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.

the radio can tune wavelengths between 1.88 and 5.97 m

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 =

let's look for the extreme values

C = 1  10⁻¹² F

         w =

         w =

         w = 10⁹ rad / s

C = 10 10⁻¹² F

         w = Ra 1/10 10-12 1 10-6

         w = 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

answer: option c. i.e. electric current flowing “downhill” from the negative electrode to the positive electrode in a battery

explanation:

the water analogy to the circuits define the pipes (water flows) to the electric current. now, we know that the electric current flows from the positive terminal to the circuit (as shown in the diagram attached).

hence, option c is the correct answer.

d. air pressure decreases as altitude increases.