Derive an expression for the resonant frequency of

DERIVE AN EXPRESSION FOR THE RESONANT FREQUENCY OF:

While R₁ will not affect the resonance frequency, it will influence the BW. This is a problem frequented in AIR and DD.

Z_r = Total impedance of total combination at resonance.

Z₁₀ = Impedance of original RLC circuit.

Z_T = Z₁₀ . R_X / Z₁₀ + R_X ; Q_T = Z_T. X_L ; BW = f₀ / Q_T

The resonance frequency is calculated 1/2π √1/LC – R²/L² . A few problems could be added.

 (b) Q of the circuit prior to the installation of ext. load.

= 2π × 4.9103 × 10⁶ × 3.5 × 10^-6 / 2.5 = 43.19

(c) Res. impedance of the circuit prior to the installation of ext. resistance

= L/CR = 3.5 × 10^-6 / 300 × 10^-12 × 2.5 = 4666.67 Ω

(d) Impedance of the res. circuit after the installation of 15 K

= 3559.32 Ω

(e) Qg of the circuit after installing the ext. resistance

= 3559.32 / 2лƒL = 32.96

(f) BW prior to the installation of ext. resistance

= 4.9103 × 10⁶ / 43.19 = 113.69 KHz

 (g) BW of the circuit after installing the ext. load

= 4.9103 × 10⁶ / 32.96 = 148.98 KHz

Suppose if 15 K Ω is reduced to 5 K,

the new ZT = 4.66667 × 5 / 4.66667 + 5 = 2.41379 K

New Q = 2.41379 × 10³ = 22.35

New BW = 2.41379 × 10⁶ / 22.35 = 219.67 KHz

In short, RX↓ Q↓ BW↑ || Very important

Ex. 7 A choke coil has a resistance of 40Ω and a Q of 5 at 1000 Hz. It is connected in parallel with a variable capacitor to a 10V 1000 Hz a.c supply. Find (a) X_c for resonance at 1000 KHz. (b) Equivalent impedance as seen by the source at resonance. (c) Values of L & Cat resonance at 1000 H_z. (d) Current drawn from the supply at resonance.

At parallel resonance, the equivalent parallel inductive reactance equal to the capacitive reactance. j208 is cancelled by -j208.

-j208 at 1000 Hz means 1 / 2πƒ × 208 → 0.76519 μF

j200 at 1000 Hz means 0.03183.

Resonance impedance = 1040 Ω

Also by cross check = L / CR = 1040 Ω

Current drawn at resonance = 10 / 1040 = 9.615 mA