00:01
Let's take a look at an lrc series circuit, but let's analyze it using complex numbers.
00:07
So on the left side here, we have a voltage source of 200 volts.
00:11
We have an angular frequency of omega equals 1 ,000 radiance per second.
00:15
We have an inductor on the top with a value of l equals 0 .5 henry's.
00:20
We have a resistor on the right with a value of r equals 400 oms.
00:24
And we have a capacitor on the bottom with a value c equals 1 .25 microfact.
00:31
Okay, so first, let's do a quick recap of complex numbers.
00:38
So complex numbers can be displayed on an x, y, axis.
00:43
But what complex numbers? the x axis is your real numbers, and the y axis are your imaginary numbers.
00:53
And your imaginary numbers all have the letter i next to them.
00:57
So if i draw a vector from the origin in the complex number plane, i'm going to have a real component.
01:05
So i'm going to have a component down here.
01:07
So let's say that that crosses the real axis at a value a.
01:12
And i have an imaginary component.
01:15
So like let's say that that crosses it at b.
01:18
So i'm going to add that i on it.
01:19
Because remember, all the values on the y axis always have an i at the end.
01:25
That means that i can actually say that this vector can be represented by the complex number, a plus b i now this actually looks very similar to how we find the the impedance of a lrc circuit because we have a vector here and then we have values on the x and y axis so our vector here is going to be z right and this is going to be our resistor and this value is going to be our xl minus xc all right? so if we just say that xl minus xc is equal to a generic value of like a x, we could basically, we could represent this as a complex number and say the complex value of c is going to be equal to r plus xi.
02:29
So why don't we find out what this value is actually? so what that means is we could put our resistor of 400 and our x is going to be, remember, it's going to be xl minus xc.
02:45
So why don't we find out what x, xl is first? so xl is omega -l.
02:52
So that's going to be 1 ,000 times 0 .5.
02:57
So our xl is going to be 500 oms.
03:00
Our xc is 1 over omega c so that's going to be 1 over 1 ,000 times 1 .25 and remember you got to multiply that 1 .25 times 10 to the negative 6 because microfarads the micro part of it means 10 to the negative 6 and for this i get 800 oms so what that means is that my my xl minus xc is actually going to be a value of negative 300 all right so um if i want to do my x that's going to be x l minus xc so 500 minus 800 is negative 300 which means that my complex impedance is going to be 400 minus 300 i okay now if we want to find out what our actual impedances we can actually take the absolute value of the complex impedance and the reason that we can do that is because with any complex number if we take the absolute value of it like our a plus b i that's going to be equal to the square root of a squared, the real part, plus b squared, which is the imaginary part without the i in it.
04:39
So what that means is that if we take this, the atsc -vabra complex impedance, that's going to be r squared plus x squared.
04:49
So this is exactly what we get when we're normally doing these without complex numbers for like impedance.
04:58
So like this matches that exactly.
05:00
Exactly.
05:01
So that means that we get the square root of 400 squared, and then we'll get plus negative 300 squared.
05:11
And that's going to give us 500 oms.
05:14
So we used our complex impedance to be able to find out what our actual impedance is.
05:23
Now let's take a look at our complex current, because that's going to help us to get our complex voltages across each circuit.
05:31
So our complex current is going to be equal to our complex voltage divided by our complex impedance, right? and we know that our voltage is given as 200 volts, and there's no imaginary part to it because there's no i.
05:52
So i can just write this as just 200 on the top.
05:55
And on the bottom, we can use our complex impedance of 400 minus 300 eye.
06:01
Now, if you ever have a i in the denominator of a function to like simplify, you got to get that i out, which means i got to multiply by the conjugate.
06:16
So i'm going to multiply both the top and bottom by 400 plus 300 i.
06:23
So this will also be 400 plus 300 i.
06:28
Right and if i do that multiplication then like what that means is on the top here i'm actually going to get a a value of i'm going to have 880 ,000 basically plus 6 000 000 i on the top and then on the bottom here anytime you're multiplying conjugates then you basically get the first number squared.
07:05
So 400 times 400, 16 ,000.
07:10
And then like you're going to get plus the, sorry, minus the middle number squared because this is actually the reverse of doing the difference of two squares formula.
07:22
So i'm going to get minus 900 and don't forget to multiply the i's together, which is 900 i squared.
07:30
Sorry, not 900, 900 ,000, or sorry, 90 ,000 i squared.
07:40
Okay.
07:42
So like what i can do here is first i could take off all these zeros because everything has like four zeros on it.
07:52
So i could just turn this into 8 plus 6i on the top and i can turn this into 16 minus 9 i squared.
08:00
So like that's the first thing.
08:04
The second thing is i can take advantage of the fact that i squared is negative 1.
08:09
So i have 16 times 9, sorry, minus 9 times negative 1, which is the same thing as 16 plus 9, which is 25.
08:21
So i get 8 plus 6i over 25.
08:24
And normally you have a complex number, you like want to divide both the a, term, both the real part and the imaginary part, but buyer 25...