (magnitude of) the voltage across the resistor should be $0.3033 |V_s|$. The series current is thus $|I| = 0.3033|V_s|/200$ and it follows that $$|Z| = \frac{|V_s|}{|I|} = \frac{|V_s|}{0.3033|V_s|/200}=\frac{200}{0.3033}$$ $\endgroup$ – Alfred Centauri Mar 18 '18 at 13:05

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