Current flow through the circuit,
I = V/Z
= 400/10
= 41.5 A
Power factor = 8/10
= 0.8
After connecting a capacitor bank, the reactive power component of current changes but the active power component of current is unchanged.
I1*cosφ1 = I2*cosφ2
41.5*0.8 = I2*0.9
I2 = 36.88 A
Q1 (before connecting capacitor) = √3*VL*IL*sinφ1
= √3*415*41.5*sin(36.86)
= 26.847 kVAR
Q2 (after connecting capacitor) = √3*VL*IL*sinφ2
= √3*415*36.88*sin(25.84)
= 15.867 kVAR
kVAR supplied by capacitor bank = Q1 - Q2
= 26.847 - 15.867
= 10.98 kVAR
EXAMIANS