Hello my name is Sejer Garcia. 
I'm a UPC student. 
And we're going to work on the second 
part of our impetus problem. 
Please refer to the results obtained in 
the previous problem. 
Alright so we've already looked at the 
circuit, in the previous problem. 
Where we have a time varying voltage. 
100 ohm resistor. 
A milifire capacitor, a 2 milifire 
capacitor, and a 5 milihairme inductor. 
So, in the previous problem, what we id 
is given a frequency of 60 hertz, and 
then taking that into radience per 
second, we have 377 radience per second. 
Given this information, we were able to 
simplify this problem into their 
independent impedances. 
And what we ended up having was an 
impedance for the resistor, equal to 100 
ohms, and impedance for both capacitors 
in series equal to negative j, 3.98 ohms. 
And an impedance for the inductor equal 
to j, 1.89 ohms. 
So let's say We were able to simplify our 
circuit into this one with little boxes, 
and let's say we were giving a voltage 
source equal to 10 cosine 377 times t 
plus 60 degrees volts, and they ask us to 
find the current that goes through each 
box. 
So that would be ir, this would be ic, 
and this would be il. 
So what we want to take from this problem 
is that even though we are working with 
impedances for different elements, our 
laws, our kickoff laws and our Ohms law 
are still going to apply to this... 
To our analyses. 
So what we, the way we're going to work 
this is exactly the same as we would work 
any other, any other problem with, let's 
say resistances. 
So, in the previous problem, we saw that 
we were able to combine these three 
impedances to find the equivalent 
impedance. 
That is in parallel with the source. 
If we were to have resistances we could 
do exactly the same thing. 
So we found out that the Cerulean 
impedance between these three elements is 
equal to a hundred minus j 3.52 Ohms. 
And this here by combining all of these 
three impedances we can find that, we can 
see that that's the current that's 
going to flow through the equivalent 
impedance is equal to the current that's 
going to flow through the resistors, so 
this is IR right here, and as I mentioned 
earlier. 
Our Ohms law still can apply, V is equal 
to I times R. 
And by using this relationship we can 
calculate IR directly from our two 
values. 
So, to find IR this is going to be equal 
to our voltage source, VT. 
Over our impedance, Z equal. 
So substituting our values, we know that 
the voltage source is equal to 10 times 
cosine 377 times T plus 60 degrees. 
And our impedance is 100 minus J times 
3.52 ohms. 
So once we get to this step, we can see 
that it's kind of weird to divide these. 
So, because, first of all, all we have 
for the voltage is an amplitude and an an 
angle here, and all we have for our 
impedance is expressed as complex 
numbers. 
So what we want to do is, use phasers for 
both cases such that are equal, and we 
can perform this calculation normally. 
So, for the voltage source, to combine 
this into phasers, we have our magnitude 
which is 10, and then our angle which is 
16 degrees. 
And then, to combine to transform our 
impendance we can use a calculator, or 
you can use our, our formulas for 
converting complex numbers into polar, 
for polar numbers and we get 100 angle 
2.02 degrees. 
So by making this division you end up 
having that the current that flows 
through our impedance right here. 
Are equal in impedance, is going to be 
equal to 100 angle 57.98 degrees, and 
this is milliamps. 
So we've already found the current that's 
going to flow through our resistor right 
here. 
Now, We can go back to our original 
circuit knowing that we know knowing that 
we have this current right here. 
And as you can see these two elements are 
in parallel. 
So we can perform a, a current, current 
divide, current divider, sorry, current 
divider as we've always done with our 
resistances. 
So let's, let's just number our steps 
here. 
Let's say That this was our first step. 
Second step. 
Third, and the next one's going to be the 
fourth step. 
So next let's find the current that flows 
through inductor. 
IL, right here. 
And like I say, we're going to use a 
current divider. 
Alright? 
So our formula for IL is going to be our 
impedance for the capacitor, which is 
negative J 3.98 over the sum of both. 
1.89 times our source. 
And our source in this case is the 
current that's going to flow through the 
resistance which we already found, times 
a hundred angle 57.98 degrees. 
Times 10 to the minus three, because 
remember that units are important. 
So, once we've performed this 
calculation, remembering that in order to 
perform this, you can either input these 
number directly into a calculator, or you 
can convert these polar, I'm sorry, these 
complex. 
the numbers into polar coordinates, and 
then make the calculation. 
We end up having a value equal to 190 
angle 57.98 degrees, and this is in 
milliamps, too. 
Keep in mind that our angle is the same 
as The angle for inductors, inductor 
current is the same as the angle for the 
resistor current and this is because we 
can see it, we can see it right here. 
We can see that by adding these two 
numbers in the denominator we end up 
having a value equal to a constant times 
j but the j's are not going to cancel 
out. 
So this number ends up being just a real 
number, a constant. 
Which does not affect at all our angle. 
So both, both currents are going to have 
the same angle. 
Now once we have our current that flows 
through resistor and current that flows 
through the conductor, we can easily 
calculate the current that flows through 
the capacitor, like I said at the 
beginning again, just because. 
 >> We can, our, our [INAUDIBLE] still 
apply for these type of circuits. 
So, our next step is to calculate IC. 
And we know that the current that goes 
through IR is going to be equal to the 
current that goes through the capacitor 
plus the current that goes through the 
inductor. 
And if we Solve for IC we get IC is equal 
to IR minus IL and we already know these 
two values because we calculated in the 
previous steps and the final values equal 
to minus 90 angle 57.98 degrees and it is 
also in And as I explained earlier, the 
angles still the same just because, by, 
when you add two numbers, the angles are 
not affected together. 
Are not affected, sorry, by each other. 
So, just to go over these steps, when we 
have elements and they're, we're not 
given their impedances, the first thing 
to do is calculate their individual 
impedances as we did in the previous 
column. 
Once we have this simplified circuit with 
the little boxes and their individual 
impedances, given a source in this case a 
voltage source, we can then start 
applying our Kirkoff's Law and our Ohm's 
Law as we've always done, in order to 
find anything that's asked for us. 
In this case, the individual currents 
were asked So we start by combining all 
of them just to find this first current 
right here. 
The current that goes to the resistor. 
Then we can apply a current divider as 
we've always done in order to find the 
current that goes to the conductor. 
We could've also done it for the 
capacitor first. 
In this case we did it for the conductor. 
And once we have two out of three values, 
we can just use Kirkoff's law in order to 
solve for our final current. 
Thank you.