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.
