Electronic – LR oscillators (no capacitor)

The inductor starts discharging the voltage across the capacitor. At some point later that voltage is zero but the current is not zero. The energy in the capacitor due to its initial voltage is totally transferred to the inductor in the form of current.

Now we have the reverse situation. That inductor energy starts pumping back into the capacitor and, after the same period of time, the energy is all taken back to the capacitor in the form of a reversed voltage. The cycle repeats.

Go look up the energy equations for a capacitor and inductor and try and visualize this. Also look up the base equations that relate current and voltage to inductors and capacitors respectively.

You can also think of it like a flywheel tethered with an elastic rope. You can turn the flywheel so as to build up tension in the rope. Then let go. The flywheel starts turning the opposite way and at some time later the elastic rope has no tension in it. Because the flywheel has mass it continues spinning but is slowed down by the rope re tensioning. Eventually, the flywheel stops but, because there is energy in the rope the flywheel starts building up speed in the opposite way.

The equations are pretty much the same for this mechanical analogy and the LC circuit.

Assuming a unit impulse current is applied at \$t=0\$, Laplace transform analysis gives:

\$I_C=-\omega \:sin(\omega t)\$

\$I_L=\omega \:sin(\omega t)\$

and voltage across combination:

\$V=\frac{1}{C}\:cos(\omega t)\$

where \$\omega =\frac{1}{\sqrt{LC}}\$

L and C currents are sinusoidal and have 180deg phase difference, so there's zero overall current flowing into the combination, but there's a sinusoidal current of \$\omega\:sin(\omega t)\$ circulating through L and C. Also, there's a (co-)sinusoidal voltage across the combination. At \$t=0\$, the capacitor is charged instantaneously to \$V=\frac{1}{C}\$ by the unit impulse current, hence the cosine voltage function.