Question

Two impedances Z₁ and Z₂ when connected separately across a 230-V, 50-Hz supply consumed 100 W and 60 W at power factors of 0.5 lagging and 0.6 leading respectively. If these impedances are now connected in series across the same supply, find : 

(i) total power absorbed and overall p.f. 

(ii) the value of the impedance to be added in series so as to raise the overall p.f. to unity

Answer 1

Inductive Impedance V₁ I cos φ₁ = power; 230 × I₁ × 0.5 = 100 ; I₁ = 0.87 A 

Now, I₁² R₁ = power or 0.87² R₁ = 100; R₁ = 132 Ω ; Z₁ = 230/0.87 = 264 Ω 

X_L = √(Z²₁ - R²₁) = √(264² - 132²) = 229Ω

Capacitance Impedance I₂ = 60/230 × 0.6 = 0.434 A ; R₂ = 60/0.4342 = 318 Ω

Z₂ = 230/0.434 = 530 Ω ; X_C = √(530² - 318²) = 424Ω (capacitive)

When Z₁ and Z₂ are connected in series 

R = R₁ + R₂ = 132 + 318 = 450 Ω; X = 229 − 424 = − 195 Ω (capacitive)

Z = √(R² + X²) = √(450² + (-195)²) = 490 , 230 / 490 0.47 A

(i) Total power absorbed = I²R = 0.472 × 450 = 99 W, cos φ = R/Z = 450/490 = 0.92 (lead) 

(ii) Power factor will become unity when the net capacitive reactance is neutralised by an equal inductive reactance. The reactance of the required series pure inductive coil is 195 Ω.