00:01
For part a, we'd like to find the reactances of the inductor and capacitor.
00:06
So, xl is equal to j.
00:08
Omega l, which is 20 times 20 times 10 to the negative 3, and we get 0 .4 oms.
00:14
Xc is 1 over j.
00:16
Omega c, which is 1 over j .20 times 5 times 10 to the negative 3, which is negative 10 oms.
00:27
So for part b, we're asked to find the supply voltage in the phaser domain.
00:32
We're given that the voltage is 25 square root of 2 sign of 20t.
00:37
So in the phaser domain, we take this and divided by square root of 2 times lo.
00:43
So we have 25 square root of 2 divided by square root of 2 lo, and we get 25 lo.
00:51
Now question part 3 asks the total impedance seen by the voltage source.
00:56
So we need to take a look at all the our options.
01:00
So we'll have negative j times 10 plus j times 0 .4.
01:04
That's equal to negative j 9 .6.
01:09
So now we need to look at 15.
01:11
So we'll have 15 times negative j 9 .6 divided by 15 minus j 9 .6...