00:01
In the first item of this problem, we have to do a sketch of the phaser diagram for that circuit.
00:07
We begin by remembering that the cosine of the phase angle is equal to the resistance divided by the impedance.
00:14
And for this case, the impedance is given by the resistance squared plus the inductive reactance squared.
00:22
And so the cosine of the phase is equal to r divided by the impedance.
00:28
Now, remember that the inductive reactance is equal to 2 pi times the frequency times the inductance.
00:37
Then the cosine of the phase is plugging the values the problem gave to us, 105 divided by the square root of 105 squared plus 2 pi times 60 times 22 .5 times 10 to the minus 3, is a millie prefix squared and this is approximately 0 .997.
01:12
Then the phase is equal to the inverse cosine of 0 .997 which is approximately 4 .4 degrees.
01:28
Then our phasor diagram is as follows.
01:31
We have two axis and three phasers...