Charging and Discharging of a capacitor in an LC circuit

  • #1

Main Question or Discussion Point

In an LC circuit, the capacitor that is initially charged to a finite value starts to discharge cross the inductor, initially the current increases and the inductor opposes it, but as the current is supplied against the back emf, due to the discharging of the capacitor, wont it reduce the value of current flowing in the circuit and cause the inductor to suport the current flow and help completely discharge the capacitor. How is it possible that when the capacitor is fully discharged, when the current is zero to have the entire energy to be stored in the induction, when the current is zero, there is no magnetic field and the how can the inductor store energy in that case.
 

Answers and Replies

  • #2
gneill
Mentor
20,795
2,773
There is still a magnetic field when the capacitor reaches zero charge. In fact the current is then at a maximum through the inductor, and its magnetic field magnitude is also at maximum. The inductor keeps the current flowing and begins re-charging the capacitor (but with the opposite polarity) until the magnetic field has entirely collapsed. All the energy is then stored in the capacitor again and the second half of the cycle begins just as the first did, only with the opposite direction of current flow.
 
  • Like
Likes tech99 and Nikhil Rajagopalan
  • #3
anorlunda
Staff Emeritus
Insights Author
8,585
5,472
Your LC circuit would make an oscillator. Energy oscillates between the L and the C. Both V and I swing positive and negative. Peaks of current flow are 90 degrees out of phase with peaks of voltage. In a real world circuit, there is nonzero R, so the oscillation eventually dies out.
 
  • Like
Likes Nikhil Rajagopalan
  • #4
Thank you gniell, How can the current be maximum when the charge is depleting, When the charge is draining out of the capacitor, the graph of current is expected to be falling down exponentially?
 
  • #5
gneill
Mentor
20,795
2,773
By that point the capacitor is not driving the current, the inductor is.
 
  • #6
Thank you gniell, at what point does the hand over take place? Initially, the capacitor drives the current flow, against the back emf generated by the inductor. How does the current and the back emf vary here?
 
  • #7
anorlunda
Staff Emeritus
Insights Author
8,585
5,472
90-450x249.gif


Think of the blue line as voltage, the dotted line as current, and the red/green shaded part is power (flow of energy).
 

Attachments

  • Like
Likes gneill and Nikhil Rajagopalan
  • #8
29,557
5,884
When the charge is draining out of the capacitor, the graph of current is expected to be falling down exponentially?
This is not correct. When the charge is draining out of the capacitor the current is increasing. The current reaches its peak when the charge reaches 0. Then the current starts to decrease as the charge goes negative.
 
  • #9
Thank you anorlunda, I highly appreciate your help. Is there a resource I can refer to which explains the details of the process.
 
  • #10
Thank you Dale. So is the exponential decrease in current only a characteristic of an RC discharge? In case of a pure capacitor shorted, does the discharge current obey any other function?
 
  • #11
29,557
5,884
There is Wikipedia
https://en.m.wikipedia.org/wiki/LC_circuit

And the farside site at UT
http://farside.ph.utexas.edu/teaching/315/Waves/node5.html

I am sure there are other good sites too

Thank you Dale. So is the exponential decrease in current only a characteristic of an RC discharge? In case of a pure capacitor shorted, does the discharge current obey any other function?
Yes, exponential decay comes from a RC circuit not a LC circuit. A short is just a very small R, so it behaves as a RC circuit with a very short time constant.
 
  • #12
gneill
Mentor
20,795
2,773
An RC circuit is an example of a first order circuit as it contains just one reactive component (the capacitor). It can be "solved" with a first order differential equation, and exhibits the exponential decay that you mention. An LC circuit is an example of a second order circuit as it employs two reactive components (L and C). It can be solved using a second order differential equation, and in general (when resistance is also present) it will exhibit both an exponential decay *and* some form of oscillation function depending upon whether the resistance makes the circuit underdamped (oscillation), critically damped or overdamped (no oscillation but not strictly exponential in shape.

You might want to look at the wikipedia page for the LC circuit. It shows the math.

edit: Ah! Dale scooped me on the wikipedia reference!
 
  • #13
Thank you Dale and gniel . I am fine with the mathematical analysis of the LC circuit. My difficulty is in visualizing and comprehending the rise and fall of charge, current and potentials across the components in a logical sequence. As of how when charge is draining out from a reservoir, the rate of charge being shipped out is increasing when the potential which should drive the charge is decreasing.
 
  • #14
Initially, when the charges start flowing out of the capacitor, the current is trying to increase from zero. So the back emf kicks in. So now, we have the charges flowing under the influence of (Voltage across capacitor - Back EMF), from here on, is it correct to assume that the rise in current is delayed and when does it attain the peak and what would the peak value be.
 
  • #15
gneill
Mentor
20,795
2,773
Perhaps think about an analogous example from mechanics: the mass-spring system. You begin by putting some potential energy into the system (stretch the spring). When you release the mass it begins to accelerate at a rate determined by the spring force and inertia of the mass. When the equilibrium length of the spring is reached the mass is moving at its fastest and begins driving energy back to the spring at the greatest rate. It passes the equilibrium point and continues to drive up the potential energy in the spring.
 
