Charging and Discharging of a capacitor in an LC circuit

[Nikhil Rajagopalan]

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, won't 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.

 

[gneill]

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.

 

[anorlunda]

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.

 

[Nikhil Rajagopalan]

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?

 

[gneill]

By that point the capacitor is not driving the current, the inductor is.

 

[Nikhil Rajagopalan]

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?

 

[anorlunda]

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

 

[Dale]

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.

 

[Nikhil Rajagopalan]

Thank you anorlunda, I highly appreciate your help. Is there a resource I can refer to which explains the details of the process.

 

[Nikhil Rajagopalan]

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?

 

[Dale]

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.

 

[gneill]

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!

 

[Nikhil Rajagopalan]

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.

 

[Nikhil Rajagopalan]

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.

 

[gneill]

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.

 

[Dale]

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.

 

[Nikhil Rajagopalan]

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?

 

[Nikhil Rajagopalan]

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.

 

[jbriggs444]

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.

 

[Nikhil Rajagopalan]

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.

 

[jbriggs444]

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).

 

[Nikhil Rajagopalan]

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.

 

[Nikhil Rajagopalan]

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.

 

[jbriggs444]

A resistor resists current. An inductor resists changes in current.

 

[gneill]

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.

 

[Nikhil Rajagopalan]

Thank you jbriggs444, so, the final period of discharging and the initial stage of the second round of charging are the highest current times. I am wondering how the capacitor manages to discharge itself completely exactly when the current attains the maximum.

 

[jbriggs444]

You understand that the maximum point on a graph is also the point where its first derivative is zero?

The current is increasing right up until the capacitor is completely discharged because, until then, the capacitor is not yet fully discharged.

 

[Nikhil Rajagopalan]

Is it true to say that during the entire time of discharging, the back emf works against the flow of charge. and the current is always increasing.

 

[Nikhil Rajagopalan]

You understand that the maximum point on a graph is also the point where its first derivative is zero?

Yes jbriggs444, yes i do. So, when the current reaches maximum, the back emf should come down to zero.

 

[jbriggs444]

Is it true to say that during the entire time of discharging, the back emf works against the flow of charge. and the current is always increasing.

The back EMF is given by the product of the inductance times the rate of change of current over time. Yes, as long as current is increasing, back EMF works to oppose that increase. In the absence of any other circuit elements, the back EMF is also proportional to the remaining charge on the capacitor.

As long as there is charge left on the capacitor, current will still be increasing.

 

[Nikhil Rajagopalan]

So, initially, there is very high back emf. Then, the current is increasing but, the rate of increase of current is decreasing and hence the back emf keeps decreasing. So the capacitor potential does lesser work on a charge moving across the inductor as the work it does on the previous one.

 

[Nikhil Rajagopalan]

Thank you jbriggs444. How we prove the fact that back emf is propotional to the charge remaining on the capacitor?

 

[Nikhil Rajagopalan]

Is it that applying Faraday's Law on the circuit, we get q/C = -Ldi/dt (where q is the charge on the capacitor)= Back EMF. I guess the propotionality can be understood here. So is it right to believe that the back emf at any instant is equal to the voltage across the capacitor.

 

[Dale]

the only reason I see for current to be increasing is because the current was initially zero

That isn’t how inductors work. The current increases because the voltage is positive. That is what inductance means.

Any change in current is opposed by the inductor.

I never liked this phrasing. An inductor establishes a proportionality between voltage and change in current. If there is a voltage it makes current change, if the current changes it makes a voltage.

Any change in current is opposed by the inductor.
1.How does the current increase from there after.

As long as the voltage is positive the current will increase.

How does the maximum value of current correspond to a situation where there is hardly any charge on the capacitor

A maximum is where a function stops increasing and starts decreasing, so the maximum current is when the current stops increasing and starts decreasing. Since an inductor makes current increase in response to a positive voltage and decrease in response to a negative voltage the maximum current will necessarily be when the voltage goes from positive to negative. This is when the charge on the capacitor goes through zero.

and then it has to decrease defenitely because the charge on the capacitor is depleting.

No, this is not relevant. There is nothing about depleting the charge on a capacitor that limits the current.

 

[Nikhil Rajagopalan]

Thank you very much Dale. Does the fact that current increases for a positive voltage correspond to a no resistance situation only. That the charge carriers are accelerated. Because if there is resistance, the shouldn't the current settle down to a constant value - V/R.

 

[Nikhil Rajagopalan]

I have seen simulators showing steady increase in current when a pure inductor is connected across a DC source. and I guess i could mathematically prove that di/dt there is a constant that is equal to V/L that implies linear growth in current.

 

[gneill]

Thank you very much Dale. Does the fact that current increases for a positive voltage correspond to a no resistance situation only. That the charge carriers are accelerated. Because if there is resistance, the shouldn't the current settle down to a constant value - V/R.

