How to check for short circuit in coil

I have 70 turns of enameled copper coil, 1.25 mm diameter and 25.4 m in total.

I wound it on top of eacher and soaked the whole thing in a thermally conductive compound. What I fear is that by pulling the coils while winding them, I scraped some of the enamel away. I am not sure how resistant it is. That would cause some of the coils to come in contact with each other thus short circuiting the whole thing.

How can I check for this?

Should I see a fluctuating resistance on a multimeter? Is there any obvious tell?

• Resistance changes may be too small to measure easily. One shorted turn will reduce the resistance by 1/75 of an already low resistance. But if you are equipped to measure its inductance, a shorted turn (tight round this coil but not necessarily even connected to it) should cause a significant drop in inductance.
– user16324
Commented Jul 9, 2016 at 9:37

Your copper wire, based on the measurements given, would have a theoretical resistance of:

$$R=\frac{25.4\mathrm{m}}{\pi \times (0.5\times1.25\times10^{-3}\mathrm{m})^2}\times1.7\times10^{-8} \Omega\cdot\mathrm{m} \approx 0.352\Omega$$

Measuring this resistance accurately may be difficult with the multimeter you have on hand and would have required measuring the resistance of the fresh wire before winding. If your multimeter is not accurately calibrated and/or is not accurate at this resistance, you can't use the theoretical calculation to 'see' a short inside your coil. If you can repeatably measure a value from the coil, you could unwind the whole coil and compare the loose wire resistance to the coil resistance, but that sounds difficult with your thermal compound.

If your are using this coil for fairly low voltage applications (<60VDC) and don't need a lot of reliability out of it, you probably didn't scrape the enamel in two places such that a short inside the coil has formed. If you scraped the wire a lot or need to use it for a long time, you would probably want to make a new coil.

To answer your other questions: you probably won't see a fluctuation in resistance on the multimeter; that would require an intermittent short circuit, which is fairly unlikely and would be difficult to accurately detect. Unfortunately, there is not an obvious 'tell' that there is a short circuit in a coil like this.

• Isn't the resistivity of copper $1.7 \times 10^{-8}$? Also I have to divide $1.25$ by $2$ right? Commented Jul 10, 2016 at 21:59
• What do you mean with your comment about the voltage? Is a short only becoming apparent at high voltages? I am using a current controlled power supply, 8 A and something like few or less than a couple of V. Commented Jul 10, 2016 at 22:04
• @SuperCiocia you have to use $. With the corrections, this makes it even more difficult to check for 'wrong' resistances due to the lower desired resistance. 8A will get you some voltage, so that could be used to back-calculate if your power supply has an accurate current meter and you use your DMM in volts. Commented Jul 11, 2016 at 2:09 A shorted turn in an inductor will present as either very low inductance or as having very low "Q" factor - or both. Measuring the inductance is fairly simple and easy. You can either calculate the expected inductance and see if the measured value is close to that or wind a similar size coil but with fewer turns. You can then calculate the expected inductance of your big coil based on the actual measurement of the test coil with few turns. Ask if you need help with any of these. • Hm, wouldn't one shorted turn just act as if the coil had one less turn? I'm not questioning the factuality of the statement but "very low inductance" makes it sound like it's changing by a huge factor. If that's true, maybe you could add a short explanation of why one shorted turn upsets the whole coil? – pipe Commented Jul 12, 2016 at 0:46 According to what other answers state, you could measure inductance since I suppose you have compute the number of turns in the coil to achieve a concrete inductance value. A simple way to do it is attaching in series (or parallel) a capacitor and look for the resonant frequency. Given the nominal value of the capacitor, you could figure out the value of the inductor as: L = \frac{1}{4\pi^2 f^2_0 C}  Where$f_0\\$ is the resonant frequency and C the nominal value of the capacitor.

If you detect that there has been a noticeable drop in inductance, a short circuit in the turns may have occurred.