# Does quickly shorting out an inductor in series with a capcitor create a small emp?

Agnostic
Does quickly shorting out an inductor in series with a capacitor create a small emp?

If both elements are ideal and the capacitor is initially charged, you should have an undamped sinusoid, due to the change of energy stored in the electric field of the capacitor and the magnectic field of the inductor.
Foe real elements you will have a damped sinusoid.

Mentor
Agnostic said:
Does quickly shorting out an inductor in series with a capacitor create a small emp?
What are you shorting out? What are the initial conditions of the inductor and capacitor?

Agnostic
berkeman said:
What are you shorting out? What are the initial conditions of the inductor and capacitor?

Sorry for the lack of clarity.

A highly charged ideal physics capacitor and inductor in series by means of an ideal physics wire.

Mentor
Agnostic said:
Sorry for the lack of clarity.

A highly charged ideal physics capacitor and inductor in series by means of an ideal physics wire.
You still haven't specified the initial conditions. Do you mean that you have an inductor connected in series with a capacitor, and the capacitor has an initial voltage Vo on it, and the initial inductor current Io is zero? And then you complete the circuit with a switch to "short them out"?

Agnostic
berkeman said:
You still haven't specified the initial conditions. Do you mean that you have an inductor connected in series with a capacitor, and the capacitor has an initial voltage Vo on it, and the initial inductor current Io is zero? And then you complete the circuit with a switch to "short them out"?

A capacitor connected in series to an inductor. At t=t.x, the capacitor has been discharged as much as possible into the inductor.

Not switching a switch to short it out instantly. But shorting it out rather quicly, saying it takes like x nanoseconds to short it out.

Let me see if I understand. You mean you have an ideal oscillating circuit and, when the current in the inductor is maximum, you open the circuit, making the current go instantaneously to zero?
Since the voltage in the inductor is $$L\frac{di}{dt}$$, making $$dt = 0$$ causes the voltage go to infinity.
A physical inductor has associated to it a winding resistance and a winding capacitance, so in practice you will have a high frequency, high voltage damped oscillation.
If the amplitude of the voltage is high enough, the air in the open gap can be ionized and a spark appears. This is observable when you switch off a fluorescent illumination circuit.

Mentor
SGT said:
Let me see if I understand. You mean you have an ideal oscillating circuit and, when the current in the inductor is maximum, you open the circuit, making the current go instantaneously to zero?
Since the voltage in the inductor is $$L\frac{di}{dt}$$, making $$dt = 0$$ causes the voltage go to infinity.
A physical inductor has associated to it a winding resistance and a winding capacitance, so in practice you will have a high frequency, high voltage damped oscillation.
If the amplitude of the voltage is high enough, the air in the open gap can be ionized and a spark appears. This is observable when you switch off a fluorescent illumination circuit.
Or else maybe he means that he has the LC circuit you describe, but then he really does short out the inductor when it has maximum current going through it. This is a pretty benign short, since the voltage across the inductor is zero when the current is not changing (at the top of its sine wave current waveform). With that shorting wire in place across the inductor (and across the capacitor too, since the L and C are connected together), the current in the inductor will keep circulating through the inductor and wire, and will ramp down slowly with the L-R time constant, where the R is the resistance of the inductor windings and the shorting wire.

Open circuiting a current-carrying inductor is definitely more interesting than shorting it out. :rofl: