00:03
In this problem, we have an lc circuit with a variable capacitance.
00:13
The minimum capacitance of this capacitor is going to be 10 pico -pharats, and the maximum capacitance of the capacitor is going to be 365 picofarhertz.
00:26
What we have to have, we need to find in part a, we need to find what is the ratio between the maximum frequency, of that oscillating circuit over the minimum frequency.
00:40
Now, how do we find this ratio? now, the frequency of an lc circuit is given by 1 over 2 pi times 1 over the square root of the inductance times capacitance, and that means that the frequency, like if the only variable is the capacitance, the frequency is proportional to 1 over the square root of the capacitance.
01:02
So the bigger the capacitance, the smaller the frequency.
01:06
And that means that the maximum frequency correspond to the minimum capacitance, and the minimum frequency correspond to the maximum capacitance.
01:20
Then having said that, we find that the ratio of the maximum frequency over the minimum frequency is going to be equal to the square root of the ratio between the maximum capacitance over the minimum capacitance.
01:40
And that is the square root of 365 picofarats over 10 picofarad.
01:48
And so this ratio is equal to 6.
02:00
Now in part b, we are told that that ratio is actually too big for our purposes.
02:06
And what we actually want is we actually want for the minimum frequency to be equal to 0 .54 megahertz and the maximum frequency equal to 1 .60 megahertz.
02:26
And in order to achieve these frequencies, we're going to add a capacitor in parallel.
02:40
Sorry, we have to add a capacitor in series.
02:49
And we need to find how, what is the, what is the capacitance of this capacitor that is going to make our frequencies take these values.
03:01
Okay, so first we need to find what is the ratio of frequencies that we actually need.
03:07
So we have that f max over f mean is going to be equal to 1 .60 over 0 .54.
03:16
So our ratio of frequencies is 2 .96...