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What happens when you put inductors and capacitors in the circuit?Something cool-and it’s actually important.
You can make many different types of inductors, but the most common type is a cylindrical coil-a solenoid.
When the current passes through the first loop, it generates a magnetic field that passes through the other loops.Unless the amplitude changes, the magnetic field won’t really have any effect.The changing magnetic field generates electric fields in other circuits.The direction of this electric field produces a change in electric potential like a battery.
Finally, we have a device with a potential difference proportional to the time rate of change of the current (because the current generates a magnetic field).This can be written as:
There are two things to point out in this equation.First, L is the inductance.It only depends on the geometry of the solenoid (or whatever shape you have), and its value is measured in Henry’s form.Second, there is a minus sign.This means that the change in potential across the inductor is opposite to the change in current.
How does the inductance behave in the circuit?If you have a constant current, then there is no change (direct current), so there is no potential difference across the inductor-it acts as if it doesn’t even exist.If there is a high-frequency current (AC circuit), there will be a large potential difference across the inductor.
Likewise, there are many different configurations of capacitors.The simplest shape uses two parallel conductive plates, each with a charge (but the net charge is zero).
The charge on these plates creates an electric field inside the capacitor.Because of the electric field, the electric potential between the plates must also change.The value of this potential difference depends on the amount of charge.The potential difference across the capacitor can be written as:
Here C is the capacitance value in farads-it also depends only on the physical configuration of the device.
If current enters the capacitor, the charge value on the board will change.If there is a constant (or low frequency) current, the current will continue to add charge to the plates to increase the potential, so over time, the potential will eventually be like an open circuit, and the capacitor voltage will be equal to the battery voltage (or power supply).If you have a high-frequency current, the charge will be added and taken away from the plates in the capacitor, and without charge accumulation, the capacitor will behave as if it doesn’t even exist.
Suppose we start with a charged capacitor and connect it to an inductor (there is no resistance in the circuit because I am using perfect physical wires).Think of the moment when the two are connected.Assuming there is a switch, then I can draw the following diagram.
This is what is happening.First, there is no current (because the switch is open).Once the switch is closed, there will be current, without resistance, this current will jump to infinity.However, this large increase in current means that the potential generated across the inductor will change.At some point, the potential change across the inductor will be greater than the change across the capacitor (because the capacitor loses charge as the current flows), and then the current will reverse and recharge the capacitor.This process will continue to repeat-because there is no resistance.
It is called an LC circuit because it has an inductor (L) and a capacitor (C)-I think this is obvious.The potential change around the entire circuit must be zero (because it is a cycle) so that I can write:
Both Q and I are changing over time.There is a connection between Q and I because current is the time rate of change of charge leaving the capacitor.
Now I have a second-order differential equation of charge variable.This is not a difficult equation to solve-in fact, I can guess a solution.
This is almost the same as the solution for the mass on the spring (except in this case, the position is changed, not the charge).But wait!We don’t have to guess the solution, you can also use numerical calculations to solve this problem.Let me start with the following values:
To solve this problem numerically, I will break down the problem into small time steps.At each time step, I will:
I think this is pretty cool.Even better, you can measure the oscillation period of the circuit (use the mouse to hover and find the time value), and then use the following method to compare it with the expected angular frequency:
Of course, you can change some of the content in the program and see what happens-go ahead, you won’t destroy anything permanently.
The above model is unrealistic.Real circuits (especially long wires in inductors) have resistance.If I wanted to include this resistor in my model, the circuit would look like this:
This will change the voltage loop equation.There will now also be a term for the potential drop across the resistor.
I can again use the connection between charge and current to get the following differential equation:
After adding a resistor, this will become a more difficult equation, and we can’t just “guess” a solution.However, it should not be too difficult to modify the above numerical calculation to solve this problem.In fact, the only change is the line that calculates the second derivative of charge.I added a term there to explain resistance (but not first order).Using a 3 ohm resistor, I get the following result (press the play button again to run it).
Yes, you can also change the values ​​of C and L, but be careful.If they are too low, the frequency will be very high and you need to change the size of the time step to a smaller value.
When you make a model (through analysis or numerical methods), you sometimes don’t really know whether it is legal or completely fake.One way to test the model is to compare it with real data.Let us do this.This is my setting.
This is how it works.First, I used three D-type batteries to charge the capacitors.I can tell when the capacitor is almost fully charged by looking at the voltage across the capacitor.Next, disconnect the battery and then close the switch to discharge the capacitor through the inductor.The resistor is only part of the wire-I don’t have a separate resistor.
I tried several different combinations of capacitors and inductors, and finally got some work.In this case, I used a 5 μF capacitor and a bad-looking old transformer as my inductor (not shown above).I am not sure about the value of the inductance, so I just estimate the corner frequency and use my known capacitance value to solve for 13.6 Henry’s inductance.For the resistance, I tried to measure this value with an ohmmeter, but using a value of 715 ohms in my model seemed to work best.
This is a graph of my numerical model and the measured voltage in the actual circuit (I used a Vernier differential voltage probe to obtain the voltage as a function of time).
It’s not a perfect fit-but it’s close enough for me.Obviously, I can adjust the parameters a bit to get a better fit, but I think this shows that my model is not crazy.
The main feature of this LRC circuit is that it has some natural frequencies that depend on the values ​​of L and C.Suppose I did something different.What if I connect an oscillating voltage source to this LRC circuit?In this case, the maximum current in the circuit depends on the frequency of the oscillating voltage source.When the frequency of the voltage source and the LC circuit are the same, you will get the maximum current.
A tube with aluminum foil is a capacitor, and a tube with a wire is an inductor.Together with (diode and earpiece) these constitute a crystal radio.Yes, I put it together with some simple supplies (I followed the instructions on this YouTube video).The basic idea is to adjust the values ​​of capacitors and inductors to “tune” to a specific radio station.I can’t get it to work properly-I don’t think there are any good AM radio stations around (or my inductor is broken).However, I did find that this old crystal radio kit works better.
I found a station that I can hardly hear, so I think my self-made radio may not be good enough to receive a station.But how exactly does this RLC resonant circuit work, and how do you get the audio signal from it?Maybe I will save it in a future post.
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