00:01
Hello students we are given the following circuit.
00:04
Using the nodal analysis we need to find out the current i0.
00:09
From here we can see that the value of angular frequency omega is 2 radiance per second.
00:18
Now using the formula x c equals to 1 upon j omega c and xl equals to j omega l, we write all the values of the inductors and capacitors in the complex form.
00:44
Thus, f1 equals to 0 .25 parrots which gives us xc1 equals to minus 2 j oms.
00:58
F2 equals to 0 .25 parrots which gives us xc1 equals to minus 2 j oms.
01:02
5 parrots which gives us xc2 equals to minus j oms.
01:11
L1 equals to 2 h which gives us x l1 equals to 4 j oms and l2 equals to 1 henry which gives us x l2 equals to 2 j oms.
01:33
Now we are given the voltage equals to 8 sine 2 t plus 30 degrees volt.
01:42
Writing it into poloform, we write it as v equals to 8, that is the amplitude, an angle that is the phase 5 volts...