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I am trying to measure the current in dependence of time in turning on and off an inductor.

If I choose an inductor with low inductivity (36 mH) it works for turning on (almost) as expected, but for turning off it doesn't.

Here is my setup:

enter image description here

and here is are the results (red curve for turning on, black curve for turning off):

enter image description here

I measured the resistance of the current path for turning off with a voltmeter to about 10 ohms. In the case of turning on, the resistance is also approximately 10 ohms as you can see from the current curve (in the case 5,2 volts).

What did I do wrong and how can I fix it?

Edit

Here are details of the setup:

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  • $\begingroup$ Could you clarify which parts you think are not correct? $\endgroup$ – K_inverse Sep 6 '18 at 9:39
  • $\begingroup$ I expect that in turning off it is also an exponential curve with the same time constant as in the case of turning on, but clearly the experimental result looks different. $\endgroup$ – Julia Sep 6 '18 at 9:49
  • $\begingroup$ Could you edit your first figure (the experimental setup) with label, say which one is inductor, which one is the voltmeter ...etc and share other parameters in your setup? $\endgroup$ – K_inverse Sep 6 '18 at 9:57
  • $\begingroup$ @K_inverse: Does my edit make the setup clearer? $\endgroup$ – Julia Sep 6 '18 at 10:48
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You need to provide a path for your current when the switch is open. See picture for details. At step 1, you have current. At step 2, your current dies out. So at step 3, when you expect your current, it is not there.

If you fix the circuit as in step 4,5,6, you are fine. enter image description here

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