A factory uses 2.22 kW of electric power, which is provided by a 60 Hz ac genera

Question

A factory uses 2.22 kW of electric power, which is provided by a 60 Hz ac generator with an rms voltage of 485 V. The factory uses this power to run a number of high-inductance electric motors.
The factory's total resistance is R = 25 ohms and its inductive reactance is 45 ohms.

1. What is the power factor?
2. What is the rms current used by the factory?
3. What is the total impedance?
4. What capacitance, connected in series with the power line, will increase the plants power factor to unity?
5. If the power factor is now unity, how much current is needed to provide the 2.22 kW of power needed for the factory operation?

Explanation / Answer

a) given resistance R = 25 inductive resistance is XL= 45.0 the total impedance of the plant is       Z =R2+XL2 plug in the values for Z b) the power factor is given as    cos = R/Z plug in the values for cos c) here given rms voltage is Vrms = 485 V. the rms current is given as Irms = Vrms/Z plug in the values for Z d) here the impedance of the circuit changes by     Z' = R2+(XL-XC)2 here Xc = 1/C where = 2f               = 2*60Hz                 = 120 s-1 so XC = 1/120C the new power factor is cos = R/Z'                   here cos = 1     so R = Z'        R = R2+(XL-XC)2    simlifying we get     XL = XC 45 = 1/120C simlify & calculate for C d) thje average power consumed by the plant is    Pave = Irmscos     given     Pave = 2.22 kW            = 2220W         here cos =1      Irms =Pave/Vrms*cos       45 = 1/120C simlify & calculate for C d) thje average power consumed by the plant is    Pave = Irmscos     given     Pave = 2.22 kW            = 2220W         here cos =1      Irms =Pave/Vrms*cos       plug in the values for Irms

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