Question

A 60 Hz sinusoidal voltage v = 141 sin ω t is applied to a series R-L circuit. The values of the resistance and the inductance are 3 Ω and 0.0106 H respectively. 

(i) Compute the r.m.s. value of the current in the circuit and its phase angle with respect to the voltage. 

(ii) Write the expression for the instantaneous current in the circuit. 

(iii) Compute the r.m.s. value and the phase of the voltages appearing across the resistance and the inductance. 

(iv) Find the average power dissipated by the circuit. 

(v) Calculate the p.f. of the circuit.

Answer 1

V_m = 141 V; V = 141/ √2 100 = V 

∴ V = 100 + j0 

X_L = 2π × 60 × 0.0106 = 4Ω. Z = 3 + j 4 = 5 ∠ 53.1° 

(i) I = V/Z = 100 ∠ 0°/5∠ 53.1° = 20 ∠ − 53.1° 

Since angle is minus, the current lags behind the voltage by 53.1 

(ii) I_m = √2 x 20  = 28.28; 

∴ i = 28.28 sin (ωt − 53.1°) 

(iii) V_R = I_R = 20 ∠ − 53.1° × 3 = 60 ∠ − 53.1° volt. 

V_L = jIX_L = 1 ∠ 90° × 20 ∠ − 53.1° × 4 = 80 ∠ 36.9° 

(iv) P = V_I cos φ = 100 × 20 × cos 53.1° = 1200 W. 

(v) p.f. = cos φ = cos 53.1° = 0.6.