Question

A television channel is assigned the frequency range from 54 MHz to 60 MHz. A series RLC tuning circuit in a TV receiver resonates in the middle of this frequency range. The circuit uses a 21 pF capacitor.
A. What is the value of the inductor?B. In order to function properly, the current throughout the frequency range must be at least 50% of the current at the resonance frequency. What is the minimum possible value of the circuit's resistance?Express your answers to two significant figures and include the appropriate units.

Answer 1

Answer:

A.) L = 0.37 μH  B) 7.61 Ω

Explanation:

A) At resonance, the circuit behaves like it were purely resistive , so the reactance value must be 0.

So, the following condition must be met:

ω₀*L = 1/ (ω₀*C) ⇒ ω₀² = 1/LC

We know that, for a sinusoidal source, there exists a fixed relationship between the angular frequency ω₀ and the frequency f₀, as follows:

ω₀ = 2*π*f₀

⇒ (2*π*f₀)² = 1/(L*C)

Replacing by the givens (f₀, C), we can solve for L:

L = 1 /((2*π*f₀)²*C) = 1/(2*π*57*10⁶)² Hz²*21*10⁻¹² f = 0.37 μH

b) At resonance, the current can be expressed as follows:

I₀ = V/Z = V/R

We need to find the minimum value of R that satisfies the following equation:

I = 0.5 I₀ = 0.5 V/R = V/Z

⇒ 0.5/R = 1/√(R²+X²)

Squaring both sides , we have:

(0.5)²/R² = 1/ (R²+X²)

⇒ 0.25 (R²+X²) = R² ⇒ R² = X² / 3

We need to find the value of R that satisfies the requested condition througout the frequency range.

So, we need to find out the value of the reactance X in the lowest and highest frequency, as follows:

Xlow = ωlow * L - 1/(ωlow*C)

⇒ Xlow = ( (2*π*54*10⁶)*0.37*10⁻⁶) - 1/ ((2*π*54*10⁶)*21*10⁻¹²) = -14.81Ω

Xhi = ωhi * L - 1/(ωhi*C)

⇒ Xhi = ( (2*π*60*10⁶)*0.37*10⁻⁶) - 1/ ((2*π*60*10⁶)*21*10⁻¹²) = 13.18Ω

For these reactance values, we can find the corresponding values of R as follows:

Rlow² = Xlow²/3 = (-14.81)²/3 = 75Ω² ⇒ Rlow = 8.55 Ω

Rhi² = Xhi² / 3 = (13.18)²/3 =  56.33Ω² ⇒Rhi =  7.61 Ω

The minimum value of R that satisfies the requested condition is R= 7.61ΩΩ.