https://file.aiquickdraw.com/mathbot/file/1730910215585f0723o5n.jpg

The given image demonstrates the calculation of the apparent power (\( S_{R(3\text{-}\phi)} \)) in a three-phase system, given a real power and a power factor. Here’s a breakdown of the calculations shown:

1. **Definition of Apparent Power:**
   - The apparent power (\( S_{R(3\text{-}\phi)} \)) in a three-phase system can be derived from the real power (\( P_{R(3\text{-}\phi)} \)) and the power factor (\( \cos \theta \)).
   - The formula used here is:
     \[
     S_{R(3\text{-}\phi)} = \frac{P_{R(3\text{-}\phi)}}{\cos \theta}
     \]
   - This relationship shows that apparent power is the real power divided by the power factor.

2. **Substituting Values:**
   - The real power \( P_{R(3\text{-}\phi)} \) is given as \( 800 \, \text{MW} \) (megawatts).
   - The power factor (\( \cos \theta \)) is \( 0.8 \).
   - Plugging these values into the formula:
     \[
     S_{R(3\text{-}\phi)} = \frac{800 \, \text{MW}}{0.8} = 1000 \, \text{MVA}
     \]
   - The unit MVA (megavolt-amperes) is used for apparent power in this case.

3. **Apparent Power as a Phasor:**
   - In the lower part of the image, the apparent power is represented in polar form with an angle.
   - \( \cos^{-1}(0.8) \) gives the angle \( \theta \), which is approximately \( 36.87^\circ \).
   - Thus, \( S_{R(3\text{-}\phi)} \) is expressed as:
     \[
     S_{R(3\text{-}\phi)} = 1000 \, \text{MVA} \angle 36.87^\circ
     \]
   - This represents the apparent power as a complex quantity with magnitude \( 1000 \) MVA and phase angle \( 36.87^\circ \).

4. **Conversion to Rectangular Form:**
   - The phasor \( 1000 \angle 36.87^\circ \) can also be expressed in rectangular form using trigonometric identities:
     \[
     S_{R(3\text{-}\phi)} = 1000 \times \left( \cos(36.87^\circ) + j \sin(36.87^\circ) \right)
     \]
   - This breaks down to a real part (representing the active power) and an imaginary part (representing the reactive power):
     \[
     S_{R(3\text{-}\phi)} = 1000 \times \left(0.8 + j \times 0.6 \right)
     \]
     where \( \cos(36.87^\circ) = 0.8 \) and \( \sin(36.87^\circ) = 0.6 \).

In summary, the calculation shows the conversion of real power (in MW) to apparent power (in MVA) by adjusting for the power factor. It then expresses the apparent power as both a magnitude and an angle (polar form) and as a sum of real and imaginary components (rectangular form).

Would you like more details on any part of this process or have specific questions?

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### Related Questions:
1. What is the significance of using MVA instead of MW in power calculations?
2. How does power factor affect apparent power in an electrical system?
3. Why do we convert power into polar and rectangular forms?
4. What does the angle \( 36.87^\circ \) represent in the context of electrical power?
5. How would a change in power factor to 0.9 alter the apparent power calculation?

### Tip:
When calculating apparent power, always ensure your units are consistent, and carefully interpret power factor as it directly impacts both the magnitude and angle of the resulting power vector.

Explain the calculation of apparent power in a three-phase system using real power and power factor.

Explore the calculation of apparent power in a three-phase electrical system. Understand the conversion of real power in MW to apparent power in MVA, the impact of power factor, and how to express power in both polar and rectangular forms. Suitable for college-level electrical engineering students.

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