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.]

Solution

Frequency of radio, f₁ = 800 kHz = 800 × 10³ Hz,
Frequency tuned to, f₂ = 1200 kHz = 1200 × 10³ Hz
Inductance of LC circuit, L = 200 μH = 200 × 10^–6 H

We know that the resonant frequency is given by

$f = \frac{1}{2 π \sqrt{LC}} or f^{2} = \frac{1}{4 π^{2} LC}$

$\Rightarrow$ $C = \frac{1}{4 π^{2} {Lf}^{2}}$

Therefore, $C_{1} = \frac{1}{4 π^{2} {Lf}^{2}} = \frac{49}{4 (22)^{2} \times 200 \times 10^{- 6} {(800 \times 10^{3})}^{2}}$

$C_{1} = 197.73 pF .$

Similarly,

$C_{2} = \frac{1}{4 π^{2} {Lf}_{2}^{2}}$
$= \frac{49}{4 (22)^{2} \times 200 \times 10^{- 6} {(1200 \times 10^{3})}^{2}} F$

$C_{2} = 87 . 88 pF .$

Hence, the range of the variable condenser is from 87.88 pF to 197.73 pF.

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