Question

A resistor of 10 Ω is connected in series with an inductive reactor which is variable between 2 Ω and 20 Ω . Obtain the locus of the current vector when the circuit is connected to a 250-V supply. Determine the value of the current and the power factor when the reactance is (i) 5 Ω (ii) 10 Ω (iii) 15 Ω .

Answer 1

As discussed in, the end point of current vector describes a semi-circle whose diameter (Fig.) equals 

V/R = 250/10 = 25 A and whose centre lies to right side of the vertical voltage vector OV.

I_max =  250/√(10² + 2²) = 24 A; θ = tan-1(2/10) =   113°;

I_min = 250/√(10² + 20²) = 11.2 A; θ = tan-1(20/10) = 63.5°

(i) θ₁ = tan^–1 (5/10) = 26.7°, p.f. = cos 26.7° = 0.89 

I = OA = 22.4 A

(ii) θ₂ = tan^–1(10/10) = 45°, p.f. = cos 45° = 1;

I = OB = 17.7 A

(iii) θ₃ = tan^–1 (15/10) = 56.3; p.f. = cos 56.3° = 0.55;

I = OC = 13.9 A.