00:01
Let's take a look at the current as it goes through a lrc parallel circuit.
00:05
So we have our circuit here.
00:07
We have a voltage source on the left with a voltage of 100 volts.
00:10
We have a 200 -oom resistor with a current ir going through it.
00:15
We have a 2 -henry inductor with our l with a current il going through that.
00:20
And we have a 0 .5 micro -farrad capacitor with a ic going through that.
00:26
So first, why don't we take a look at these occurrence as a function of the angular frequency omega? because in a ac circuit, our voltage and our currents are like always alternating at a angular frequency omega.
00:43
So let's start with our resistor.
00:46
So resistors actually don't depend on omega at like all.
00:50
So the current through our resistor is just going to be v over r.
00:53
It's just going to be a constant value.
00:57
If we look at our inductor, that's going to be v over the reactants in the inductors.
01:03
That's xl.
01:04
And we know for our capacitor, our current through there is going to be our voltage over the reactants in the capacitor, xc.
01:12
So these three things, the structure of them is very similar.
01:17
But now we can start to plug in some of our values.
01:23
Our resistor current, that's going to be our voltage of 100.
01:27
Over our resistance of 200.
01:30
So that's just going to be 0 .5 amps.
01:33
For our inductor, the reactance of an inductor is just omega times l.
01:40
So we have v over omega -l.
01:43
So that would be 100 times omega -times 2.
01:48
So that means that this will be 50 over omega will be our formula for the current through the inductor.
01:58
And like that will be in amps as well.
02:00
And for the capacitor, our reactants is one over omega -c.
02:05
And we know if we're dividing by a fraction, it's the same thing as multiplying by the reciprocal.
02:11
So this is v omega -c, which again, we'd plug in our v of 100 and our c of 0 .5.
02:19
And now this is micro -farad.
02:22
So we want to make sure that we're multiplying that times 10 to the negative 6.
02:27
And this is actually going to give us, if we put that in the calculator, we're going to have 0 .5e6 times 100.
02:35
This is actually going to give us 5 times 10 to the negative 5 omega amps.
02:43
So there's a couple of things that we want to notice here.
02:47
This is linear, right? this is like y equals mx plus b, where this value is our slope, our 5, e like negative 5 is our slope.
02:57
In like omega is, is like our, you know, our independent variable.
03:04
Our resistor current, because it's a constant, there's just going to be horizontal line because it's not really going to change with like omega, right? and this is going to be very similar to like the graph of 1 over x, which looks like this.
03:27
Okay.
03:28
That's why it's important to, like, remember there was apparent.
03:31
Functions from your algebra class.
03:36
Okay, so now let's graph this and see how it looks.
03:40
So that means if we graph this in terms of omega, so we have our omega here, and then this is going to be our current, we're going to have a straight line, right? so we'll do that in this color.
03:56
Whoops.
03:57
In this color.
03:59
So this will be iir, just a horizontal line.
04:02
Through 0 .5, right? we're also going to have a linear line as well, right? and this is going to be our our ic.
04:14
And we know that it's going to go up into the right because even though 5 times 10 to the negative 5 is a pretty small number, still a positive slope.
04:23
And then for our inductor current, it's going to look like this.
04:28
It's going to go down here.
04:31
And it's going to have as a and totes that both y equals zero and x equals zero.
04:37
So this is going to be our v of l.
04:39
So like this is a graph of like all of our currents, you know, as a function of omega.
04:48
Also important to a look at our omega values at the extreme point...
