To achieve the outcome shown in the image, follow these steps in the context of electrical power systems involving apparent, real, and reactive power in a three-phase system:

1. Calculate Apparent Power $S_{R(3-\phi)}$:

   • The given formula for apparent power $S_{R(3-\phi)}$ is: $S_{R(3-\phi)} = \frac{P_{R(3-\phi)}}{\cos \theta}$
   • Here, $P_{R(3-\phi)} = 800 \, \text{MW}$ (real power) and $\cos \theta = 0.8$ (power factor).
   • Substitute these values: $S_{R(3-\phi)} = \frac{800 \, \text{MW}}{0.8} = 1000 \, \text{MVA}$

2. Determine the Angle $\theta$:

   • The power factor $\cos \theta = 0.8$ can be used to find the angle $\theta$ using the inverse cosine: $\theta = \cos^{-1}(0.8) \approx 36.87^\circ$

3. Express $S_{R(3-\phi)}$ in Polar Form:

   • Represent $S_{R(3-\phi)}$ with a magnitude and angle: $S_{R(3-\phi)} = 1000 \, \text{MVA} \angle 36.87^\circ$

4. Convert to Rectangular Form:

   • To convert from polar to rectangular form, use $S = |S| \times (\cos \theta + j \sin \theta)$: $S_{R(3-\phi)} = 1000 \times (\cos(36.87^\circ) + j \sin(36.87^\circ))$
   • By calculating $\cos(36.87^\circ) \approx 0.8$ and $\sin(36.87^\circ) \approx 0.6$, we get: $S_{R(3-\phi)} = 1000 \times (0.8 + j 0.6) = 800 + j 600 \, \text{MVA}$

Would you like further details or have any questions?

Related Questions

1. What is the significance of power factor in electrical systems?
2. How is reactive power calculated from real power and apparent power?
3. Why do we represent apparent power in polar and rectangular forms?
4. How does power factor angle affect the reactive component of power?
5. What role does complex power play in analyzing three-phase systems?

Tip

Understanding power factor and its angle is crucial for analyzing power systems, especially in optimizing load and reducing energy loss.