Question

An inductive circuit of resistance 2 ohm and inductance 0.01 H is connected to a 250-V, 50-Hz supply. What capacitance placed in parallel will produce resonance ? Find the total current taken from the supply and the current in the branch circuits.

Answer 1

at resonance C = L/Z²

Now, R = 2Ω, X_L = 314 × 0.01 = 3.14 Ω ; Z = √(2² + 3.14²) = 3.74 Ω 

C = 0.01/3.742 =714 × 10⁻⁶ F = 714 μF ; 

I_RL = 250/3.74 = 66.83 A 

tan φ_L = 3.14/2 = 1.57 ; φ_L = tan⁻¹ (1.57) = 57.5º 

Hence, current in R-L branch lags the applied voltage by 57.5º

∴ I_C = V/X_C = V/1/X_C = ωVC = 250 × 314 × 714 × 10⁻⁶ = 56.1 A

This current leads the applied voltage by 90º. 

Total current taken from the supply under resonant condition is 

I = I_RL cos φ_L = 66.83 cos 57.5º = 66.83 × 0.5373 = 35.9 A 

(or I = V/(L/CR))