00:01
The variable capacitor at the center of its range.
00:11
Variable capacitor at the center of its range is c1 is equal to 2 plus 6x by 2.
00:27
So it comes out to be 4 nanofarid.
00:29
So its value is 4 into 10 to the power of minus 9 ferret.
00:35
So the resonant frequency is given as f0 is equal to 1.
00:40
2 pi under root lc so from here we can write l is equal to 1 upon 4 pi square f0 square c now let us substitute the values here so it comes out to be 1 over 4 pi square the value of f not is 720 kilohertz it means 720 into 10 to the power 3 huds so, we can write 720 into 10 to the power 3 square.
01:19
The value of capacitance is 4 nanofarad that is 4 into 10 x2 power minus 9.
01:25
So on calculating the value of l comes out to be 12 .2 into 10 to power minus 6 in v.
01:33
So this is the answer for e part.
01:35
Let us solve for b part.
01:42
Required inductance so from a part we can see that required inductance l is equal to 12 .2 into 10 to power minus 6 in d so from the available inductors the one which can be used to make this configuration are l1 is equal to 5 into 10 to power minus 6 in re and l2 is equal to 7 .2 into 10 to 0 .2 x to power minus 6 in the series combination of l1 and l2 will give series combination of l1 and l2 will give ls is equal to l1 plus l2 so on adding we are getting it as the required inductance so, we can say that series combination is the answer.
02:55
Now let us solve for the c part.
03:01
The capacitance is varying from c1 is equal to 2 into 10 to power minus 9 ferret to to c2 is equal to 6 into 10 to power minus 9 ferret.
03:15
And the value of l is 12.
03:18
0 .6 into 10 raise 2 power minus 6 henry.
03:22
So the range of resonant frequencies in order to find the range of resonant frequencies we can write f02 is equal to 1 upon 2 pi under root l c2...