Answer:
The answer is below
Explanation:
[tex]\theta_1=cos^{-1}0.28=73.74^o\ lagging\\\\S_1=125\angle 73.74^o=35\ kW+j120\ kVAR\\\\S_2=10\ kW-j40\ kVAR\\\\S_3=15\ kW[/tex]
Total power = P = 35 kW + 10 kW + 15 kW = 60 kW
Total kVAR = 120 kVAR - 40 kVAR = 80 kVAR
[tex]Total\ apparent \ power =S= S_1+S_2+S_3=(35+j120)+(10-j40)+(15)\\\\S=60\ kW+j80\ kVAR=100\angle 53.13^o\\\\Current(I)=\frac{S^*}{V^*} \frac{100000\angle -53.13^o}{1400\angle0}=71.43\angle-53.13^o\\ \\Power\ factor (PF)=cos(53.13)=0.6\ lagging\\\\The \ new\ power\ factor\ is\ to \ be\ 0.8[cos^{-1}0.8=36.87^o], hence\ since\ the\ total\ \\real\ power(P)= 60\ kW, the\ capacitor\ kVAR(Q_c)\ is:\\\\Q_c=60tan(53.13)-60tan(36.87)=80-45=35\ kVAR\\\\[/tex]
[tex]C=\frac{Q_c}{wV^2} =\frac{35000\ VAR}{2\pi*60\ Hz*1400\ V}=47.38\ \mu f[/tex]
New current (I') = [tex]\frac{S'^*}{V^*}=\frac{60000-j45000}{1400}=53.57\angle-36.87^o[/tex]
Current reduce from 71.43 A to 53.57 A
