Suggested languages for you:

Americas

Europe

Q100PE

Expert-verified
Found in: Page 864

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# 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 2.50pF capacitor? (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?

1. The inductance needed to produce this resonant frequency is $1.31\mu H$.
2. The capacitance at this frequency is $1.66pF$.
See the step by step solution

## Step 1: Definition of capacitor

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.

## Step 2: Write the formula of frequency in a series circuit

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)

## Step 3: Find the inductance needed to produce this resonant frequency(a)

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,

$L=\frac{1}{4{\pi }^{2}×{\left(8.8×{10}^{7}Hz\right)}^{2}×\left(2.50×{10}^{-12}F\right)}\phantom{\rule{0ex}{0ex}}=1.31×{10}^{-6}H\left(\frac{1\mu H}{{10}^{-6}H}\right)\phantom{\rule{0ex}{0ex}}=1.31\mu H$

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

## Step 4: Calculate the capacitance of the frequency(b)

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

$C=\frac{1}{4{\pi }^{2}×{\left(1.08×{10}^{8}Hz\right)}^{2}×\left(1.31×{10}^{-6}H\right)}\phantom{\rule{0ex}{0ex}}=1.66×{10}^{-12}F\left(\frac{1pF}{{10}^{-12}F}\right)\phantom{\rule{0ex}{0ex}}=1.66pF$

Therefore, the capacitance at this frequency is 1.66 pF .