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Well, I have the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When the switch is closed for a long time, the current that is flowing trough the resistor \$R_1\$ and the coil is given by:

$$\text{I}_\infty=\frac{\text{V}}{\text{R}_1}=\frac{10}{100}=\frac{1}{10}\tag1$$

With that given I can solve what happens when I open the switch. The current trough the coil will stay in the same direction as when the switch is closed. Now, at the moment when I open the switch the voltage across the coil \$\text{V}_{\text{Y}_1}\$, is given as:

$$\text{V}_{\text{Y}_1}=-\text{V}\cdot\frac{\text{R}_1+\text{R}_2}{\text{R}_1}\cdot\exp\left(-\frac{\text{R}_1+\text{R}_2}{\text{L}}\cdot t\right)\tag2$$

So, the voltage over the coil will start (when I open the switch) at \$-\text{V}\cdot\frac{\text{R}_1+\text{R}_2}{\text{R}_1}=-10\cdot\frac{100+100}{100}=-20 V\$, which is (absolutely seen) higher than the supply voltage. How is that possible? Or is the formula \$(2)\$ wrong?

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There is nothing wrong.

Inductor will store some energy during the time time the switch is closed, which is given by

\$E_L = \frac{1}{2}LI^2\$

=380μJ in your case.

When the switch opens, inductor will develop some voltage on its terminals in order to keep the current constant. Obviously, over time the stored energy will dissipate on the resistors.

Remember that

  • Capacitors cannot change their voltage instantly, but can change their current instantly
  • The exactly opposite is true for inductors. Their current cannot change instantly, but their voltage can

We take advantage of this effect in Boost Converters (google it) in order to step up a DC voltage

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  • \$\begingroup\$ So, the voltage calculation is right? It exceeds the supply voltage (absolutely seen) \$\endgroup\$ – yuiop May 18 '19 at 14:17
  • \$\begingroup\$ I believe that your calculations are alright. I haven't done them myself, but there is nothing "weird" with \$V_L(0) > V_{DC}\$. It is a known effect that sometimes is to our benefit (boost converters) and sometimes not (controlling a DC motors with a switching mode converter). \$\endgroup\$ – thece May 18 '19 at 14:21
  • \$\begingroup\$ Yes I understand, thanks for your answer. Can you take a look at my newest question (electronics.stackexchange.com/questions/439158/…), maybe your answer here is also right in my new one. \$\endgroup\$ – yuiop May 18 '19 at 14:22
  • \$\begingroup\$ It's exactly the same effect. No need for separate answer. \$\endgroup\$ – thece May 18 '19 at 14:23
  • \$\begingroup\$ That was what I thought after your answer. But in my equation I linked I asked how they get the -90-volts \$\endgroup\$ – yuiop May 18 '19 at 14:25

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