Question

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 radio wave.]

Answer 1

The range of frequency (v) of a radio is 800 kHz to 1200 kHz.

Lower tuning frequency, v₁ = 800 kHz = 800 × 10³ Hz

Upper tuning frequency, v₂ = 1200 kHz = 1200 × 10³ Hz

Effective inductance of circuit L = 200 μH = 200 × 10⁻⁶ H

Capacitance of variable capacitor for v₁ is given as:

`"C"_1 = 1/(ω_1^2 "L")`

Where,

ω₁ = Angular frequency for capacitor C₁

`= 2pi  "v"_1 = 2pi xx 800 xx 10^3  "rad s"^-1`

∴ `"C"_1 = 1/((2pi xx 800 xx 10^3)^2 xx 200 xx 10^-6)`

`= 1.9809 xx 10^-10`

F = 198.1 pF

Capacitance of variable capacitor for v_2,

`"C"_2 = 1/(ω_2^2 "L")`

Where,

ω₂ = Angular frequency for capacitor C₂

`= 2pi  "v"_2 = 2pi xx 1200 xx 10^3  "rad s"^-1`

∴ `"C"_2 = 1/((2pi xx 120 xx 10^3)^2 xx 200 xx 10^-6)`

= 88.04 pF

Hence, the range of the variable capacitor is from 88.04 pF to 198.1 pF.