A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 μH, 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.]
The range of frequency (ν) of a radio is 800 kHz to 1200 kHz.
Lower tuning frequency, ν1 = 800 kHz = 800 × 103 Hz
Upper tuning frequency, ν2 = 1200 kHz = 1200× 103 Hz
Effective inductance of circuit L = 200 μH = 200 × 10−6 H
Capacitance of variable capacitor for ν1 is given as
$C_{1}=\frac{1}{\omega_{1}^{2} L}$
Where,
$\omega_{1}=$ Angular frequency for capacitor $C_{1}$
$=2 \pi v_{1}=2 \pi \times 800 \times 10^{3} \mathrm{rad} \mathrm{s}^{-1}$
$\therefore C_{1}=\frac{1}{\left(2 \pi \times 800 \times 10^{3}\right)^{2} \times 200 \times 10^{-6}}$
$=1.9809 \times 10^{-10} \mathrm{~F}=198.1 \mathrm{pF}$
Capacitance of variable capacitor for $v_{2}$,
$C_{2}=\frac{1}{\omega_{2}^{2} L}$
Where,
$\omega_{2}=$ Angular frequency for capacitor $C_{2}$
$=2 \pi v_{2}=2 \pi \times 1200 \times 10^{3} \mathrm{rad} \mathrm{s}^{-1}$
$\therefore C_{2}=\frac{1}{\left(2 \pi \times 1200 \times 10^{3}\right)^{2} \times 200 \times 10^{-6}}$
$=88.04 \mathrm{pF}$
Hence, the range of the variable capacitor is from 88.04 pF to 198.1 pF.