Mumbai University > FE > Sem 1 > Basic Electrical and Electronics Engineering

Marks: 2 M

Year: May 2013, Dec 2014

Given:

$pf = 1 W1 = 7kW \\ pf = 0.707 (lagging) \\ \text{When the power factor is unity} \\ W1 = W2 \\ W2 = 7kW \\ Power = W1 + W2 = 7 + 7 = 14kW – 1 \\ \text{When the power factor is changed to 0.707 lagging} \\ Pf = \cos (ф) = 0.707 \\ ∴ ф = \cos-1 (0.707) \\ \boxed{ф = 450} \\ ∴ \tan ф = \sqrt3 W1 – W2/ W1 + W2 \\ ∴ \tan (450) = \sqrt3 W1 – W2 / 14 \\ \boxed{1= \sqrt3 W1-W2/14} \\ W1-W2 = 14/\sqrt3 = 8.082 kW -2 \\ \text{Solving eq. (1) & (2)} \\ W1 = 11.041 kW \\ W2= 2.959 kW$