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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.]
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Solution
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 v2,
`"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.
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