A radio can tune over the frequency range of a portion of MW broadcast band: (800kHz to 1200kHz). If its LC circuit has an effective inductance of 200μH, what must be the range of its variable capacitor?
[Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radiowave.]
Exercise
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The range of frequency (v) of a radio is 800kHz to 1200kHz.
Lower tuning frequency, $v_1 = 800kHz = 800 \times 10^3Hz$
Upper tuning frequency, $v_2 = 1200kHz = 1200 \times 10^3Hz$
Effective inductance of circuit $L = 200\mu H = 200 \times 10^{-6}H$
Capacitance of variable capacitor for $v_1$ is given as:
$\text{C}_1=\frac{1}{\omega_1^2\text{L}}$
Where,
$\omega_1=$ Angular frequency for capacitor $C_1$
$=2\pi\text{v}_1=2\pi\times800\times10^3\text{rad}\text{ s}^{-1}$
$\therefore\ \text{C}_1=\frac{1}{(2\pi\times800\times10^3)^2\times200\times10^{-6}}$
$= 1.9809 \times 10^{-10}F = 198.1\ pF$
Capacitance of variable capacitor for $v_2,$
$\text{C}_2=\frac{1}{\omega_2^2\text{L}}$
Where,
$\omega_2=$ Angular frequency for capacitor $C_2$
$=2\pi\text{v}_2=2\pi\times1200\times10^3\text{rad}\text{ s}^{-1}$
$\therefore\ \text{C}_2=\frac{1}{(2\pi\times1200\times10^3)^2\times200\times10^{-6}}$
= 88.04 pF
Hence, the range of the variable capacitor is from 88.04 pF to 198.1 pF
Lower tuning frequency, $v_1 = 800kHz = 800 \times 10^3Hz$
Upper tuning frequency, $v_2 = 1200kHz = 1200 \times 10^3Hz$
Effective inductance of circuit $L = 200\mu H = 200 \times 10^{-6}H$
Capacitance of variable capacitor for $v_1$ is given as:
$\text{C}_1=\frac{1}{\omega_1^2\text{L}}$
Where,
$\omega_1=$ Angular frequency for capacitor $C_1$
$=2\pi\text{v}_1=2\pi\times800\times10^3\text{rad}\text{ s}^{-1}$
$\therefore\ \text{C}_1=\frac{1}{(2\pi\times800\times10^3)^2\times200\times10^{-6}}$
$= 1.9809 \times 10^{-10}F = 198.1\ pF$
Capacitance of variable capacitor for $v_2,$
$\text{C}_2=\frac{1}{\omega_2^2\text{L}}$
Where,
$\omega_2=$ Angular frequency for capacitor $C_2$
$=2\pi\text{v}_2=2\pi\times1200\times10^3\text{rad}\text{ s}^{-1}$
$\therefore\ \text{C}_2=\frac{1}{(2\pi\times1200\times10^3)^2\times200\times10^{-6}}$
= 88.04 pF
Hence, the range of the variable capacitor is from 88.04 pF to 198.1 pF
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