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I've the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

The switch was 'forever'closed and a certain time \$t=0\$ I open the switch. I tried to find the energy of the coil when I opend the switch over all time that the switch is open:

$$\text{E}_\text{total}=\int_0^\infty\text{P}_\text{L}\left(t\right)\space\text{d}t=\int_0^\infty\text{V}_\text{L}\left(t\right)\cdot\text{I}_\text{L}\left(t\right)\space\text{d}t=$$ $$\int_0^\infty\text{I}_\text{L}'\left(t\right)\cdot\text{I}_\text{L}\left(t\right)\cdot\text{L}\space\text{d}t=\lim_{\text{n}\to\infty}\left[\frac{1}{2}\cdot\text{L}\cdot\text{I}_\text{L}\left(t\right)^2\right]_0^\text{n}=$$ $$\lim_{\text{n}\to\infty}\left(\frac{1}{2}\cdot\text{L}\cdot\text{I}_\text{L}\left(\text{n}\right)^2-\frac{1}{2}\cdot\text{L}\cdot\text{I}_\text{L}\left(0\right)^2\right)\tag1$$

And I found that that is equal to:

$$\text{E}_\text{total}=-\frac{\text{LV}^2}{2\text{R}_1^2}\tag2$$

Where \$\text{V}\$ is the source voltage.

But that means that the energy in the coil is NEGATIVE when I open the switch? Is that possible or logical? I do not see why.

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  • \$\begingroup\$ I'm sorry, I'm mistaken... ill remove my comments \$\endgroup\$ – Huisman Jun 15 '19 at 12:25
  • \$\begingroup\$ @Huisman Haha done! \$\endgroup\$ – Topil Jun 15 '19 at 12:27
  • \$\begingroup\$ The reason the coil energy is negative is because the coil is sourcing power. If you do the same for the resistors, their energy is positive, because they're dissipating it. \$\endgroup\$ – Huisman Jun 15 '19 at 12:28
  • \$\begingroup\$ @Huisman But the capacitor is that too, but it gave positve energy. \$\endgroup\$ – Topil Jun 15 '19 at 12:29
  • \$\begingroup\$ The sign of the current reverses for the capacitor as it is decharging. That should yield negative energy as well. \$\endgroup\$ – Huisman Jun 15 '19 at 12:32
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The reason the coil energy is negative is because the coil is sourcing power.
If you do the same for the resistors, their energy is positive, because they're dissipating it.
The energy for the capacitor is negative as well, because the sign of the current reverses for the capacitor as it is decharging.

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I am not sure I 100% understand what you want to achieve with your integral calculation, or at least, it does not exactly match your explanation.

The energy in the coil right when the switch is opened, at \$t=0^+\$, can be calculated with the usual \$E_{0^+}=\frac{1}{2}LI_{0^+}^2\$, and this leads to your result, with a plus sign though.

Your integral, as you wrote it, can be used to calculate the energy that is absorbed by the inductor between \$t=0\$ and \$t\to+\infty\$. Notice the word absorbed: since the inductor is not absorbing any energy, but rather emitting it, your result is less than zero, and since it is equal to the total stored energy at \$t=0\$ we can conclude that there will be no energy on the inductor when time approaches infinity.

Just so you know, the sign you are getting is right because you are using the passive sign convention, where positive current flows into the positive terminal, and positive power is absorbed by your device.

If you use the active sign convention instead, then you get a positive energy.

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