Question

A manufacturing plant uses 2.49 kW of electric power provided by a 60.0 Hz ac generator with an rms voltage of 470 V . The plant uses this power to run a number of high-inductance electric motors. The plant's total resistance is R = 30.0 Ω and its inductive reactance is XL = 42.0 Ω .

A) What is the total impedance of the plant?

B) What is the plant's power factor?

C) What is the rms current used by the plant?

D) What capacitance, connected in series with the power line, will increase the plant's power factor to unity?

E) If the power factor is unity, how much current is needed to provide the 2.49 kW of power needed by the plant?

F) Compare the current found in part E with the current found in part C. (Because power-line losses are proportional to the square of the current, a utility company will charge an industrial user with a low power factor a higher rate per kWh than a company with a power factor close to unity.)

Answer 1

A) imedence Z =(R² + X² ) =(30² + 42² ) =51.6ohm

B)  plant's power factor cos = R/_Z = 30/51.6 = 0.581

C) true power = 3 VIcos

2.49 kW =3 x 470 V x I x   0.581

I= 5.264A

D) 3488kVAr

to find value of capacitance, connected in series with the power line, will increase the plant's power factor to unity determine the reactive power

reactive power =3 VIsin  

= cos^-10.581 =54.48⁰

sin =0.814 V=470 I=5.264A

reactive power =3488kVAr capacitance is required

E) I= 3.058 Ampere

2490 =3 x 470 x I

I= 3.058 Ampere