# Question: A motor has an output of 6 kW, an efficiencyof 75% and a power factor of 0.64 lagging when operated from a 250 V, 60 Hz supply. It is required to raise the power factor to 0.925lagging by connecting a capacitor in parallel with the motor. -Free Course Hero Question Answer.

Question Description: A motor has an output of 6 kW, an efficiencyof 75% and a power factor of 0.64 lagging when operated from a 250 V, 60 Hz supply. It is required to raise the power factor to 0.925lagging by connecting a capacitor in parallel with the motor.

Determine

- (a) the current taken by the motor,
- (b) the supply current after power factor correction,
- (c) the current taken by the capacitor,
- (d) the capacitance of the capacitor and
- (e) the kVAR rating of the capacitor.

**Course Hero** **Answer & Explanation**:

a). The current taken by motor = 50 A.

b). The supply current after power factor correction = 34.59 A.

c). The current taken by capacitor = 25.28 A.

d). The capacitance of a capacitor = 268.2 micro farad.

e). The KVAR rating of capacitor = 6.32 KVAR.

*Given data; Pout = 6KW 2 = 75% P. f = Cos$ = 0- 64 lagging = 250V f = 60HZ cost2 = 0. 925 lagging a) . current taken by motors Is efficiency = QUEPUL xloo Input 75 = 6 x 100 Input – v Input power = 8 KW $1 = 50.21 Pin = VI, cost Is = I 3 8 X 10 – 250 X I, X 0. 64 800 0 I = 2 50 X 0 . 6 4 I = 50 A*

Image transcription text

D). The supply cussent after powerfaction correction cos $2 = 0. 925 lagging Ic $ 2 = cos (0. 925 ) 4 2 = 2 2.33" M I s = I ‘ + Ic I = 50 A 90 $2 = 22:33 Horizontal component of II (OB ) = I Cos $ 1 = 50.21’ = 5ox cos ( 50. 21 ) A II = 32 A Horizontal component of Is ( OB ) = Ig Cos $2 = 0. 925 19 By equating both components ( as they are equal) 32 = 0.925 19 32 Is = 0 . 925 I. = 34. 59 A