I've the following circuit:
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.
Best Answer
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.