00:05
Hi, hello everybody, this is selma.
00:08
Today i would be presenting a solution for this problem on how to get the equivalent impedance for such circuit.
00:26
So before we start, let's put some solid background about some properties that we would use to solve this problem.
00:37
Problem.
00:39
First that the serious impedance, if we have serious impedance, for example, these two components, the resistor of 2 -oam and the capacitor of negative j2 -oam, are serious.
00:54
So when they are serious, i add up their impedances.
01:01
So it's z -1 plus z -2.
01:03
If they are parallel, so it's adding two fractions where their z equivalent would be z1 times z2 over z over their summation which is z1 plus z so these are the two main processes we use to get the equivalent sometimes in our circuits we don't have this combination but we have a shape that's in between it's either a store or a delta so actually a star in most cases looks like a pie section, like the one that we have here.
01:49
And the star or the y looks like a t section.
01:56
For us to break down this shape in order to solve it with the normal series in parallel methods, we have to break this to from a delta to a star or to a y.
02:13
So how do we break that down? there is this formula that from each note i see the shared impedances and i divide by the summation of the three impedances of my delta.
02:33
Now that we know what we need, let's go and see how to solve our question.
02:40
So with these kind of questions, first, i would divide my impedances into sections.
02:52
So let's take the easiest ones that are in series with each other.
03:00
So here we have z1, this one is z2, that one is z3, set 4, z5, and z6.
03:28
Let's get the impedances of each one of them.
03:48
So we have that z1, which be 1 minus j.
03:57
Set 2 would be 2 minus 2 j.
04:04
Set 3 is just 1 ome.
04:10
Z 4 is 1 plus j.
04:17
5 is 2 minus 2 j.
04:22
And 6.
04:27
We have 1 minus j ome.
04:35
Okay.
04:37
So now the first thing to do, is to take z4, z2, and z3, and convert it into a y shape.
04:48
So how do we do that? as we said, we have two common impedances between a node.
04:56
So first we need to define our three nodes.
04:59
Let's name them, a, b, and c.
05:10
Now, the equivalent would look something like a t shape.
05:15
For now, i would make it in its conventional, way to do it as a y shape.
05:25
So i will name each impedance, new impedance, by the node it's coming from.
05:31
So this one is za, that one is zb, and this one is zc.
05:41
So now za, this is the node a.
05:46
It's the combination of z2 and z4.
05:51
So if you go down, we would define that za is z2 times z4.
05:59
Over their summation, which is 2 plus 3 plus 4.
06:13
So can you now say similarly how to get zb? so yes, as it's obvious that the common two impedances from the node b is z2 and z3.
06:33
So it's set 2 multiplied by z3 over their summation.
06:38
So z -b is 2 times 3 over their summation.
06:58
And similarly, zc is going to be z -4 as shown here.
07:10
The common here is z -4 and z -3, common by the node c in this delta section.
07:18
So z -4 times z -3 over their summation.
07:33
Now my new circuit would look like that...