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 $200\mu H$, what must be the range of its variable capacitor ?

Solution: The range of frequency (v) of a radio is 800KHz to 1200 KHz.
Lower tuning frequency, $v_1=800KHz=800\times {10}^{3}Hz$
Upeer tuning frequency, $v_2=1200KHz=1200\times {10}^{3}Hz$
Effective inductance of circuitL=200$\mu H=200\times {10}^{-6}H$
Capacitance of variable capacitor for $v_1$ is given as,
$=\frac{1}{{\omega }_1^{2}L}$
Where,
${\omega }_1=$ Angular frequency for capacitor $C_1$
$=2\pi v_1=2\pi \times 800\times {10}^{-3}\text{rad}{s}^{-1}$
$\therefore C_1=\frac{1}{{\left(2\pi \times 800\times {10}^3\right)}^2\times 200\times {10}^{-6}}$
$=1.9809\times {10}^{-16}F=198.1pF$
Capacitanc e of variable capacitor for $v_2$.
$=\frac{1}{{\omega }_2^{2}L}$
Whre,
${\omega }_2=$ angular frequency for capacitor $C_2$
$=2\pi v_2=2\pi \times 1200\times {10}^3\text{rad}{s}^{-1}$
$\therefore C_2=\frac{1}{{\left(2\pi \times 1200\times {10}^3\right)}^2\times 200\times {10}^{-6}}$
=$88.04pF$
Hence, the range of the variable capacitor is from 88.04pF to 198.1 pF.