Chapter 22, Problem 31P

To determine

The capacitance

Answer to Problem 31P

The answer is $327pF$ .

Explanation of Solution

Given info:

Capacitance of AM radio, $C_1=2800pF$ and frequency $f_1=550kHz$

Frequency of another end, $f_2=1610kHz$

Formula used:

Formula for the resonant frequency of an LC circuit

  $f=\frac{1}{2\pi \sqrt{LC}}$

Where, $f$ is frequency

  $L$ is inductance and $c$ is capacitance

Calculation:

The ratio of two frequencies $f_1,f_2$ in terms of two capacitors $C_1,C_2$ (assuming a constant inductance):

  $F^2={\left(\frac{1}{2\pi }\right)}^2\left(\frac{1}{LC}\right)$

  ${\left(\frac{f_2}{f_1}\right)}^2=\frac{C_1}{C_2}$

Substitute all the given values:

  ${\left(\frac{1610kHz}{550kHz}\right)}^2=\frac{2800pF}{C_2}$

  $\Rightarrow C_2=327pF$

Conclusion: The required capacitance is $327pF$ .