00:01
In this problem, we're given a differential equation for a rlc circuit and some initial values and asked to solve it for a different range of situations.
00:13
I will show you how to convert it into a standard problem and answer the first few parts.
00:20
And then that should give you enough information to be able to solve for the last few parts.
00:27
So let's first start by writing down the differential equation.
00:30
So we have l di dt plus ri plus 1 over c, the integral from 0 to t of i t.
00:52
In the question it uses t, but technically in an integral you can use a dummy variable, and i want to use that to distinguish from this t, dt prime, and this is all equal to some voltage function, e of t.
01:11
And we're told that i of zero is zero, and q of zero is 2c.
01:27
Assuming here that q represents the charge as a function of time, and i here represents the current through the circuit, we can transform this into a standard second order differential equation in terms of the charge rather than having to deal with this awkward integral.
01:43
So if we recall that the current as a function of time is simply the rate of change of the charge in the system as a function of time or dq by dt.
01:56
This means that di by dt is simply d squared q by dt squared.
02:10
I is the first derivative and now if we look at the integral from zero to t, i t prime dt prime, well, this is simply equal to the integral from zero to t of dq by dt prime dt prime, which by the fundamental theorem of calculus will just give us q of t minus q q naught.
02:47
And we know from our initial values that q naught is two times c.
02:51
And so we're left with the differential equation l d squared q dt squared plus r times dq dt plus 1 over c times q of t minus 2, 2 because that's 2c divided by 2c.
03:33
All of this is equal to e of t.
03:39
So now this is the sort of standard form for a non -homogeneous second order differential equation.
03:45
So now for the different conditions given below we can look for solutions for q and we have to remember that it is the the question is asking for the function i of t, so we can solve for q of t and then take the derivative to give us i of t.
04:10
So let's look at the first case.
04:12
So a, we're given l equals 0, we're given r equals 100, and we're given c equals 10 to the minus 3, and we're given the function e of t.
04:36
So e of t we are told is 100 for 0 is less than or equal to t is less than 1 and e of t is equal to 0 for t greater than or equal to 1.
05:05
So that gives us some interesting ranges of behavior.
05:10
The biggest simplifying factor of this part is the fact l equals zero means the second derivative disappears.
05:17
And so we're just going to be left with a first order linear equation.
05:21
So i'm going to leave r and c as variables for now to make my life easier.
05:25
Easier.
05:26
So we're looking at r dq dt plus q over c minus 2.
05:45
Now let's first consider the case that 0 is less than or equal to t is less than 1 and then that means e of t is going to be 100 and so that's the solution we're looking for, that's the range where we're looking for our solutions and we know that in this regime the function has a value of 2c at 0, q0 is 2c, and so we can go ahead and try and solve this equation.
06:30
What have we got? well let's rearrange range it so that we have dq dt on its own so we have r q dt is equal to 102 minus q over c c and then we can get dq dt on its own by dividing by 1 over r dividing by r 102 minus q over c.
07:23
So we have a standard first order differential equation which we can solve a few ways.
07:29
You could use an integrating factor or in the case i'll do here just by separation of variables.
07:36
So let me rearrange by separating the variables.
07:39
So now we have the integral of, well, let me pull out another factor of 1 over c here, actually.
07:49
So if i pull out a factor of c, then this becomes c here.
07:54
And then we lose the c on the denominator.
07:59
And so we're left with the integral of dq over 102c minus q is equal to the integral of dt over rc.
08:26
And we're running out of space, so i will clear the board and continue.
08:34
So as i was saying, we now have this separated version of the first order differential equation, and we just calculate these integrals.
08:44
So we can see that on the left hand side, we just have a logarithmic integral.
08:52
And so this, when you do the integral, will be minus the log of q minus one hundred and two c.
09:04
And then on this side, we just have the integral of dt, which will just give us t over rc.
09:15
I can't forget, we have a plus constant, which i'm going to call capital a.
09:19
And so now we can rewrite this.
09:26
So let me write this as the logarithm of q minus 102c is equal to minus t over rc minus a.
09:43
Exponentiate this so i get q of t is equal to 102c plus e to the minus t over rc minus a.
10:06
And now if i let b equal e to the minus a, then i can rewrite this as q of t is equal to 102 c plus b e to the minus t.
10:28
Yes, t, t over rc.
10:35
And so b is unknown, but we can use our initial values which we have as q of 0 is equal to 2c so we know that q of 0 should be equal to 2c and so we should have that this is equal to 102 c plus b because b e to the 0 is 1 and so so i subtract 102 from this side, and so i get that b is equal to minus 100 c.
11:33
And so for 0 is less than equal to t is less than 1, we have the solution q of t is equal to 102c minus 100c e to the minus t over rc and now you can go ahead and plug in your values for rc and you would get the exact factors there but i'll leave that as an exercise for you.
12:13
Now the second part of this problem is that e of t is is zero when t is greater than or equal to one.
12:20
So let's consider the equation again.
12:23
So we have l zero, so we have r dq dt plus one over cq minus two equals zero.
12:36
So we have, so for now t greater than or equal to one dq dt plus one over c times q minus two.
12:53
Minus two is equal to zero, exactly, yep.
13:04
And so we can do this again...