00:01
To solve the first item, we have to calculate what is the phase, and only then we can draw a phase or diagram.
00:08
And to calculate the phase, we have to do the following.
00:11
Remember that the tangent of the phase is equal to the difference between the inductive and capacitive reactances divided by the resistance.
00:22
So, before calculating the phase, we have to calculate what are the reactances.
00:27
We will do this on the other side.
00:31
Beginning by the reactance of the inductor, we have the following.
00:34
So the reactance of the inductor is equal to 2 pi times the frequency times the inductance.
00:41
Plugging in the values that were given by the problem, we have 2 pi times 60 times 22 .5 times 10 to the minus 3.
00:53
And this is approximately 8 .482 oms.
01:01
And now we calculate the reactance of the capacitor, which is equal to 1 divided by 2 pi times the frequency times the capacitance.
01:10
Plagging in the values given by the problem, we have 1 divided by 2 pi times 60 times 32 .2 times 10 to the minus 6.
01:26
And this is approximately 8 .2 .38 oms.
01:36
And then we can calculate what is the tangent of the phase.
01:40
So the tangent of the phase is equal to 8 .482 minus 8 .82 .388.
01:55
Divided by 105 and then the phase is equal to the inverse tangent of 8 .482 minus 82 .382 divided by 105 which is approximately minus 35 .14 degrees.
02:20
Now we can proceed to draw our phasor diagram.
02:26
Our reference axis.
02:28
We will have three phasers, one for the resistor, one for the inductor, and one for the capacitor, and also another one for the source.
02:36
So there is a total of four phasors.
02:39
Note that the resistance is bigger than both reactances.
02:44
So our biggest phaser will be the resistance phaser.
02:49
So the resistor phaser will be this one.
02:56
And now, now we can draw either the inductor phaser or the capacitor phaser.
03:02
We will choose to draw the inductor phaser first...