00:01
So here i'll be deriving an expression for the charge as a function of time in a series rlc circuit that is being driven by an ac battery, oscillating voltage source.
00:16
The important thing to remember is that the current through all the components is the same.
00:23
And the charge, what we're looking at is, first of all, we could be thinking about the charge on one side of the capacitor plates, but it's important to remember that dq by dt gives us the amount of charge passing a certain spot in the circuit as a function of time, and that is defined to be the series current in the circuit.
00:55
So the first thing to do is to develop the differential equation that governs the circuit.
01:01
And to do that, you can simply use kirchhoff's loop law.
01:05
In which case the battery voltage, which is e0, sign of omega -t, is the sum of the voltages across each of the components.
01:17
According to faraday's law, that would be the l, the i, by dt on the inductor.
01:24
According to oms law, that's r times i across the resistor.
01:31
And then finally, just q over c for the capacitor.
01:35
And this is where we would use that the current is the first derivative of the charge changing in time.
01:47
And so, d .i .d .t.
01:52
Is the second derivative of the charge changing in time.
02:00
And so our final differential equation is l, second derivative of cube with respect to time, plus r, d .q by d .t.
02:11
Plus q over c is equal to e0, sine of omega t.
02:21
That is a differential equation involving charge.
02:26
And it's important to recognize that this is the same form of the equation that you would get for a damped -driven oscillator.
02:42
Another set of terms that you would here describe this equation is it is a second -order linear diffi -q with constant coefficients.
03:02
That is a mouthful, but it just means the inductance, the resistance, and the capacitance are all constants.
03:13
And the way i'm going to handle this is there is the differential equation way of handling this, which makes a lot of sense.
03:23
And it turns out that the q, because it is a linear equation, can be written as the sum of two separate solutions.
03:36
Let me write this out to begin with.
03:38
It is the sum of what's called the homogeneous solution that i'll call q sub h plus a particular solution, which i'll call q sub p.
03:53
In the vernacular of physics or physical science, the homogeneous equation is called the transient solution, and the particular is known as the steady state, and we'll see why they are given those particular interpretations.
04:15
So the way you would solve this is to find each of those solutions separately, and then simply add them together.
04:24
And put in for your initial conditions on the charge once you have everything together.
04:33
We won't get to that point because we don't have any initial conditions selected.
04:45
Okay, so let me give myself some room and we will get blasting on the homogeneous solution.
04:55
By homogeneous, we're going to set the right -hand side of the equation, equal to zero.
05:01
So we're looking at l second derivative of cube with respect to time, plus r, dq by d t, plus q over c is equal to zero.
05:17
And the way i'd like to solve this is with the characteristic equation approach, sometimes called the auxiliary equation.
05:26
But you assume a exponential solution, q is equal to say, e to the lambda t, what you call the, constant out in front of time.
05:37
You can call it almost anything.
05:39
I'll call it lambda for now.
05:42
And we simply want to take the first derivative with respect to time of that assumed solution.
05:52
By the way, the reason why that solution works is because you have a second order equation with constant coefficients.
06:03
The second derivative just brings the lambda out a second time.
06:09
If we put all that back in, it's a good way to solve a differential equation, is assume a solution and then put it back in.
06:22
So then we would have l, lambda squared, e to the lambda t, plus r, e to the lambda t, plus e to the lambda t, whoops, r lambda.
06:39
We fit in a factor of lambda.
06:46
And then e to the lambda t, 1 over c is equal to 0.
06:56
And because the exponential can't be zero in general, we can simply cancel e to the lambda t out on each term.
07:08
And what we're left with then is a quadratic equation, which we can solve for lambda.
07:21
There will be two roots.
07:23
And so we get lambda is equal to minus the b constant, which is r plus or minus the square root of r squared minus 4 coefficient l divided by c.
07:50
C is capacitance, not c as in quadratic coefficient.
07:55
And then we have divided by 2l.
08:01
So there are two roots.
08:02
There's a positive root, and it's kind of awkward to keep writing this down, and there is a negative root.
08:35
Just like mechanically, you have a situation where the nature of that square root will govern the behavior of the transient solution, and will also give us some interesting things about the frequency response of the source.
08:56
But here we are going to just describe how the solution depends on the nature of that square root.
09:09
So if the square root of r squared minus 4lc is positive, you have an over -damped situation.
09:23
And then the circuit will not resonate.
09:25
At a particular frequency.
09:28
If that square root is zero, you have what's called critical damping, and if it should be negative, then you have some ability to oscillate because you would have a complex solution on an exponential, and that would be an oscillating solution.
09:59
So if that square root is negative, where root is negative, that's underdamped.
10:13
But in any case, your homogeneous solution, beg your pardon, can be written as a sum of two exponentials, lambda plus times t plus b, e to the lambda minus t.
10:41
That may be the easiest way to write it, with the lambdas being these nasty looking things.
10:48
But the first part, the minus r over 2l, indicates that both solutions decay, both parts of qh is a function of time decay exponentially.
11:13
And that's why this is often called the transient qh of t.
11:19
Will go to zero no matter what the root is doing as t approaches infinity.
11:28
The only way that is not true is if the resistance in the circuit is zero.
11:37
There's always some resistance.
11:43
So now we'll take a look at the particular solution.
11:54
That will oscillate according to the battery.
12:19
And we want to pick up that that oscillating term, which is called the steady state simply because the circuit will slosh back and forth with its charge at the same frequency as the drive, but as we'll see, it's going to be out of phase.
12:36
So the way i'd like to handle this is kind of similar to what we did with a characteristic equation, but with some caveats.
12:45
So there are no arbitrary amplitudes.
12:57
So the homogeneous solution, the a and the b, depend on initial conditions.
13:19
And the other difference is that we're going to assume a solution to particular of time is equal to alpha e to the j omega t.
13:33
And so there is no ambiguity in either the alpha or the omega -t.
13:43
Those are both determined by not only the drive, but the differential equation...