If there is any resistance at all the current will settle down to a final value of zero. Check out the wikipedia entry on the RLC circuit and pay particular attention to the three cases for damping (underdamped, critically damped, and overdamped).

 

[Nikhil Rajagopalan]

So is this fair to put it this way.
When a fully charged capacitor is connected to an inductor, It is just like connecting a potential source to an inductor except for the fact that the capacitor as a power source has a potential difference which is propotional to the charge it holds and this potential source will have a reduce in its potential difference as it keeps dicharging.
To begin with, the fully charged capacitor at its highest potential difference causes a high increase in current from zero. as this current flows, the capacitor is in fact discharging, which means the potential difference is getting lesser. This in turn means that the rate of increase in current is going to be lesser (q/c gets lesser) and thus the increase in current decreases until there is no charge left in the capacitor. This is the time where the current in the circuit has reached its maximum owing to successive increments with decreasing magnitude(since potential difference across the capacitor was continuously decreasing).

 

[gneill]

To begin with, the fully charged capacitor at its highest potential difference causes a high increase in current from zero

Make that "rate of increase". The current will follow a sine curve, starting at zero and heading towards a maximum. If you look at a sine curve you can tell the "rate of increase" by the slope of the curve as time progresses.

EDIT: You can get all of this graphically if you'd plot the capacitor voltage and circuit current on the same graph. Math can be visualized through graphs.

 

[Nikhil Rajagopalan]

Thank you gniell, I understand that. it should be high increase in unit time or rate of increase.

 

[Nikhil Rajagopalan]

I believe I have an understanding that I can work on. gniell, Dale, jgbriggs444, anorlunda, I cannot thank you enough for your precious time you took out for me. You taught me a lot of physics and a lot of values that anyone who works in education should have. I couldn't thank you enough for the help. The light you gave me shall be shared the same way.

 

[gneill]

Here's another graphical summary of what is going on. It doesn't include the fancy energy curves that @anorlunda 's plot had, but hey, apparently I'm far more lazy than he is

Done with Mathcad: (See: lazy!)

 

[Dale]

Does the fact that current increases for a positive voltage correspond to a no resistance situation only.

The current will increase any time that there is a voltage across the inductor. In a RLC circuit some of the voltage will go across the resistor, so you will have a lower rate of increase.

 

[Nikhil Rajagopalan]

Thank you so much gniell, the way you helped me, you have earned lazy rights for the rest of your life. I can understand how the LC oscillator works. It was mostly not being able to understand how a potential difference across a pure inductor works in contrast with a resistor.

 

[Nikhil Rajagopalan]

The current will increase any time that there is a voltage across the inductor. In a RLC circuit some of the voltage will go across the resistor, so you will have a lower rate of increase.

In the presence of a resistor, for damping to happen, something should cut into the current in every cycle so that it eventually dies out. When i wrote down the equation it becomes di/dt = (q/LC) - iR , so the rate of increase is reduced by the potential drop and the discharging. How does it not just result in the flattening of the curve but also reduction in amplitude.

 

[gneill]

The resistor is dissipating energy (as heat). The current and voltage amplitudes are measures of the energy stored in the circuit in the magnetic field of the inductor and electric field of the capacitor. Dissipate the energy and you reduce the amplitudes.

 

[Nikhil Rajagopalan]

Thank you gniell. I understand the idea of the energy dissipation. I am trying to understand how the damping can be thought of graphically. The peak values of two half cycles in a damped oscillation should be asymetric, the later being smaller than the former. How does that continuous decrease in peak with every half cycle fit with the equation di/dt = (q/LC) - iR . Defenitely, di/dt will be a lesser value due to the presence of the term iR. But wouldn't that still be symmetric. with two half cycles.

 

[Nikhil Rajagopalan]

Is it because, a lesser I max cannot store as much charge in the capacitor as it used to have. making the next charging quarter cycle less taller and this continues.

 

[gneill]

I understand the idea of the energy dissipation. I am trying to understand how the damping can be thought of graphically. The peak values of two half cycles in a damped oscillation should be asymetric, the later being smaller than the former. How does that continuous decrease in peak with every half cycle fit with the equation di/dt = (q/LC) - iR .

Your equation doesn't look right to me. It should be second order. Can you show how you got it? You can't mix I and Q as separate variables in the same equation when I = dQ/dt. So at the very least your dI/dt should become $d^2 Q/dt^2$.

Solve the differential equation and you will be able to see, graphically by plotting the resulting expressions for the charge or current as a function of time, how it goes.

I encourage you to look at the math on the Wikipedia page where the effect of damping is presented in detail. They also have a graph shoing the effect of damping.

 

[Nikhil Rajagopalan]

The equation I tried was from applying Faraday's Law to the LCR circuit.

i - Current in the circuit.
Q - Maximum charge in capacitor
q - Charge in the capacitor at any instant
i = d/dt (Q-q)

According to Faraday's Law, iR - q/C = E_induced
iR - q/C = -Ldi/dt
Ldi/dt = q/C - iR