# how would inductor react if I inject current impulse into parallel LC tank?

What would happen if I inject current impulse into parallel LC tank? How would current flowing through inductor look like over time?

If we look at the delta function in the S-domain (laplace transform), energy is uniformly distributed over s-domain. This means, delta function cannot be just treated as high frequency signal.

This means, some of current impulse will flow through capacitor, and some will flow through inductor.

Since they are LC tanks (let's assume they are ideal LC tanks), then oscillation will occur. This case, which will react to this oscillation first: inductor? or capacitor? would inductor start dumping current into capacitor first? or would capacitor start dumping current to inductor first?

Thanks,

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.

Injecting a current impulse onto a L-C parallel tank is like hitting a bell with a hammer. If nothing was going on before (voltages and currents were 0), then this will start the tank ringing.

The inductor can't change it's current instantly, but the capacitor can (I'm assuming we're talking theoretical ideal components here). The impulse will have no immediate affect on the inductor's state.

A current impulse on a capacitor causes a step change in voltage. Depending on whether this step adds or subtracts from the existing voltage, it either adds or subtracts energy from the system.

For example, if you happen to hit the system with +1 V when the capacitor has -1 V on it and the current is 0, then you've just removed all the energy from the system and now everything will just sit there at 0. On the other hand, if the voltage was +1 V and the current 0, you've now doubled the voltage and quadrupled the energy of the system. The amplitudes of the voltage and current sines will be twice what they were before.

If it's an impulse, then it will all flow in the capacitor. The indctor cannot change its current instantaneously.

Best way to characterize a current impulse is the total charge dumped (integral of current.dt). Basically, you end up the same as if you connected a charged capacitor (ith Q = dumped charge) across an inductor -- that starts the ringing.

What would happen if I inject current impulse into parallel LC tank? How would current flowing through inductor look like over time?

If you injected a current impulse into a parallel capacitor and inductor then all that impulse of current will flow through the capacitor. None will flow thru the inductor but, gradually things will change.

At this point (and based on the current being an impulse) I can assume that following that impulse there will be an open circuit from the driving source. I can make this assumption because you said "inject current impulse" and this assumes that the only source is that impulse of current and thereafter there exists an open circuit from the source.

This means, some of current impulse will flow through capacitor, and some will flow through inductor.

No, absolutely wrong. Current will only flow into the capacitor because an inductor will reject changes in current in those few sub-fempto seconds and the capacitor will take ALL the current and form a voltage across its plates.

So, the cap will take the energy from the impulse and become charged to some voltage. The inductor will barely have noticed anything had happened if you want to put humanesque viewpoint on things.

Shortly afterwards (a few more fempto seconds) the source is out of the equation and the charge in the capacitor becomes sapped by the inductor and forever, until the end of time, there will be a perfect oscillating intechange of energy between capacitor and inductor forming a sinewave of frequency $\dfrac{1}{2\pi\sqrt{LC}}$