Charging and Discharging of a capacitor in an LC circuit

[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

 

[sophiecentaur]

The barrier i find in understanding it is in any case of capacitor discharging, the current is suppoused to decrease,

Why is it "supposed to decrease"? ***You are trying to impose your (inaccurate) intuition. An L and C, together not behave like an R and C. An RC circuit follows an exponential law. An L and C follow a sinusoidal variation in time; the energy, initially stored in the Capacitor does not get dissipated but passes into the Inductor and back again and so on. As the Voltage across the Capacitor approaches Zero, it changes faster and faster and (as it can hardly suddenly change direction or remain at zero) it carries on into the negative region.
I have to point out that an arm waving description of a process like this is really not adequate. There are some things for which a simple verbal description is just not enough - which is why Science can be very difficult. That's something people just need to accept. This link describes what happens and (with Maths, of course) derives an equation that describes the waveforms. Notice, the whole mathematical description takes up very little space because it is concise and cannot be misinterpreted - unlike so many other attempts at descriptions..

*** Explanation of why this idea is wrong: There is a constant amount of energy in the circuit, sloshing back and forth between the L and C.
Energy in a Capacitor is CV²/2
Energy in an Inductor is LI²/2
As V_c reduces, the energy drops faster and faster because of the squared term. This means that the Current I_L has to be Increasing to a Maximum as V_C passes zero.
With a CR circuit, the current drops as the V_C drops; the Energy decays.