Problem 24.9

24.9.
From equation 24.18, we see that
$f_{0} = \frac{1}{2\pi \sqrt{LC}}$
Solving for the capacitance, we have
$C = \frac{1}{4 \pi^{2} L f_{0}^{2}}$
Plugging in the value of the inductance, we have
$C = \frac{1}{(7.90 \times 10^{-5} H) f_{0}^{2}}$
The range of the FM band is from 88 MHz to 108 MHz, so the range that the capacitance that must be available is given by
C₁₀₈ = 1.09 pF
C₈₈ = 1.64 pF
So, we must have a variable capacitor that can varry between 1.09 pF and 1.64 pF in order to tune the entire FM band.