00:01
Hi there, so for this problem, we are told that an rlc series circuit has a resistance that is given and that is equal to 2 .75 oms.
00:18
The inductor, the inductance, it has a value of 95 micro henries, and the capacitor, a capacitance of 72 .10.
00:35
0 .5 micro ferrets.
00:42
So the question for par...
00:46
So we are asked to find the circuit's impedance in alms with a frequency that is known and that frequency is 105 hertz.
00:59
So for this, we use the usual equation for the impedance that states that the impedance is equal to the square root of the resistance to the square, and this plus the inductance reactance minus the capacitive reactance, and all of that to the square, the square root of this.
01:28
So we just need to simply substitute the values that we are given.
01:32
The resistance is 2 .75 oms, and that to the square plus the inductance, the inductance, the inductance, the inductance, the inductance is defined as two, times pi times the frequency times the inductance.
01:52
And this minus the capacitive reactance, that is 1 divided by 2 times pi times the frequency times the capacitance, and all of that to the square.
02:02
So we just need to simply substitute the values into this expression right here.
02:08
So let me just substitute that in here right away.
02:12
So we start with the inductive, the inductance, which is 2 times pi times the frequency.
02:19
That we are given that is 105 hearse, and this times the inductance, which is 95, micro, which means 10 to the minus 6, henry's, and this minus 1 divided by 2 times pi times the frequency, 105 hertz times the capacitance, which is equal to 72 .5 times 10 to the minus 6, pharacts.
02:56
So all of these to the square, the square root of all of this.
02:59
So using our calculator, we obtain an impedance of 21 .025 oms.
03:13
So that's a solution for the first part of this problem.
03:20
And now we pass to part b of this problem.
03:27
And the question in this part is about to find the circuit impedance in ums at a frequency.
03:38
Now the frequency is 6 .5 times 10 to the 3 because it's kilohertz.
03:49
So we need to use the same equation as before, this equation right here.
03:54
But the only thing that we need to change is the frequency.
03:59
So let me just put this in here and let's just change the frequency.
04:07
So the frequency is, as i said, 6 .5 times 10 to the 3 hertz.
04:19
The same in here.
04:26
6 .5 times 10 to the 3 hertz.
04:29
So using our calculator, in this case, the impedance is equal to so the impedance in this case is 4 .48 oms.
04:44
So that's a solution for part b of this problem...