00:01
In this question, we are given some information about the rfc circuit.
00:08
The resonant frequency, f0, is 2 ,000 over pi, and then we are given that at omega greater than omega -0, xl is 12 oms, and xc is equal to 8 oms.
00:31
So we need to find the inductance and capacitance in this circuit.
00:38
So to solve for this question, first we write out a few things.
00:46
So omega -0 is equal to 2 pi f -0.
00:52
This is 2 -py times 2 ,000 over pi.
00:58
So we have 4 ,000 is equal to 1 over square root l -c.
01:04
The product of lc is 1 over 4 ,000 square and then xl is omega l equals to 12 oms and xc is equal to one over omega c which is 8 oms so we can see that omega is 1 over 8c and putting the omega 1 over 8c into the formula for omega l okay so we have omega l equals to l over 8 c which is 12 right so l equals to l equals to 96c so we are going to put we have two equations okay lc equals to 1 with 4 ,000 square and l is 96 c so we have a sub 2 into 1...