Chapter 22, Problem 33P

To determine

The range of values must the inductance have to tune station form $88MHz$ to $108MHz$ .

Answer to Problem 33P

The answer is $2.59nH\le L\le 3.89nH$ .

Explanation of Solution

Given info:

Capacitance of FM radio tuning circuit, $C=840pF=840\times {10}^{-12}F$ (fixed)

Primary Frequency of station, $f_1=88MHz=88000000Hz$

Secondary Frequency of station, $f_2=108MHz=108000000 Hz$

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

Calculation for primary inductance, $L_1$

  $f=\left(\frac{1}{2\pi }\right){\left(\frac{1}{L_{1}C}\right)}^{\frac{1}{2}}$

Substitute the given values:

  $88000000Hz=\left(\frac{1}{2\pi }\right){\left(\frac{1}{L{1}_{}\left(840\times {10}^{-12}F\right)}\right)}^{\frac{1}{2}}$

  $L_1=3.89\times {10}^{-9}H=3.89nH$

Calculation for secondary inductance, $L_2$

  $f=\left(\frac{1}{2\pi }\right){\left(\frac{1}{L_{1}C}\right)}^{\frac{1}{2}}$

Substitute the given values:

  $108000000 Hz=\left(\frac{1}{2\pi }\right){\left(\frac{1}{L{2}_{}\left(840\times {10}^{-12}F\right)}\right)}^{\frac{1}{2}}$

  $L_2=2.59\times {10}^{-9}H=2.59nH$

Conclusion: The range of values must the inductance have to tune station form $88MHz$ to $108MHz$ is $2.59nH\le L\le 3.89nH$ .