0
$\begingroup$

I'm trying understand the physics of simple inductor-capacitor circuits such that there is just an inductor L and a cacpacitor C and a switch.

Imagine first that the capacitor is fully charged and the switch is then closed.

I do not understand why the current increases from an initial low value as the charge difference between the plates DECREASES because this is in direct contradiction to how a capacitor discharges in isolation.

I know the solution lies in the inductor being present but I can't seem to follow the physics of cause and effect to understand it properly.

Any illumination would be appreciated.

$\endgroup$
0
$\begingroup$

An inductor "resists" changes in the current through the energy required to build up the magnetic field. Much like a capacitor "resists" changes in voltage through the energy stored in the electric field between the plates.

If you connect an inductor and a resistor in series, you will get a charge/discharge curve for the current, as with a capacitor+resistor circuit, when you connect to a constant voltage/short circuit the circuit.

The slow build up of current through the inductor together with the decreasing voltage over the capacitor as more current is drawn is what gives the characteristic sinusoidal current. Incidentally the inductor's "resistance" to changes in current is also what makes the capacitor discharge "overshoot" beyond zero voltage, and thus continuing the sinusoidal current and (phase shifted voltage).

$\endgroup$
0
$\begingroup$

The instant the switch is closed the current in the circuit $I$ is zero because the changing current in the circuit $\frac{dI}{dt}$ induces an emf in the inductor $L\frac {dI}{dt}$ which opposes the exactly "opposes" the voltage across the capacitor $V = \frac QC$.

As time progress the voltage across the capacitor decreases and so must the the emf induced by the inductor $L\frac {dI}{dt}$ which in turn means that the rate of change of current in the circuit must decrease.

$L\frac {dI}{dt} = CV$ so as the voltage across the capacitor decreases so does the rate of change of current in the circuit whilst at the same time the current is increasing.

When you discharge a capacitor through a resistor the resistor does not induce an emf in opposition the the voltage across the capacitor, indeed as you have stated the current almost instantaneously reaches a maximum value and the decreases.


You will note that I used the words "almost instantaneously" because in the real world the capacitor and resistor circuit would also have inductance as the circuit is a loop.
This very, very small inductance would prevent the current having a finite value at the start but in practice you would not notice that because the change from zero current to maximum current would take place over a time which was much, much shorter than the time constant of capacitor-resistor circuit, $CR$.

$\endgroup$
0
$\begingroup$

Actually, LC Circuit is cause of LC oscillation,

When you apply kvl you get. $Q/C=Ldi/dt$, and $dQ/dt$ so after double differentiating we get

$Q/C=Ld^2Q/dt^2$, which look quite same as shm equation $a=-w^2x$,so we get $w=1/(LC)^1/2$, so energy oscillate,

At any instant instant energy will. Be equal. To. Field. =energy stored in capacitor +energy stored in inductor. Like we have in simple pendulum shm, I think if you know how actually inductor work you can easily answer your own question now.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.