00:01
All right, so in this problem, calculate the value of zab in the network of figure given to us, right? but this is given to us in the delta form, right? so delta circuit is given to us.
00:18
We can also transform it into a star form, right? so this is a circuit with a star form.
00:29
So i have assumed l1, l2, l3.
00:32
Right so this is a star form right okay so now um first of all we will see what all things could be resolved first okay so equivalent impedance of the ceq which is in this diagram you can see so we will just calculate the equivalent capacitance right so this is basically ceq okay so now we can see that minus j 9 om and minus j 9 om is in series and this is this equivalent is parallel in parallel with minus j9 oh right so you can write it as like this equation okay so when you've solved this we will be getting minus j18 into minus j9 upon minus j in 18 minus j 9 which will be giving us c equivalent as minus j 6 o so now impedance of inductance l1 can be calculated as what this could be calculated as delta to start from transformation where you can say j6 into j6 will be this is divided by j6 plus j6 plus j6 right so this will be after solving we will be getting l1 as j2 o all right similarly for l2 we will calculate the impedance so this will be again the same we will be getting l2 as j2 o itself okay so now okay so now we will just calculate for l3 similarly so here this will be again l3 will be equals to j2 om itself, isn't it? so this we have got the value of impedance of inductance.
02:58
Okay.
02:59
So now we will just calculate for r1, r2 and r3 here.
03:05
Okay.
03:08
So for calculating that, we will just multiply it by a 20 into 20.
03:14
So this is just a basic calculation.
03:18
Right so this is all the delta transformation delta to star transformation uh methodology you can see right so here it becomes what it becomes r1 equals to 20 into 20 upon 20 into 20 plus 10 that will be equals to 400 by 50 that is 8 om so which is r1 and we have r1 2 that is equals to 10 into 20 upon 20 plus 20 plus 10 equals to 200 by 50 that is equals to 4 own okay now similarly for r 3 we can just calculate 10 into 20 upon 20 plus 20 plus 10 that is 200 upon 50 which is giving us r 3 as equals to 4 o okay so now you can just easily transform this again the circuit right so circuit transformation becomes circuit transformation becomes like this okay i'll draw that so this is a terminal and we have one inductor and we have here one more inductor then we had one c equivalent here right and again this is drawn with one more inductor right and again this is series with resistance then we have one more resistor here right okay so this is okay so now total things are there this is j2 j2 oh this is j2 again j2 this is 8 oom this is minus j6 o this is 4 o okay this is 4 okay so what all things can be resolved further so we know that this is in you can see that this uh one resistor one capacitor and one inductor is in series with each other and here one more series combination we have found of right okay so now let us suppose this is z1 and let us suppose this is z2 therefore z 1 will be equal to 8 plus j 2 can you write this because they are in series and z 2 equals to 4 plus j 2 minus j 6 which is equals to 4 minus j 4 minus j 4 oh, okay.
06:48
So this we have found of equivalent.
06:53
Right...