00:01
In this problem on the topic of alternating current circuits, we are given an lc circuit which consists of a capacitor, c, which has a rating of 2 .5 microferds, and an inductor l, which is a rating of 4 millie henries.
00:14
The capacitor is fully charged using a battery and then connected to the inductor.
00:19
The circuit is opened and a resistance r is inserted in series with the inductor and capacitor, and the capacitor is again fully charged using the same battery and connected to the circuit.
00:30
We find that the angular frequency of the dammed oscillations in the rlc circuit is 20 % less than the angular frequency of the oscillations in the lc circuit.
00:40
And using this information, we want to find the resistance of the resistor.
00:43
The time after the capacitor is reconnected in the circuit that the amplitude will, that the amplitude of the dam current will be 50 % of the initial amplitude.
00:55
And we want to find how many complete damp oscillations will have occurred in that time.
01:00
Now for the oscillation frequency, we have the frequency of the rlc circuit to be 0 .8th of the frequency for the lc circuit.
01:15
And we know the angular frequency of the rlc circuit, omega rlc, is equal to the square root of omega -0 squared minus r over 2ll.
01:32
Squared and this is equal to 0 .8 omega not which means that omega not squared minus r over 2 l squared is equal to 0 .64 omega not squared and so rearranging and solving for the resistance we get r to be 1 .2 .2 times l omega -0, which is 1 .2 times the square root of l over 1 .2, rather, times l over the square root of l c, and this is 1 .2 times the square root of l over c.
02:38
And so putting our values in, we can find this resistance.
02:42
This is 1 .2 times the square root of the inductance l given to be 4 times 10 to the minus 3 henrys divided by the capacitance c which is 2 .5 times 10 to the minus 6 ferrets...