Voltage, real reactive power question

In summary, a 33 kV single phase generator is supplying a customer with a load through a cable with an impedance of j2 ohms. The customer will experience a voltage of 33 kV if a rated current of 500 A is drawn at a power factor of unity. The generator will provide both active and reactive power, but the exact values are not given. To find the customer voltage, the total impedance of the circuit must be calculated using the V = IZ equation and then the impedance of the load can be subtracted. It is not clear if the question is asking for the voltage at the end of the cable
  • #1
debwaldy
38
0

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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  • #2
Are you trying to solve this using RMS Values ?
 
  • #3
we have been told to assume that all voltage values given are rms voltage values so we don't need to worry about that...
 
  • #4
Would it be possible to just say that P = S and hence the voltage at the customer end will also be 33kV?
 
  • #5
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
 

1. What is voltage and how is it measured?

Voltage is a measure of the electrical potential difference between two points in a circuit. It is measured in volts (V) using a voltmeter.

2. What is the difference between real power and reactive power?

Real power is the actual power that is consumed by a circuit and is measured in watts (W). Reactive power is the power that is stored and released by inductive and capacitive elements in a circuit and is measured in volt-amperes reactive (VAR).

3. How is reactive power related to voltage and current?

Reactive power is related to voltage and current through the power factor, which is the cosine of the angle between the voltage and current in an AC circuit. A higher power factor means that less reactive power is being used.

4. Why is it important to consider reactive power in a circuit?

Reactive power is important because it affects the efficiency and stability of a circuit. It can cause voltage drops, which can lead to equipment damage and power outages. It also affects the overall power consumption of a system.

5. How can reactive power be reduced?

Reactive power can be reduced by using power factor correction techniques, such as adding capacitors or using synchronous generators with automatic voltage regulators. It can also be reduced by using more efficient equipment and minimizing the length of power lines.

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