  • Like
Likes sophiecentaur
  • #16
29,557
5,884
the potential which should drive the charge
In this circuit the potential doesn’t drive the current, it drives the change in the current. You appear to be thinking of the inductor as if it were a resistor, but it isn’t.
 
  • Like
Likes Nikhil Rajagopalan
  • #17
Thank you gniell and Dale. The inductor here can only respond to change in current and when the switch is turn on, there is charge coming out of the capacitor which initiates a current and inductor readily opposes this change from zero producing a back EMF, how does this back emf affect further discharging? How will the discharging happen in this case. will the charges flow faster or slower?
 
  • #18
I sincerely apologize to Dale gniell and anorlunda about my inability to grasp it quickly. and i strongly appreciate the help and it means a lot to me. I think it is partly because i am unable to phase my question well.

Let me put it this way. When the charge is draining out of the capacitor, the only reason I see for current to be increasing is because the current was initially zero in the circuit and now since the charges come out of the capacitor, the current is building up to a non zero value from zero. Any change in current is opposed by the inductor.
1.How does the current increase from there after.
2.How does the maximum value of current correspond to a situation where there is hardly any charge on the capacitor or any potential difference across the capacitor. Where is that charge situated at that point.
 
  • #19
jbriggs444
Science Advisor
Homework Helper
2019 Award
8,764
3,527
It may be helpful to think by analogy with a mass on a spring.

The capacitor is like a spring. The higher the capacitance, the weaker the spring.
The inductor is like a mass. The higher the inductance, the greater the mass.
The potential across the capacitor is like the force from a spring.

Charging a capacitor is like displacing the mass away from its neutral point. You can either give it a positive charge (compressing the spring) or a negative charge (extending the spring). As you put more charge into the capacitor, the resulting potential difference increases (the spring resists more strongly as you compress it). When the capacitor is fully charged, you open a switch (latch the mass into place with the spring compressed).

When you close the switch, current begins to flow (you release the latch and the mass begins to move). The rate of change in the current is inversely proportional to the inductance (the acceleration is inversely proportional to the mass of the mass). It is directly proportional to the potential difference (it is proportional to the compression of the spring). The resulting current begins to deplete the capacitor (the spring begins to relax).

When the capacitor is fully discharged, the current has reached its maximum value (the mass is moving at its maximum speed).

Current keeps flowing (the mass keeps moving). The capacitor is recharged (the spring is stretched). The rate of change of the current becomes negative (the mass slows down while stretching the spring).

It is simple harmonic motion.

Edit: Missed seeing @gneill propose the identical analogy.
 
Last edited:
  • Like
Likes Nikhil Rajagopalan
  • #20
jbriggs444, Thank you for the kindness in taking the effort to explain it in detail. The barrier i find in understanding it is in any case of capacitor discharging, the current is suppoused to decrease, but here, it has to first increase because it is starting from zero and the inductor makes the increase slower, and then it has to decrease defenitely because the charge on the capacitor is depleting.
1.So how does the point of this maximum current coincide with a point where there is maximum current.? Because if current has to be high, it should be in a situation where there is a high amount of charge flowing per second. How will that be during the final stage of discharging? Shouldn't the completely discharged situation in the capacitor coincide with zero current, at the end of the decrease following the initial increase of current.
2.What effect does the induced emf have on this process. Is there a tug off between the induced emf and the potential difference across the capacitor Also is it theoretically wrong to say that the induced emf is set up across the inductor. Is it the induced emf of the entire circuit.
 
  • #21
jbriggs444
Science Advisor
Homework Helper
2019 Award
8,764
3,527
The barrier i find in understanding it is in any case of capacitor discharging, the current is suppoused to decrease
The current (the speed of the mass on the spring) is supposed to increase all the way to the neutral point (capacitor discharged/spring relaxed).

At the neutral point, all of the original potential energy in the capacitor (or spring) now manifests in the inductor (or as kinetic energy of the mass).
 
  • #22
Thank you jbriggs444 . I attribute that increase in speed to the fact that work is being done on the mass by the restoring force in the direction of displacement. In this case, how can current or number of charges flowing per second increase if the charges are flowing along the gradient and the gradient is decreasing as this happens. I believe logic behind my question is probably flawed by missing the role of the induced emf and the inductor in this context.
 
  • #23
Hope I can put it this way, If the capacitor was across a resistor, the current would have been zero just after the capacitor was fully discharged. What differences does an inductor replacing the resistor do the the discharging process of the capacitor. How is it so different that when the capacitor is fully discharged, the current in the circuit is maximum.
 
  • #24
jbriggs444
Science Advisor
Homework Helper
2019 Award
8,764
3,527
A resistor resists current. An inductor resists changes in current.
 
  • Like
Likes Nikhil Rajagopalan
  • #25
gneill
Mentor
20,795
2,773
In this case, how can current or number of charges flowing per second increase if the charges are flowing along the gradient and the gradient is decreasing as this happens
In the spring, how can the velocity increase as the driving force tends to zero near the equilibrium point? The thing is, the force does not actually reach zero until exactly at the equilibrium point. The same is true for the capacitor voltage.
 
  • Like
Likes Nikhil Rajagopalan
Top