Chapter 22, Problem 32P

To determine

The capacitance

Answer to Problem 32P

The answer is $1.5pF$ .

Explanation of Solution

Given info:

Inductance, $L=1.8\mu H=1.8\times {10}^{-6}H$

Frequency of station, $f=96.1MHz=96.1\times {10}^{6}Hz$

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:

Rearrange the formula: $f=\frac{1}{2\pi \sqrt{LC}}$ . So, $f=\left(\frac{1}{2\pi }\right){\left(\frac{1}{LC}\right)}^{\frac{1}{2}}$

Substitute the given values:

  $96.1\times {10}^{6}Hz=\left(\frac{1}{2\pi }\right){\left(\frac{1}{\left(1.8\times {10}^{-6}H\right)C}\right)}^{\frac{1}{2}}$

  $C=1.5\times {10}^{12}F=1.5pF$

Conclusion: The required capacitance is $1.5pF$ .