Question

A coil of resistance 20 Ω and inductance 200 μH is in parallel with a variable capacitor. This combination is in series with a resistor of 8000 Ω. The voltage of the supply is 200 V at a frequency of 10⁶ H_z . Calculate (i) the value of C to give resonance (ii) the Q of the coil (iii) the current in each branch of the circuit at resonance.

Answer 1

The circuit is shown in Fig.  

X_L = 2πfL = 2π × 10⁶ × 200 × 10⁻⁶ = 1256 Ω 

Since coil resistance is negligible as compared to its reactance, the resonant frequency is given by

f = 1/2π√(LC)

∴ 10⁶ = 1/(2π√(200 x 10^-6 x C)

(i) ∴ C = 125 μF

(ii) Q = 2μfL/R = (2π x 10⁶ x 200 x 10^-4)/20 = 62.8

(iii) Dynamic resistance of the circuit is = L/CR = (200 x 10^-6)/(125 x 10^-12 x 20) = 80, 000Ω

Total equivalent resistance of the tuned circuit is 

= 80,000 + 8,000 = 88,000 Ω 

∴ Current I = 200/88,000 = 2.27 mA 

p.d. across tuned circuit = current × dynamic resistance 

= 2.27 × 10⁻³ × 80,000 = 181.6 V

Current through inductive branch

= 181.6/√(10² + 1256²) = 0.1445A = 144.5mA

Current through capacitor branch

= V/(1/ωC) =  ωVC = 181.6 × 2π × 10⁶ × 125 × 10⁻¹² = 142.7 mA