00:01
Here we will find the v -thavenine.
00:02
Here we will say that omega is given as 10, value of omega is given as 10, and 0 .5 henry is given as j -umega -l, this can be written as j -5, and 10mpharid, this can be written as minus j into 10.
00:23
Firstly, we have to find j -n.
00:26
To find z -n, we will consider this circuit.
00:29
To find z -n, we will consider this circuit.
00:32
Here we will write that 1 plus 2 v node is equal to vs by j5 plus bs 10 minus j into 10, where we will say v node is equal to 10 bs divided by 10 minus j into 10.
00:52
After we will solve, we will get 1 plus 19 bs divided by 10 minus j to j.
00:59
Is equal to bs by j5.
01:02
Here it will solve the value for bs.
01:06
It will solve value for vs.
01:10
This can be written as minus 10 plus j10 by 21 plus j2.
01:16
If we have to find zn, we will find z8...