Voltage, real reactive power question

AI Thread Summary
The discussion revolves around calculating the voltage experienced by a customer connected to a 33 kV single-phase generator through a cable with j2 ohms of impedance. Participants debate whether the customer voltage is simply 33 kV or if it requires calculating the total impedance and voltage drop. The importance of using RMS values is emphasized, clarifying that all voltage values are assumed to be RMS. The conversation also touches on the relationship between active and reactive power, with participants attempting to derive the load impedance from the given parameters. Ultimately, the complexity of the circuit and terminology is acknowledged, highlighting the need for precise calculations.
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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