Reoccurring Error in RLC Series Circuit Problems.

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SUMMARY

The forum discussion centers on the recurring errors encountered in calculating active and reactive power in RLC series circuits. The user identifies that while the magnitudes of the power values are accurate, the signs are incorrect, attributing the issue to potential miscalculations in phasor arguments or absolute power. The user provides a detailed equation involving complex power calculations, specifically using the formula S=V.I*=(V.ejα).(I.ejβ)*, which highlights the relationships between voltage, current, and their respective phase angles. The discussion emphasizes the importance of correctly computing phasors and understanding the phase relationships in RLC circuits.

PREREQUISITES
  • Understanding of RLC series circuit fundamentals
  • Knowledge of complex power and phasor representation
  • Familiarity with electrical engineering concepts such as active and reactive power
  • Proficiency in using complex numbers in circuit analysis
NEXT STEPS
  • Study the calculation of complex power in RLC circuits using S=V^2/Z.ejθz
  • Learn about phasor analysis and its application in AC circuit analysis
  • Explore the significance of phase angles in determining power factor
  • Review examples of common mistakes in calculating active and reactive power in RLC circuits
USEFUL FOR

Electrical engineering students, circuit design professionals, and anyone involved in analyzing RLC circuits and their power characteristics will benefit from this discussion.

buildingblocs
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Hi All,

In doing some practice RLC series circuit question, I have been obtaining incorrect values for active and reactive power. The magnitudes I obtain are correct however the sign of the value is not. I have attributed this to obtaining an incorrect argument for the absolute power, or computing phasors incorrectly. I have attached an image of the reasoning behind my problem solving however I must be fundamentally wrong somewhere.
 

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S=V.I*=(V.ejα).(I.ejβ)*= (V.ejα).(I.e − jβ)= V.I.ej(α − β) = VIcos(α − β) + j.VIsin(α − β)
I*=conjugated electric current α=θv ; β=θv-θz ;α-β=θz
Then S=V^2/Z.ejθz
 

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