Question

LC-circuit of the radio receiver consists of variable capacitor (Cmin= 1 pF, Cmax=10 pF) and inductor

with inductance 1 µH. Determine the wavelength range of this radio receiver.

Answer 1

the radio can tune wavelengths between 1.88 and 5.97 m

Step-by-step explanation:

The signal that can be received is the one that is in resonance as the impedance of the LC circuit.

X = X_c - X_L

X = 1 / wC - w L

at the point of resonance the two impedance are equal so their sum is zero

X_c = X_L

1 / wC = w L

w² = 1 / CL

w =
`\sqrt{(1)/(CL) }`

let's look for the extreme values

C = 1 10⁻¹² F

w =
`\sqrt{(1)/( 1 \ 10^(-12) \ 1 \ 10^(-6)) }`

w =
`\sqrt{1 \ 10^(18)}`

w = 10⁹ rad / s

C = 10 10⁻¹² F

w =
`\sqrt{(1)/(10 \ 10^(-12) \ 1 \ 10^(-6)) }`Ra 1/10 10-12 1 10-6

w =
`\sqrt{0.1 \ 10^(18)}`Ra 0.1 1018

w = 0.316 10⁹ rad / s

Now the angular velocity and the frequency are related

w = 2π f

f = w / 2π

the light velocity is

c = λ f

λ = c / f

we substitute

λ = c 2π/w

we calculate the two values

C = 1 pF

λ₁ = 3 10⁸ 2π / 10⁹

λ₁= 18.849 10⁻¹ m

λ₁ = 1.88 m

C = 10 pF

λ₂ = 3 10⁸ 2π / 0.316 10⁹

λ₂ = 59.65 10⁻¹ m

λ₂ = 5.97 m

so the radio can tune wavelengths between 1.88 and 5.97 m