Question

IP Tuning a Radio A radio tuning circuit contains an RLC circuit with R = 4.5 and L = 3.0 pH

Answer 1

R = 4.5 Ω; L = 3.0 x 10^-6 H

Part A

Resonant frequency, f₀ = 86 x 10⁶ Hz

Resonant frequency, f₀ = 1/2π (LC)^1/2

Capacitance, C = 1/4π² Lf² = 1/4 x 3.14 x 3.14 x 3.0 x 10^-6 x (86 x 10⁶ )² = 1.1 x 10^-12 F

Capacitance, C = 1.1 x 10^-12 F

Part B

The impedance is minimum at the resonance. Any change in capacitance will disturb the resonance and the impedance will be increased.

· Increases

Part C

Impedance, Z = {R² + (X_C – X_L)²}^1/2

At resonance, X_C = X_L

Impedance at resonance, Z = R = 4.5 Ω

Part E

New capacitance, C1 = C + 1% of C = 1.01C = 1.01 x 1.1 x 10^-12 F = 1.11x 10^-12 F

Capacitive reactance, X_c = 1/2πfC = 1/2 x 3.14 x 86 x 10⁶ x 1.11x 10^-12 = 1668.1 Ω

Inductive reactance, X_L = 2πfL = 2 x 3.14 x 86 x 10⁶ x 3.0x 10^-6 = 1620.2 Ω

Impedance, Z = {4.5² + (1668.1 – 1620.2)²}^1/2 = 48.1 Ω