When the capacitor is fully charged there is a potential difference between its poles and that creates a current.

This current would create a magnetic field that is changing in the Inductor (because the current changes due to the capacitor), creating an EMF in the circuit.

Wouldn't this inductor's emf counteract the discharging capacitor and actually charge it? / stop the capacitor from fully discharging?

From what I understand as the current from the capacitor decreases exponentially, the EMF created by the inductor would increase exponentially.

At some point the EMF of the inductor should be bigger than the voltage difference between the poles of the capacitor - and that may not be when the capacitor is fully discharged.

Can someone explain why then the capacitor does fully discharge before the Inductor starts creating a current in the opposite direction?


Wouldn't this inductor's emf counteract the discharging capacitor and actually charge it? / stop the capacitor from fully discharging?

The inductor doesn't care about what the charge state of the capacitor is. All it cares about is how quickly the current through it is changing, and it generates a back-voltage according to the equation V=L*dI/dt. You can think of an inductor as giving "momentum" to the current. If the current is zero, then it wants to keep the current zero. If the current is non-zero, it wants to keep the current at that same non-zero value. If the current is increasing, it generates a counter-voltage acting in the opposite direction to the current flow.

The analogy I like to use is a circuit of water pipes in which inductors are represented by a heavy propellor in a water pipe. If water flow is suddenly turned on, the heavy propellor initially resists the flow of water. But over time the propellor spins faster in response to the water flow. If the water flow past the propellor is then reduced, the heavy propellor resists the decrease in water flow because it is now spinning fast and tries to continue pushing the water through the pipe. This is analogous to how an inductor resists changes in the electrical current flow through it.

Using this water circuit analogy, a capacitor can be represented as a section of pipe which has a rubber membrane stretched across the inside of it. If you push water into one end of this pipe section, the rubber membrane stretches and creates a back pressure resisting attempts to push more and more water into it. If you then stop applying water pressure to that side of the pipe section, the rubber membrane springs back to its flat, equilibrium position, pushing the water back out the same side of the pipe as you were trying to push the water in. This is analogous to how a capacitor "pushes back" with a back-voltage when you push electrical charge into a capacitor.

If you make a closed electrical circuit with this heavy propellor (which represents the inductor) and the rubber-membrane pipe section (which represents the capacitor), then you should be able to see how a resonant water oscillation in the circuit can be set up. Imagine the rubber membrane pipe section being "charged" by forcing water into one side. When you release the applied pressure, water will flow past the heavy propellor, which will then speed up and try to maintain a constant water flow past it. However, as the water flows past the heavy propellor and into the other side of the rubber membrane pipe section, the rubber membrane goes to its equilibrium position and then starts getting stretched in the opposite direction. Eventually, the back-pressure becomes so large that the direction of water flow is reversed and the cycle happens all over again.

In summary, with this analogy we have the following:

electrical current      <-> water flow
voltage                 <-> water pressure
inductor                <-> heavy propellor
capacitor               <-> rubber membrane pipe section

Hopefully, visualizing things this way can give you an intuitive grasp of how a capacitor and inductor work together to form a resonant circuit.

  • $\begingroup$ This is a great explanation of what happens. So when the current is increasing the inductor is slowing it down, but when the current is decreasing the inductor is speeding it up. $\endgroup$ – Ulad Kasach Nov 4 '15 at 1:15
  • $\begingroup$ @Vlad K - That's basically right. If the current is increasing then the inductor is generating a counter-voltage to try to keep it from increasing. If the current is decreasing then the inductor is generating a counter-voltage to try to keep it from decreasing. It's similar to what the heavy propellor in a water pipe tries to do with water flow. Both the inductor and the heavy propellor try to maintain the status quo for the electrical current and water flow, respectively. $\endgroup$ – user93237 Nov 4 '15 at 1:28

Just a simple answer - there's nothing to dissipate energy. If there were a resistor in the circuit, it would dissipate energy as heat. Inductors and capacitors don't dissipate energy. The energy just sloshes back and forth between being stored in the magnetic field, and being stored in the electric field. It's just like a spring-mass system, where energy goes back and forth between being stored in the spring, and being stored in kinetic energy of the mass.

  • $\begingroup$ That part I understood, but I was confused as to why an equilibrium between the Inductor due to the current/emf its changing magnetic field induces and the current the capacitor creates. $\endgroup$ – Ulad Kasach Nov 4 '15 at 1:10
  • 3
    $\begingroup$ @VladK, if there is a voltage across an inductor, the inductor current must be changing; if the there is a current through a capacitor, the capacitor voltage must be changing. To reach an equilibrium, the voltage across and current through each must be zero at the same time. But this would mean zero energy and, as Mike points out, this is impossible if there is no dissipation in the circuit and there is initial energy (and you've specified that there is initial energy stored in the capacitor). $\endgroup$ – Alfred Centauri Nov 4 '15 at 1:15

The inductor never creates a current in the opposite direction. An inductor creates an EMF to counteract the changing B field(Lenz law). The B field is changing because the current in the inductor is changing. So effectively, the inductor resists changes in current. So initially, the capacitor tries to discharge strongly but is slowed down by the inductor. Once the capacitor is able to drive current, that current doesn't want to stop. So once the capacitor is completely discharged current is still flowing since the inductor is resisting a change in current. Eventually this current reverses the charge on the capacitor which slows to current down until it is 0. At this point, if no energy is loss, the capacitor is charged with exactly the opposite charge since all the energy at the beginning and end of this half cycle and is on the capacitor and is equal. So then the reverse happens, your back to the beginning and the cycle repeats.

  • $\begingroup$ But wouldn't the EMF the inductor creates cause a current of its own? If it is just the changing magnetic field just inducing a current in the wire, what effect does the EMF established even have on the circuit? $\endgroup$ – Ulad Kasach Nov 4 '15 at 0:52
  • $\begingroup$ The EMF the inductor creates slows the charge pushed off the capacitor down. As the charge moves through the inductor, it has to go up a potential ramp(the induced EMF) losing kinetic energy until it comes to rest on the opposite end of the ramp. $\endgroup$ – Shane P Kelly Nov 4 '15 at 2:29

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