The lowest frequency in the FM radio band is 88.0 MHz. (a) What inductance is needed to produce this resonant frequency if it is connected to a 2.50pFcapacitor? (b) The capacitor is variable, to allow the resonant frequency to be adjusted to as high as 108 MHz. What must the capacitance be at this frequency?

A capacitor is a two-terminal electrical component that may store energy in the form of an electric charge. It is made up of two electrical wires separated by a certain distance. The space between the conductors can be filled with the vacuum or a dielectric, which is an insulating substance.

Remember that in a series circuit with a coil and a capacitor, the resonant frequency is given by

$f=\frac{1}{2\pi \sqrt{LC}}$…………….(1)

The coil's inductance can be expressed as follows by rearranging the above equation (1), such that,

$L=\frac{1}{4{\pi }^2{f}^{2}C}$…………….(2)

Substituting the given data in equation (2) we can get,

$$\begin{gathered} L=\frac{1}{4{\pi }^2\times {\left(8.8\times {10}^{7}Hz\right)}^2\times \left(2.50\times {10}^{-12}F\right)} \\ =1.31\times {10}^{-6}H\left(\frac{1\mu H}{{10}^{-6}H}\right) \\ =1.31\mu H \end{gathered}$$

Therefore, the inductance needed to produce this resonant frequency is $1.31\mu H$.

We can solve for the value of the capacitor at a new frequency if we know the coil inductance, using the equation (2) such that,

$C=\frac{1}{4{\pi }^2{f}^{2}L}$………………(3)

Substituting the given data in the above equation (3), we'll have

$$\begin{gathered} C=\frac{1}{4{\pi }^2\times {\left(1.08\times {10}^{8}Hz\right)}^2\times \left(1.31\times {10}^{-6}H\right)} \\ =1.66\times {10}^{-12}F\left(\frac{1pF}{{10}^{-12}F}\right) \\ =1.66pF \end{gathered}$$

Therefore,the capacitance at this frequency is 1.66 pF .