Voltage, real reactive power question

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SUMMARY

The discussion focuses on a 33 kV single-phase generator supplying a load with an impedance of j2 ohms at a rated current of 500 A and a power factor of unity. The customer voltage is calculated using the formula V = IZ, where the total impedance must be determined. The active and reactive power provided by the generator can be derived from the equations P = IV and PF = P/S = IV cos phi, leading to a comprehensive understanding of the voltage experienced by the customer.

PREREQUISITES
  • Understanding of AC circuit analysis
  • Familiarity with impedance in electrical systems
  • Knowledge of power factor and its implications
  • Proficiency in using the formulas V = IZ and P = IV
NEXT STEPS
  • Study the concept of impedance in AC circuits
  • Learn about calculating active and reactive power in electrical systems
  • Explore the implications of power factor on system performance
  • Investigate the use of RMS values in voltage and current calculations
USEFUL FOR

Electrical engineers, students in power systems, and professionals involved in generator and load analysis will benefit from this discussion.

debwaldy
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Homework Statement



A 33 kV single phase generator is supplying a customer a load through a cable with an impedance j2 ohms

a.What voltage would the customer experience if rated current of 500 A is drawn at power factor of unity at the customer end?

b. How much active and reactive power would be provided by the generator?


Homework Equations


V = IZ
P = IV

PF = P/S = IV cos phi

The Attempt at a Solution



Part a: Do I need to find the total impedance using V = IZ equation, and then calculate the impedance of the customer load by subtraction, and then sub this impedance of the load back into the V = IZ formula to find the customer voltage?

Or is the customer voltage simply 33 kV?

I am very confused by terminology?

Any help much appreciated
 
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Are you trying to solve this using RMS Values ?
 
we have been told to assume that all voltage values given are rms voltage values so we don't need to worry about that...
 
Would it be possible to just say that P = S and hence the voltage at the customer end will also be 33kV?
 
You have a circuit with three elements, a voltage source of 33kV, an impedance of j2 ohms, and another impedance which is assumed to be purely resistive. That is, the generator supplies 33kV to some network, what the customer sees, you don't know.

Equivalent Z is Z=R+2j, that is 33kV = (R+2j)*500A, which yields the load impedance. The voltage drop is then 33kV*R/(R+2j).

/M
 

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