Manufacturing Plant Power A manufacturing plant uses 2.36 kW of electric power provided by a 50.0 Hz ac generator with an rms voltage of 500 V . The plant uses this power to run a number of high-inductance electric motors. The plant's total resistance is R = 22.0 Ω and its inductive reactance is XL = 43.0 Ω .a) What is the totalimpedence 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'spower factor to unity?e) If the power factor is unity, how muchcurrent is needed to provide the 2.22 kW of power by the plant?

(a) Z = 48.3 Ω

(b) cos ∅ = 0.455

(c) Irms = 10.35 A

(d) C = 74.02 μF

(e) Irms = 4.44 A

Explanation:

Power (P) = 2.36 kW

Frequency (f) = 50 Hz

RMS Voltage (Vrms) = 500 V

Resistance (R) = 22 Ω

Inductive Reactance (XL) = 43 Ω

(a) to calculate the total impedance, use the formula:

Z = √(R² + XL²)

   = √((22)² + (43)²)

   = √2333

Z = 48.3 Ω

(b) To calculate the plant's power factor, we will use the formula:

cos ∅ = R/Z

          = 22/48.3

cos ∅ = 0.455

(c) To calculate the RMS current used by the plant, divide the RMS voltage value by the impedance of the plant.

Irms = Vrms/Z

        = 500/48.3

Irms = 10.35 A

(d) For the power factor to become unity, the inductive reactance must be equal to the capacitive reactance i.e. Xc = XL

Xc = XL

1/(2πfC) = XL

1/(2πfXL) = C

C = 1/(2π*50*43)

   = 7.402 x 10⁻⁵

C = 74.02 μF

(e) P = Vrms*Irms*cos∅

    Irms = P/Vrms*cos∅

             = 2.22 x 10³/500*1

    Irms = 4.44 A

answer is c

explanation:

the amount of current in the coil