00:01
Hi there, so for this problem we have an lc circuit that consists of a capacitor, which has a capacitance of 2 .50 micropharets, and an inductor l equals to 4 milliehendres.
00:25
So the capacitor is fully charged using a battery and then connected to the inductor.
00:36
An oscilloscope is used to measure the frequency of the oscillations in the circuit.
00:42
Next, the circuit is open and a resistor art is inserted in series with the inductor and the capacitor.
00:50
Now, we are told that the capacitor is again fully charged using the same battery and then connected to the circuit.
01:00
The angular frequency of the damped oscillations in the rlc circuit is, found to be 20 % less than the angular frequency of the oscillations in the lc circuit.
01:17
So the information that is given is that the angular frequency in the rlc circuit is equal to 0 .8, which corresponds to 20 % less than the frequency in the lc circuit.
01:37
So with that said, the first thing that we need to determine is the resistance of the resistor.
01:46
So we know that the rlc frequency is equal to the square root of the natural frequency minus r the resistance over two times the inductance and all of that.
02:13
To the square the square root of all of that, where we know that the angular frequency sub -zero, which also corresponds to the frequency of the lc circuit, is equal to one over the square root of the product between the inductance and the capacitance.
02:41
So with that information, we can obtain the resistance because we can solve for the resistance in this expression right here.
02:51
So first we substitute the condition that we have.
02:56
So we have that the angular frequency is 0 .8 the frequency for the lc or the frequency to 0 and this is equal to the square root of omega sub 0 to the square minus r over 2 times the inductance to the square.
03:23
So in here what we can do to get rid of the square root is to elevate both sides of this equation by 2.
03:32
So we obtain omega sub 0 to the square we can pass to the other side minus r over 2 times the inductance to the square and this is equal to 0 .64 omega sub 0 .0.
04:03
So with that, we can solve, keep solving.
04:16
So we will obtain 0 .36 omega -0 to the square, and that is equal to the resistance to the square divided by four times the inductance to the square.
04:31
And if we solve for the resistance, we will obtain that the resistance is equal to 1 .1.
04:41
Times the inductance times the angular frequency sub 0.
04:50
And in here we substitute that of frequency, which is 1 .2.
04:56
We obtain this and simplifying for this expression, we will obtain that this is equal to 1 .2 the square root of the inductance over the capacitance.
05:10
So to obtain the resistance, we just need to substitute those values in here.
05:16
So we will have that this is equal to 1 .2.
05:19
The square root of the inductance, which is 4 times 10 to the minus 3 hundreds, divided by the capacitance, which is 2 .5 times 10 to the minus 6 ferrets, the square root of that.
05:39
So from here we obtain a resistance of 48 oms.
05:45
Now for part b of this problem, we are asked about how long after the capacitor is reconnected in the circuit will the amplitude of the damped current through the circuit be 50 % of the initial amplitude...