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.